Talk:Pythagorean triple/Archive 4

This is an archive of past discussions about Pythagorean triple. Do not edit the contents of this page. If you wish to start a new discussion or revive an old one, please do so on the current talk page.

Archive 1

Archive 2

Archive 3

Archive 4

Archive 5

Moving parts of this to Formulas for generating Pythagorean triples

This is one of the 500 most viewed mathematics articles, so I think it is time to make some tough choices here. The first thing to do is to move, the big list of formulas out to Formulas for generating Pythagorean triples. A thorough discussion of Euclid's formula should remain in this article, including its geometrical interpretation in projective geometry. I don't know whether the connections to the modular group should stay, but these are central for what seem to be the most important studies of Pythagorean triples over the past fifty years. siℓℓy rabbit (talk) 20:47, 17 January 2009 (UTC)

The "Barning Tree"

This is a common misnomer. It should be called "Berggren's Tree" after B. Berggren. see: *Berggren, B. (1934), "Pytagoreiska trianglar", Tidskrift för elementär matematik, fysik och kemi 17: 129–139. —Preceding unsigned comment added by 72.230.147.240 (talk) 16:32, 10 July 2009 (UTC)

Can we swap a and b?

Hi, I just added a derivation of the formulae for Pythagorean triples based on the Gaussian integers. Unfortunately, I'm apt to write the complex number in the form a + bi rather than b + ai. That makes my formulae swapped relative to those given above. Does anyone mind if I exchange a and b? Proteins (talk) 21:10, 9 April 2009 (UTC)

Aren't the "Notes" and "References" sections backwards? I mean shouldn't "Notes" connected with superscripted citations be labeled "References", and "References" be labeled "Notes"? —Preceding unsigned comment added by 208.106.16.145 (talk) 03:36, 31 July 2009 (UTC)

Definition in Haskell

A pythagorian triple can be defined quite nicely in haskell:

triple :: Int -> [(Int,Int,Int)]
triple n = [(x,y,z) | x <- [0..n], y <- [0..n], z <- [0..n], x^2 + y^2 == z^2, x>0, y>0]

Could be an useful addition to the article -Benbread (talk) 16:47, 4 May 2009 (UTC)

Hmm. Pseudo code/array might be better? Not knowing Haskell, but knowing that variable types don't need to be defined specifically, I guess I don't have that much to contribute. It would seem pseudo code would apply more. Idofen (talk | contribs | March 2)

I don't think translations of simple mathematical formulae into harder-to-understand computer code are a particularly helpful thing to include in our encyclopedia articles. —David Eppstein (talk) 00:26, 5 August 2014 (UTC)

Ratio notation

While it's clear from the context that $A:B=C:D$ implies that $A:B$ is the same as $C:D$ , from what I've been around I'd see $A:B::C:D$ used for that purpose (yes, I am being excessively fussy about minor issues like this :D). Any words from others? 118.90.44.104 (talk) 11:35, 6 June 2009 (UTC)

Euler or Euclid?

The article refers to "the Euler formula". Surely "the Euclid formula" is intended. I'll change it with apologies if I'm incorrect. Aliotra (talk) 06:38, 26 June 2009 (UTC)

Nope, its Euler. However I don't have much time so it would be appreciated if you edited it for me please. —Preceding unsigned comment added by 067012732s (talk • contribs) 18:00, 19 August 2009 (UTC)

It is "Euclid's formula" for the simple generator function, but the Euler's totient function has something to say about the parabola patterns you see in the scatter plot. It is cited material.--Coastword (talk) 22:25, 21 December 2009 (UTC)

Every triple generated?

Hi. The article states that not every triple will be generated, but this seems to imply otherwise. Can anyone add some insight to this? Grj23 (talk) 04:01, 4 January 2010 (UTC)

Okay, so ^[1] points out that (12,9,15) cannot be represented in the Euclidean way. It might be worth including this counter-example in the text. They also have a way of generating all of the triples, which I might include in the other article (if it's not there already) Grj23 (talk) 04:13, 4 January 2010 (UTC)

(9,12,15) is just (3,4,5) multiplied by 3; this is well understood, don't sweat it. MaxEnt (talk) 03:15, 8 January 2010 (UTC)

Cleanup tag useless

I didn't read closely, the but article as a whole doesn't seem to contain much that surprises me. IMO the top-level clean-up tag needs to be replaced by section-level cleanup tags to justify the eyesore:dubiousness ratio. Or maybe I'll slap a cover page on my thesis with a giant red box proclaiming "this thesis contains a spelling mistake" to put my readers in the right frame of mind. MaxEnt (talk) 03:15, 8 January 2010 (UTC)

Primitives

I undid this edit by Ip 220.225.244.114. Reason in edit summary. DVdm (talk) 18:19, 31 March 2010 (UTC)

theta and beta

The theta and beta that appear as angles in the additional relationships section are undefined. 't' is also undefined. These do not appear in the reference ([2]) either.

192.249.47.195 (talk) 14:49, 5 May 2010 (UTC)

definition of coprime

In the definition at the beginning of this article, the three variables are said to be coprime, linking to the article coprime. However, the definition of coprime given in the article coprime is for two integers. It should be made clear whether or not the integers must be pairwise coprime, or coprime in the sense that a, b, and c do not have a common divisor greater than 1. It turns out in this case that one implies the other, but that's a theorem, not a definition. —Preceding unsigned comment added by 96.37.136.123 (talk) 22:58, 27 May 2010 (UTC)

c < b x √2

Is it not true that c is less than √2 x b? Myrvin (talk) 15:38, 11 July 2010 (UTC)

Not necessarily. The formulas used in the article for primitive triples are c=m²+n² and b=2mn. Now c < √2×b if c² < 2b², or m⁴ + 2m²n² + n⁴ < 8m²n². For given n, this is violated for sufficiently large m. For example, (m, n) = (4, 1) gives (a, b, c) = (15, 8, 17). Also, c need not be less than √2×a : a counterexample is (a, b, c) = (3, 4, 5). Duoduoduo (talk) 16:40, 17 July 2010 (UTC)

Ah, sorry. I should have added: if b is the larger of a and b. It's weird for me to see a triple like (15, 8, 17); to me that's always (8, 15, 17). But I wasn't precise enough. Given that proviso, my contention still holds Myrvin (talk) 06:38, 18 July 2010 (UTC)

Yes. In fact this is true for all non-isosceles right triangles (integer-sided or otherwise) and indeed for all acute triangles as well. c > b x √2 would imply c² > 2b² = b² + b² ≥ a² + b² so c² > a² + b², which is true if and only if the triangle is obtuse. And c = b × √2 would imply c² = 2b² ≥ a² + b², which is consistent with the Pythagorean formula only if a = b — the isosceles case, which can never occur with Pythagorean triangles. Duoduoduo (talk) 22:50, 18 July 2010 (UTC)

So I thought. But is it citable and important enough to put in the article? By the way, for Fermat, I also think that c < ⁿ√2 x b. Myrvin (talk) 11:22, 19 July 2010 (UTC)

A counterexample for all n > 2 is (a, b, c)= (119, 120, 169).

I don't think the n=2 theorem belongs in this article, since the property is not specific to Pythagorean triples. If you had a citation, it would fit in an article on acute triangles, but Wikipedia doesn't have an acute triangle article; and I can't see any place in the triangle article where it would naturally fit. But it is an interesting observation! Duoduoduo (talk) 14:46, 19 July 2010 (UTC)

A smaller counterexample for all n > 2 is (20, 21, 29). Duoduoduo (talk) 15:50, 19 July 2010 (UTC)

A pity about n=2, it seemed important to me. For n> 2, I am confused. There are no integer triples for Fermat. I reasoned the same way. c is at its maximum when a = b (where b >= a). => 2bⁿ = cⁿ; => c <= ⁿ√2 x b. I guess that doesn't work. Myrvin (talk) 20:35, 19 July 2010 (UTC)

For your 20, 21 example, the following appears. c gets closer and closer to b as n increases.

n_a__b__c

2 20 21 29

3 20 21 25.84373728

4 20 21 24.40047274

5 20 21 23.57627517

6 20 21 23.04458706

7 20 21 22.67405357

8 20 21 22.40168712

9 20 21 22.19352337

10 20 21 22.02963561

Myrvin (talk) 10:56, 21 July 2010 (UTC)

Sides a and b in terms of m and n

In the heading "Generating a triple", Euclid's formula uses

a=m^{2}-n^{2},\ \,b=2mn,\ \,c=m^{2}+n^{2},

whereas, in the heading "Geometry of Euclid's formula", Euclid's formula uses

a=2mn,\quad b=m^{2}-n^{2},\quad c=m^{2}+n^{2}.

Although, aside from the order of a and b, most of the rest of the article does not hinge on which is correct (since a and b are either added or multiplied, never subtracted from each other), one consistent usage would be appreciated, preferably the former to reduce the amount of editing needed. − Glenn L (talk) 03:48, 17 August 2010 (UTC)

shared c ?

Are there any pairs (or groups) of triples (a,b,c) (d,e,f) such that c = f ? Equivalently, this would mean that there are some concentric circles on the plot of triples.

I don't know the answer, but would be interesting to see it listed in the properties of the pythagorean triples. —Preceding unsigned comment added by 87.194.171.29 (talk) 22:37, 25 January 2011 (UTC)

Yes, there are. We know that 3,4,5 and 5,12,13 are not the same. But both can be multiplied by a constant so that their hypotenuses are the same. The LCM of 5 and 13 is 65. This gives 39,52,65 and 25,60,65. Georgia guy (talk) 22:57, 25 January 2011 (UTC)

In general, the number of primitive Pythagorean triples of hypotenuse c is dependent on the number of prime factors of c that are congruent to 1 modulo 4. It turns out that only powers of 2 appear as these numbers. These are sometimes called "sibling triples". Examples: 33,56,65 , 63,16,65. — 184.153.109.223 (talk) 00:49, 20 September 2011 (UTC)

Proving a,b,b+1 is not a Pythagorean triple using divisibility properties

I've been studying patterns based on the divisibility properties in this article to see when using 3,4,5,7,11 is not sufficient. For example:

5,8,9 we know is wrong because exactly one of 5 and 8 must be divisible by 3, but neither 5 nor 8 is, and so 5,8,9 can be ruled out
9,28,29 is wrong because none of 9, 28, 29 is divisible by 5
15,88,89 is wrong because none of 15, 88, 88+15=103, and 88-15=73 is divisible by 7
15,92,93 is wrong because none of 15, 92, 2*15+92=122, |2*15-92|=62, 2*92+15=199, 2*92-15=169 is divisible by 11
21,200,201, however, appears to meet all the above criteria, so it needs a higher number (13 minimum) to rule it out as a Pythagorean triple. For clarification, I know you can prove it's not by calculating each side and the left side is 40441 but the right side is 40401. However, it appears to meet all the criteria for congruence modulus 3,4,5,7,11; namely, exactly one member of a given list of members must be divisible by 3, 4, 5, 7, or 11.

Anyone have an external link on what list of values related to a, b, and c must have exactly one member divisible by 13?? The properties section of this article already has the lists for being divisible by 3, 4, 5, 7, and 11. Georgia guy (talk) 17:18, 17 March 2011 (UTC)

Now, there's one additional rule I forgot. c cannot be divisible by 3, 7, or 11. So 21,200,201 can easily be ruled out by this because 201 is divisible by 3. However, checking 23,240,241, it still meets all of the above criteria if expanded to include this rule. To prove it's wrong by calculation, the left side is 58129 and the right side is 58081. However, I would still like an appropriate external link. Georgia guy (talk) 17:25, 17 March 2011 (UTC)

If you look above, you will see the idea that c < b * √2, which rules out all c = b+1 except 3,4,5. No citation though. Myrvin (talk) 19:10, 17 March 2011 (UTC)

How does it rule out 5,12,13?? Georgia guy (talk) 19:12, 17 March 2011 (UTC)

Quite right - I misunderstood what you were saying. I thought your idea WOULD rule out 5, 12,13. And my point has nothing to do with it. I'll keep quiet now. Except to say:

3 4 5
5 12 13
7 24 25
9 40 41
11 60 61
13 84 85
15 112 113
17 144 145
19 180 181
21 220 221
23 264 265
25 312 313
27 364 365
29 420 421
31 480 481
33 544 545
35 612 613
37 684 685
39 760 761
41 840 841
43 924 925

Am I missing something else? Myrvin (talk) 23:35, 17 March 2011 (UTC)

That is all correct. I know how to prove these are Pythagorean triples using calculation. You should understand the difference between:

Proving (a,b,b+1) is not a Pythagorean triple by calculation. An example: 5,8,9 is not a Pythagorean triple because the left side is 89 and the right side is 81.
Proving (a,b,b+1) is not a Pythagorean triple by modular arithmetic. An example: 5,8,9 is not a Pythagorean triple because exactly one of a,b must be divisible by 3 but neither 5 nor 8 is divisible by 3. Georgia guy (talk) 23:41, 17 March 2011 (UTC)

For additional clarification, notice these notes:

Exactly one of a, b is divisible by 3.
Exactly one of a, b is divisible by 4.
Exactly one of a, b, c is divisible by 5.
Exactly one of a, b, (a + b), (b − a) is divisible by 7.
Exactly one of (a + c), (b + c), (c − a), (c − b) is divisible by 8.
Exactly one of (a + c), (b + c), (c − a), (c − b) is divisible by 9.
Exactly one of a, b, (2a + b), |2a − b|, (2b + a), (2b − a) is divisible by 11.

Any info on what "Exactly one of (set of numbers written in terms of a,b,c) is divisible by 13??" I can't find any web page revealing what this set is. Georgia guy (talk) 23:57, 17 March 2011 (UTC)

Actually, the divisibility by 8 property rules out 23,240,241. Georgia guy (talk) 00:04, 31 March 2011 (UTC)

Properties of properties of Pythagorean triples

Can anyone find a pattern in the sequence of properties of Pythagorean triples of the following form:

Exactly one of a group of expressions written in terms of a, b, c is divisible by p, where p is an odd prime number.

The general rule is that the expressions must not have any constant terms (a + 1) is not valid) and no terms where 2 variables are multiplied (ab + c is not valid.)

Here are the properties of these properties that I know:

The total number of expressions is always (p+1)/2. (For example, if p = 7, the total is (7+1)/2 = 8/2 = 4.
The monomials a and b are always among the expressions.
The monomial c is among the group of expressions if and only if p is of the form 4n+1, where n is an integer. (For example, if p = 7, c isn't among the expressions.)
The binomials a+b and b-a are among the group of expressions if and only if p is of the form 8n+1 or 8n-1 where n is an integer.

Any other properties?? Georgia guy (talk) 20:17, 25 October 2011 (UTC)

We need to be careful here -- original research is not permitted on Wikipedia. As it is a lot of the material in this article is unsourced and may or may not be original research. Anything further added should be sourced. Duoduoduo (talk) 20:38, 25 October 2011 (UTC)

Can you triple-check what I added to being divisible by 13 to the properties section of the article?? Please do not simply remove it; please correct it if it's wrong. Georgia guy (talk) 20:44, 25 October 2011 (UTC)

No, it should be removed unless you can find a published source for it. Otherwise it's original research. —David Eppstein (talk) 20:47, 25 October 2011 (UTC)

It's not original unless I'm the first one to know it; I'm sure I'm not; I just don't know who else knows it. Georgia guy (talk) 20:55, 25 October 2011 (UTC)

Wrong. In academic areas, things are only allowed in Wikipedia if they are known to have gone through a peer-review process to get published. If you don't know who else knows it, then you don't know that it's been peer-reviewed and judged correct. Duoduoduo (talk) 22:36, 25 October 2011 (UTC)

Defining a and b

I always define a as the shorter leg and b as the longer leg. Someone, however, prefers to define a as the odd leg and b as the even leg, regardless of magnitude. Any suggestions?? Georgia guy (talk) 18:42, 31 October 2011 (UTC)\

That someone is Euclid! Euclid's formula is

a=m^{2}-n^{2},\ \,b=2m.n,\ \,c=m^{2}+n^{2}

(sometimes expressed alternatively as

b=m^{2}-n^{2},\ \,a=2mn,\ \,c=m^{2}+n^{2}

). Either way, this pegs one of the symbols a or b as even and the other as odd. That's the reason for doing it this way, and nothing is lost by having a>b sometimes and a<b sometimes. Duoduoduo (talk) 19:17, 31 October 2011 (UTC)

However, when presenting lists of triples as in the Examples section, it's easier on the eye to list the smaller leg first, as that section does. I don't think doing it different ways in the different sections, each with a purpose, will confuse anyone. Duoduoduo (talk) 19:27, 31 October 2011 (UTC)

A simple proof of the necessity

A real simple proof looks quite different.

Euclid's formular is of course valid for real numbers x,y too

a=x^{2}-y^{2},\ \,b=2xy,\ \,c=x^{2}+y^{2}

,

meaning that a² + b² = c² is satisfied. If (a,b,c) is given, positive numbers x,y uniquely can be calculated:

x={\sqrt {\frac {c+a}{2}}},\ \,y={\sqrt {\frac {c-a}{2}}}

If (a,b,c) is a primitive Pythagorean triple x=m and y=n. That's it! --178.203.183.31 (talk) 10:28, 2 January 2012 (UTC)

That's correct, and I would have put it in long ago, but I've never seen that in a source, and in the absence of a source it's original research. Do you have a source for this proof? Duoduoduo (talk) 17:24, 2 January 2012 (UTC)

But, much more, namely Fermats Last Theorem, can be proven in this way.

{\sqrt {a^{n}}}=x^{2}-y^{2},\ \,{\sqrt {b^{n}}}=2xy,\ \,{\sqrt {c^{n}}}=x^{2}+y^{2}

--88.152.231.121 (talk)+

Actually the proof is not really complete. It remains to show that x,y are integers. Indeed the real numbers x,y are only integers, if one choose a to be the odd leg. First one have to show, that one of the short legs is even, the other odd, and the is hypothenuse odd, too. First, obviously one number in {a,b,c} is even, the other two odd, if there is no common divisor. It easy to see that every odd square is of the form 4n + 1. The sum of two odd squares can't be a square, since it would not be divisible by 4. If a and c are odd, the sum and the difference are even, meaning that x² and y² are integers. These numbers must be coprime. Since the square root of the product x² times y² must be an interger, x and y must be intergers, too. But, this only true, because of the fact that they cannot have a common prime factor. --88.152.231.121 (talk) 17:20, 20 January 2012 (UTC)

I just added to the article the reference of my page with a simple proof. I had found it on my own many years ago, as a young student, but never considered to publish it as it seemed so simple I did not expect it to be unknown. So now I was surprised to not find it in this article, but to only see some more complicated proofs instead. In your above discussion you say some things around it but you did not seem to state it very clearly that there is a possible simple proof. So anyway now here it is.Spoirier (talk) 21:55, 24 November 2012 (UTC)

Can you explain why you think this link passes the Wikipedia criteria for reliable sources, original research, and links to avoid (see in particular #11)? —David Eppstein (talk) 22:12, 24 November 2012 (UTC)

Well I don't know what you think should be done about it, can you explain ? If a simple proof exists, should Wikipedia mention it or not ? The other references lead nowhere, only to names of mathematicians and journals, while the contents of the mentioned articles cannot be directly read online, as it seems. Actually I don't see the sense of mentioning "references" that one cannot easily check. If you think it is the purpose of Wikipedia to mention the best solutions to a given problem, please make the article look whatever you think the way is right to do so. In other cases, well, do whatever you like either, but then I just cannot see what Wikipedia can be here for.Spoirier (talk) 07:44, 25 November 2012 (UTC)

Sorry I cannot get the meaning of the argument "Not a reliable source" in this case ?? If it was for an obscure 3-pages proof that needs to be reviewed by some expert I would understand the point. But my proof is clear and simple. It just takes a couple of minutes to check its correctness. If you saw any possible gap there then where is it please ? Is there any problem of need of reliable source to check obvious claims such as how much is 2+3 ? Spoirier (talk) 14:48, 25 November 2012 (UTC)

Let's enter the argument, if anyone would think of any reason to dismiss my page as an "unreliable source". Skipping intro and conclusion, here is a copy of my proof:

...we can restrict the study to primitive Pythagorean triples, that is when a and b are coprimes. It is equivalent to say that a and c are coprimes.

Assume a is odd. Then a²-1=(a+1)(a-1) is multiple of 8.

If b was also odd then c²-2 would be a multiple of 8, which is impossible, as c² must either be odd or multiple of 4.

Thus b is even, c is odd, and (c+a), (c-a) are even.

Let u=(c+a)/2 , v=(c-a)/2, so that c=u+v and a=u-v

u and v are coprimes, because a and c are coprimes (any common divisor of u and v would be a common divisor of a and c). Writing b=2b' the relation a²+b²=c² gives b'²= (c²-a²)/4= uv.

uv is a square, thus has only even exponents in its prime factorization.

u,v are coprimes, thus have no common element in their prime factorization.

Thus u and v also each have only even exponents in their factorization. Thus both are squares.

In conclusion, u=m² and v=n².

So if you find any gap or unclear step there, then where is it ??Spoirier (talk) 16:05, 25 November 2012 (UTC)

Finally I just noticed now that the French Wikipedia article on this subject had given the same proof as mine since December 2005. So basically I was right to expect this proof to be already known.Spoirier (talk) 11:15, 27 November 2012 (UTC)

Problem

A long time ago (I don't remember much about this) I saw on a site online that if a^2 + b^2 = c^2 (a Pythagorean triple,) then a+b cannot be a square. This is wrong. In 9,40,41, 9+40 = 49. Was this mis-remembering something?? What should it be?? Georgia guy (talk) 17:05, 25 February 2012 (UTC)

Yes, but there are many counterexamples for this. But its really hard to find one, if c must be a square too. --88.152.231.121 (talk) 09:31, 1 March 2012 (UTC)

The only thing that occurs to me is that maybe you're remembering that the sum of the hypotenuse and the odd leg of a primitive triple cannot be a square -- it's 2m². This is pointed out in the sixth property. Duoduoduo (talk) 21:04, 25 February 2012 (UTC)

I'm searching ...

def isq (x):

 return math.sqrt(x) == math.floor(math.sqrt(x))

def fsq():

 for m in range(2,1000000):
   for n in range(1, m):
     if (isq((m*m-n*n) + 2*m*n)):
       if (isq(m*m+n*n)):
         print "counterexample found", m, n  — Preceding unsigned comment added by 88.152.231.121 (talk) 09:54, 1 March 2012 (UTC)

I cannot find a small counterexample, but the example given in the article is correct. --88.152.231.121 (talk) 10:52, 1 March 2012 (UTC)

It looks really unbelievable, but there seems to be no counterexample smaller than the one given in the article text. It demonstrates that checking millions of small numbers is still no proof, that no larger number exists having a certain property. — Preceding unsigned comment added by 88.152.231.121 (talk) 18:07, 2 March 2012 (UTC)

The search con be accelerated, since m must be odd, and n an integer multiple of 42.

 for m in range(43, 1000000, 2):
   for n in range(42, m, 42):

--88.152.231.121 (talk) 15:00, 5 March 2012 (UTC)

Properties

In the title of the section about properties it says they hold for primitive triples. Of course they do, but aren't they valid in general for P. triples? Nijdam (talk) 16:11, 29 March 2012 (UTC)

A lot of them are valid in general, but not all of them. For example, "The sum of the hypotenuse and the even leg of a primitive Pythagorean triple is the square of an odd number": if we start with a primitive example for which this is true, and then make a non-primitive example by doubling all sides, we've doubled the sum of the hypotenuse and a leg, so the result is twice a square, which is not a square. As another counterexample, to your suggestion, "Exactly one of a, b is odd; c is odd" no longer holds true if we double all sides. Duoduoduo (talk) 16:26, 29 March 2012 (UTC)

Thanks, I've noticed this myself in the mean time. Maybe we can mention separately when a property is valid for all triples. Nijdam (talk) 16:37, 29 March 2012 (UTC)

Format

The section "Generating triples" contains two fairly long equations. If the article is viewed in a window where these equations cannot fit to the left of the illustrations, the result is a ugly gap between one of them and the preceding text.

Can someone more experienced in the ways of placing graphics in Wikipedia adjust the placement of the illustrations to avoid tihs problem? --76.69.136.84 (talk) 02:38, 9 April 2012 (UTC)

Not original research

I don't currently see anything that looks like original research. Does anyone else? This must be a well-trodden area. There might always be mistakes, and maybe the order of presentation might be original, but this seems like a good article to me. Can we remove the "original research" template -- or is that meant as a warning to someone wanting to add something that is not generally known or accepted?

Some of the divisibility properties in the section "Elementary properties of primitive Pythagorean triples" (for 7, 8, 9, 11, and 13) are unreferenced, and I don't think I've seen them elsewhere. Duoduoduo (talk) 19:17, 30 April 2012 (UTC)

I've removed the article original research tag and added a section original research tag to the section Elementary properties of primitive Pythagorean triples. Foxjwill (talk) 02:59, 23 October 2012 (UTC)

C uneven

Why if A B C triple, C is always uneven? — Preceding unsigned comment added by 93.172.232.190 (talk) 10:43, 3 September 2012 (UTC)

See this [1] Myrvin (talk) 09:18, 23 October 2012 (UTC)

If C is even, then it's either all even (which is possible,) meaning that it can be reduced by dividing all of them by 2, or A and B are both odd, which is impossible because all odd squares are 1 plus a multiple of 4, and the sum of 2 odd squares will never be divisible by 4. Georgia guy (talk) 12:29, 23 October 2012 (UTC)

YOUR STATEMENT : "For each natural number n, there exist at least n different PPT's with the same leg a, where a is some natural number". I think the following is a sharper result.

Step 1 : Each odd number 2p + 1, p ε Z+, can act as the odd leg(a) of a PPT. Take m = p + 1 and n = p. Then m, n ε Z+, m > n, m and n of opposite parity and gcf(m,n) = 1, because m and n are consecutive numbers. By Euclid's formula is (m² - n², 2mn, m² + n²) = (2p + 1, 2p² + 2p, 2p² + 2p + 1) a PPT.

Step 2 : A number of the form p^A, p prime and A ε Z+ can act only once as an odd leg, because p^A = m² - n² = (m - n)(m + n) and the only allowed factorisation is m + n = p^A and m - n = 1. All other factorisations will lead to gcf(m - n,m + n) ≥ p and this together with the imparity of m and n to gcf(m,n) ≠ 1.

Step 3 : If a = p₁^A·p₂^B· · · · · p_k^T than a can 2^k-1 times act as odd leg in different PPT's. We prove this by mathematical induction. The statement is true for k = 1 (see step 2). Assume the statement is true for k and consider p₁^A·p₂^B ·····p_k^T·p_k+1^S. Without the factor p_k+1^S we have 2^k-1 factorisations in two factors. Each of the two factors in the 2^k-1 factorisations can be multiplied by p_k+1^S to give a total of 2·2^k-1 = 2^k factorisations.

Step 4 : By a given natural t one can take a = p₁·p₂·····p_t, for instance the first t odd prime numbers. One has 2^t-1 different PPT's, all with the same a and 2^t-1 ≥ t for all t ε Z+.

Note : The even leg b can be investigated in a similar way. The hypotenuse is much more difficult.

No sources.C.W. Vugs (talk) 15:38, 8 November 2012 (UTC) C.W. Vugs (talk) 12:18, 9 November 2012 (UTC)

b EVEN OR ODD I know that your article is used in Dutch schools due to the large list of properties. The exchange of even and odd is a bit confusing. In proofs around Euclid's formula b is even. In the property 20² + 21² = 29² b is odd. In a + j² = c = b + 2^k b is odd. In the property a+b a square b is even. In some relationships b = 2r+1 b is odd. In sinθ b is even. In tanθ b is even. In geometry for PPT b is odd. In parent/child relationship b is even. In Gaussian integers b is even. C.W. Vugs (talk) 17:35, 30 November 2012 (UTC) C.W. Vugs (talk) 21:18, 2 December 2012 (UTC)

DEFINITION OF A PPT. A primitive Pythagorean triple (PPT) is one in which a, b and c are pairwise coprime.Here, I think, are too many conditions and gcd(a,b) = 1 would be sufficient. Most articles on PPT's use gcd(a,b,c) = 1. If gcd(a,b) = 1 then gcd(a,b,c) = 1. If in a PPT gcd(a,b,c) = 1 then for instance gcd(a,c) = 1. Assume gcd(a,c) = g ≥ 1 then (fundamental theorem of arithmetic) the factorisation of g contains a prime factor p|g → (euclid's algorithm) p|a and p|c → p|a² and p|c² → p|(c² - a²)= b² → (euclid's lemma) p|b, because p is prime → (euclid's algorithm) gcd(a,b,c) ≥ p. Contradiction. — Preceding unsigned comment added by C.W. Vugs (talk • contribs) 21:14, 2 December 2012 (UTC)

PROOF OF EUCLID'S FORMULA In the sufficient part has to be proved that gcd(m² - n², 2mn, m² + n²) = 1 given that m, n ε Z⁺, m > n, m and n of opposite parity and gcd(m,n) = 1. Assume gcd(m² - n², 2mn, m² + n²) = g > 1. The number g can not be 2, because both m² - n² and m² + n² are odd due to the imparity of m and n. It means that the factorisation of g in prime factors contains the factor 2² or an odd prime number p. Firstly when 2² | g → 2² | (m² - n²) and 2² | (m² + n²) → 2² | 2m² and 2² | 2n² → 2 | m² and 2 | n² → (because 2 is prime) 2 | m and 2 | n → gcd(m,n) ≥ 2. Contradiction. Secondly when p (prime) | g then p | (m² - n²) and p | (m² + n²) → p | 2m² and p |2n² → (because p not a divisor of 2)p | m² and p | n² → (because p prime) p | m and p | n → gcd(m,n) ≥ p. Contradiction.C.W. Vugs (talk) 15:01, 3 December 2012 (UTC)

PROOF OF EUCLID'S FORMULA In the necessity part I do not agree with the statement : b and c have opposite parities (one is odd, one is even) and the proof that follows. What about a and c? This is the start of the b-even-or-odd-confusion. Before going into Euclid's formula we need some fundamental facts. FACT 1 : In a PPT (a,b,c) we have gcd(a,b,c) = 1 → gcd(a,b) = gcd(a,c) = gcd(b,c) = 1, see my definition. FACT 2 : Every a ε Z⁺ can be written as 4p, 4p + 1, 4p + 2 or 4p + 3. The squares become 16p² = 4(4p²), 16p² + 8p + 1 = 4(4p² + 2p) + 1, 16p² + 16p + 4) = 4(4p² + 4p + 1), 16p² + 24p + 9 = 4(4p² + 6p + 2) + 1. Thus a square is never a multiple of 4 plus 2 or 3. FACT 3 : Not both legs in a PPT can be even, because then c² and c are also even and gcd(a,b,c)≥ 2. Not both legs can be odd, because then c² = a² + b² = (2p + 1)² + (2q + 1)² = 4(p² + q² + p + q) + 2, but a square can not have this form. One leg is odd, one leg is even and we can take b is even and a is odd, because a and b are interchangeble in both a² + b² = c² and gcd(a,b,c) = 1. Summary : In a PPT is a the odd leg, b the even leg and c the odd hypotenuse, a < c and b < c. This is the most used convention.C.W. Vugs (talk) 19:15, 3 December 2012 (UTC)

I agree with the rest of this nice proof where b is even is used in the cross multiplication.C.W. Vugs (talk) 20:45, 3 December 2012 (UTC)

ELEMENTARY PROPERTIES

Please delete the heading a < b < c. I like to add some properties that can be found by solving quadratic diophantic equations using the m,n - substitution.

There doesn't exist a PPT with c = 100. There exists exactly 1 PPT with c = 41. There exist exactly 4 PPT's with b = 60. There exist exactly 2 PPT's with b = 48. There exists exactly 1 PPT with a = 31. There exist exactly 2 PPT's with a = 33. There exists exactly 1 PPT with a perimeter of 84. There exists exactly 1 PPT with an area of 84. There exists exactly 1 PPT with sin α = 11/61. There exists exactly 1 PPT with cos α = 28/53. There exists exactly 1 PPT with tan α = 33/56. In each PPT m + n is an odd number. This property enables us to make a list of the PPT's. Take m + n is 3, 5, 7, 9, ...... and split m + n = 2p + 1 in pairs (p+1, p), (p+2, p-1), (p+3, p-2), ........ , (2p-1, 2), (2p, 1). If m + n is prime all splittings can serve as pairs (m, n). If m + n is composed some pairs will have gcd(m, n) > 1 and those should be deleted.C.W. Vugs (talk) 15:04, 4 December 2012 (UTC)

SUSPICIOUS SOURSES If [2] of Mitchell, Douglas W. doesn't prove that gcd(m² - n², 2mn, m² + n²) = 1 then [2] is suspicious. If [3] of Beauregard, Raymond A contains the same part of your article on imparity of b and c then [3] is suspicious.C.W. Vugs (talk) 17:44, 6 December 2012 (UTC)

NON-SOLVABILITY OF R^2 + S^4 = T^4

Point 3 of properties states : at most one of a, b, c is a square. The article refers to infinite descent#non-solvability of r2 + s4 = t4 and so do I.

The case of y and z : I agree with that part.

The case of y and x : If y is a square and x is not a square (for instance twice a square) we are lucky because then z is a square by assumption and we can go back to case 1. But if y is a square and x = 2ab is a square we are in trouble, because then not both a and b can be squares and the constructed smaller PPT (y^½, b, a) doesn't have 2 square sides.

A PPT (x, y, z) can not have 2 sides that are both twice a square because then the other side is also a multiple of 2 and thus gcd(x, y, z) ≥ 2.C.W. Vugs (talk) 12:15, 7 December 2012 (UTC)

FERMAT'S LAST THEOREM FOR SPECIFIC EXPONENTS.

Looking for a better proof of the non-solvability I found, in my opinion, an almost correct proof in Fermat's last theorem for specific exponents. Here an answer to critical remarks of 178.203.183.31. See the talk page of that article.

N = 4 stands for the equation x⁴ + y⁴ = z⁴, that of which the writer gives the proof (not trivial at all) it doesn't have solutions in Z⁺. Much later he starts talking about PPT's. Your (178.203.183.31) N = 2 stands for a PPT (a, b, c) and you like to prove the non-existence of the properties (1) b = 2uv is a square or twice a square and (2) a = u² - v² is a square or c = u² + v² is a square. This is complicated but in simple words you like to show : a PPT (a, b, c) doesn't have 2 square sides. You are using the same method as above (see my non-solvability contribution)and this comes to a dead end.C.W. Vugs (talk) 19:32, 7 December 2012 (UTC)C.W. Vugs (talk) 12:56, 8 December 2012 (UTC)

PROPERTY : AT MOST ONE OF a, b, c IS A SQUARE

The reference is Infinite descent#Non-solvability of r2 + s4 = t4. The proof in that website is not correct. A better source is mathematical proof by Fermat. Look also at both non-solvable equations in the talk page. You need both here.C.W. Vugs (talk) 20:09, 10 December 2012 (UTC)C.W. Vugs (talk) 20:32, 10 December 2012 (UTC)

AN IMPORTANT FACT IN THE WEBSITE : PROOF OF FERMAT'S LAST THEOREM FOR SPECIFIC EXPONENTS

Under prime factorization in that article you find the following important fact. One consequence is if x^p = uv and if gcd(u, v) = 1 then u and v are themselves the p^th power of two other numbers, u = r^p and v = s^p. This fact is frequently used and I couldn't find it in other websites on Euclidean number theory. It is difficult to reach the website in question from other websites like for instance this one.C.W. Vugs (talk) 11:13, 11 December 2012 (UTC)

FERMAT : THE EQUATION X⁴ - Y⁴ = Z² DOESN'T HAVE SOLUTIONS IN Z⁺.

Step 1 : Assume the solution set is not empty then the set {X : X, Y, Z ε Z⁺ and X⁴ - Y⁴ = Z²} is a non-empty subset of the natural numbers and this has a smallest element x (well ordening principle). The triple (x, y, z) is a solution of the equation above and no triple with a smaller x is also a solution. Moreover is gcd(x, y, z) = 1, if not there would be a smaller x. Further assume gcd(x, y) = g > 1 then the factorisation of g contains a prime factor p and it follows that p|g → (euclid's algorithm) p|x and p|y → p|x⁴ and p|y⁴ → p|z² → (euclid's lemma)p|z and thus gcd(x, y, z) ≥ p. Contradiction. We claim further gcd(x² - y², x² + y²)is 1 or 2. Because assume that gcd = g > 2 then the factorisation of g contains the factor 4 or a odd prime number p. If 4|g then 4|.......(see above)→ 4|2x² and 4|2y² → 2|x² and 2|y² → 2|x and 2|y because 2 is prime. But then gcd(x, y) ≥ 2. Contradiction. If odd prime p|g then p|.......(see above) → p|2x² and p|2y² → (euclid's lemma) p|x and p|y and is gcd(x, y) ≥ p. Contradiction.

Step 2 : From here onwards we take the approach of this article and start with gcd(x² - y², x² + y²)= 2. Example x = 3 and y = 1. Not both x and y can be even because gcd(x, y) is still 1. Further x and y can not have opposite parity because then both x² - y² and x² + y² are odd and their gcd would not be 2. Hence x and y are odd, x > y and z is even. So the triple (y², z, x²) is a PPT because (y²)² + z² = (x²)² and gcd(y², z, x²) = 1 (gcd(x²,y²) = gcd(x, y) = 1). According to Euclid's formula is there a pair (d,e), d and ε Z⁺, d > e, d and e of opposite parity and gcd(d, e) = 1 with y² = d² - e², z = 2de, x² = d² + e². It follows that (xy)² = d⁴ - e⁴ and here we have a solution of Fermat's equation but 0 < d < x. Contradiction.

Step 3 : Now we look at gcd(x² - y², x² + y²) = 1. In this case both x² - y² and x² + y² are squares because their product is a square (z²) and ggd(x, y) = 1 (see the consequence of unique factorisation in this present article). Thus x² - y² = s² , x² + y² = t², t² - s² = 2y², t² + s² = 2x² (formulae 1). Due to the last formula s and t are of equal parity and not both can be even because then 4|(s² - t²) and 4|(s² + t²) → 2|x² and 2|y² and gcd(x, y) ≥ 2. Both s and t are odd and their sum and difference are even. Gcd(s, t) = 1, because gcd(s, t) = g > 1 would lead to gcd(x, y) ≥ p for a prime number p.

Step 4 : Define u = (t + s)/2 and v = (t - s)/2 and note that u and v ε Z⁺, u > v and gcd(u, v) = gcd(t, s) = 1. Not both u and v can be even since then both t and s are also even and this is impossible (see step 3). Further y² = (t² - s²)/2 = 2((t + s)/2)((t - s)/2) = 2uv (see also formulae 1) and one of the numbers u and v must have a factor 2, otherwise 2uv is not a square. Without loss of generality we take u = 2m² even and v = k² odd.

Step 5 : The triple (v, u, x) is a PPT, since v² + u² = (t² - 2ts + s²)/4 + (t² + 2ts + s²)/4 = (t² + s²)/2 = x² (see formulae 1) and gcd(v, u, x) = 1 because gcd(u,v) = 1. According to Euclid's formula is there a pair (d, e), d and e ε Z⁺, d > e, d and e of opposite parity, gcd(d, e) = 1 and (v, u, x) = (d² - e², 2de, d² + e²). It follows that u = 2m² = 2de, thus de = m² and d and e are squares because gcd(d, e) = 1. Now taking d = g², e = h² we get v = d² - e² = g⁴ - h⁴ = k² and have a solution of Fermat's equation with g = d^½ < d² + e² = x. Contradiction.

NON-SOLVABILITY OF X⁴ + Y⁴ = Z²

Step 1 : Like in Fermat's equation we start with a solution (x, y, z), gcd(x, y) = gcd(x, z) = gcd(y, z) = gcd(x, y, z) = 1 and in this case with minimal z. The triple (x², y², z) is a PPT because x², y² and z ε Z⁺, gcd(x², y², z) = 1 and (x²)² + (y²)² = z². There exists a pair (m, n), m and n ε Z⁺, m > n, m and n of opposite parity, gcd(m, n) = 1 (euclid's formula) such that (x², y², z) = (m² - n², 2mn, m² +n²). The triple (x, n, m) is a PPT because x² + n² = m² and gcd(x, n, m) = 1 since gcd(m, n) = 1.

Step 2 : Once again (euclid's formula) there is a pair (d, e), d and e of opposite parity, gcd(d, e) = 1 such that (x, n, m) = (d² - e², 2de, d² + e²). Now y² = 2mn = 4de(d² + e²) or (y/2)² = de(d² + e²). Since gcd(d, e) = gcd(d, d² + e²) = gcd(e, d² + e²) = 1 and their product is a square the 3 factors are squares : d = p², e = q² and d² + e² = r². Thus p⁴ + q⁴ = r² and we have a solution but with a smaller hypothenuse r = (d² + e²)^½ < d² + e² = m < m² + n² = z. Contradiction.84.24.10.61 (talk) 16:11, 11 December 2012 (UTC)

DIFFICULT PROPERTIES CAN NOW BE PROVED

At most one side of a Pythagorean triple (primitive or not) can be a square. The area of a Pythagorean triple (primitive or not) can not be a square. The area of a Pythagorean triple (primitive or not) can not be twice a square. C.W. Vugs (talk) 16:51, 11 December 2012 (UTC)

Today I like to talk with Duoduoduo. I, C.W. Vugs, pensioner teacher principal and lecturer, am an admirer of your website. The long list of properties makes it very usefull for students. And it is used frequently. Students can find their own original properties, there are still enough. Other websites on Euclidean number theory are full of learned slogans but without applications. Students can't do anything with those. Can you not give your e-mail address so that we can discuss one and another. My address is cwvugs@ziggo.nl . Or can you allow me to make some changes in the core of your article.C.W. Vugs (talk) 11:27, 12 December 2012 (UTC)

TRIANGLES OF PYTHAGORAS AND PLATO

I do not understand the blackboard part. I think you are mixing up side a with a given even/odd number a.

Pythagoras starts with an odd number 2p + 1 and forms (a, b, c) = (2p + 1, 2p² + 2p, 2p² + 2p + 1)

Plato starts with an even number 2q and forms (a, b, c) = (q² - 1, 2q, q² + 1)

Property : if q is even the triangle of Plato is a PPT.C.W. Vugs (talk) 09:35, 21 December 2012 (UTC)

SOME RELATIONS

In the some relations section I agree with the blackboard part. Perhaps it is better to replace θ by α.

You don't need r and another side to find the PPT. This is a serious mistake. The unknown sides of a triple can be calculated directly from the radius r of the incircle. Point no more givens. Example : r = n (m - n) = 10 with m, n ε Z⁺, m > n, m and n of opposite parity and gcd(m,n) = 1. Solutions n = 10, m = 11; n = 5, m = 7 (not applicable); n = 2, m = 7; n = 1, m = 11 (not applicable) with the PPT's (21, 220, 221) and (45, 28, 53). The first one you find as well with complicated assumptions and calculations.

In general you don't use the m,n-substitution to solve diophantic equations. See for instance the last sentence of the section. This has to be changed. If one number of a triple is known, the two others can be found using the m,n-substitution. C.W. Vugs (talk) 12:30, 23 December 2012 (UTC)

How to generate ALL Pythagorean triples

The following form:

$n(2k+d),n(2k+{\frac {2k^{2}}{d}}),n(2k+d+{\frac {2k^{2}}{d}})$

generates all Pythagorean triples, for all or any n > 0, k > 0, where d can be any factor or divisor of 2K².

P.S.: C.W. Vugs, thank you very, very much for your help and contribution! The error in my former statement was due to the fact that in my initial calculations I've used the same notation (d) for two different things: d = c - a, and d = gcd(A, B, C), where (A, B, C) = d (a, b, c).

Euclid's formula : ALL Pythagorean triples (primitive or not) are generated by k(m² – n², 2mn, m² + n²), with k, m, n ε Z⁺, m > n, m and n impair and gcd(m, n) = 1. With k = 1 we get ALL primitive Pythagorean triples.C.W. Vugs (talk) 09:48, 11 February 2013 (UTC)

Your formula generates the Pythagorean triple (6, 8, 10) by taking n = 2, k = 1 and d = 1 as well as by taking n = 1, k = 2 and d = 2. Other than in Euclid's formula is there not an 1-1-correspondence.C.W. Vugs (talk) 14:53, 11 February 2013 (UTC)

Euclid's formula (which is by far more elegant than mine) also generates redundant results. For instance, (t², m, n) and (1, mt, nt) always generate the same (a, b, c), for all and any given values of m, n, t. — Preceding unsigned comment added by 79.113.196.88 (talk) 12:30, 12 February 2013 (UTC)

Do you also happen to know any formula(s) that can generate ALL bi-pythagorean (or pluri-pythagorean) numbers ? By that I mean numbers whose square can be written as the sum of two other squares in two (or more) different ways ? — Preceding unsigned comment added by 79.113.196.88 (talk) 10:16, 12 February 2013 (UTC)

You don't obey Euclid's formula. In the case (t², m, n) you take k = t² and hopefully m and n as prescribed by Euclides. In the case (1, mt, nt) you take k = 1 but now gcd(mt, nt) ≥ t and thus against the subconditions. Your letters t, m and n are ε Z⁺, but in Euclid's formula k, m and n has many subconditions. — Preceding unsigned comment added by C.W. Vugs (talk • contribs) 18:11, 12 February 2013 (UTC)

I'm sure that similar conditions could be added to this formula as well to make it non-redundant (if one so desires). There's always room for refinement and improvement. — Preceding unsigned comment added by 79.113.196.88 (talk) 18:37, 12 February 2013 (UTC)

Becker-Sievert Formulas

$(a+b)^{2}+((-a^{2}-ab)/(a-b)+a)^{2}=(b+(-a^{2}-ab)/(a-b))^{2}$

a,b ∈ℕ ; b>a

or

$(a+b)^{2}+((a^{2}+ab)/(a-b)-a)^{2}=((a^{2}+ab)/(a-b)-b)^{2}$

a,b ∈ℕ ; b<a

or

$(ab)^{2}=((a^{2}+b^{2})/2)^{2}-((a^{2}-b^{2})/2)^{2}$ --Becker-Sievert (talk) 12:17, 18 December 2014 (UTC)

a,b ∈ℕ ; b<a both odd or both even --Becker-Sievert (talk) 14:04, 18 December 2014 (UTC)

— Preceding unsigned comment added by Becker-Sievert (talk • contribs) 13:29, November 19, 2014‎

There's no "c" in the formulas, and you don't specify the conditions for the expressions to be integers, and ... why use such a complicated formula when the one above is so much simpler. — Arthur Rubin (talk) 09:58, 20 November 2014 (UTC)

c deleted thx

By the way, if you are the "Becker-Siervert" whose formula this is, it shouldn't be in the article unless published. and the first formula above with k and d (removing the n) is simpler, due to the transformation I discovered in #Analysis below. — Arthur Rubin (talk) 16:05, 20 November 2014 (UTC)

Analysis

$a=(2k+d),b=\left(2k+{\frac {2k^{2}}{d}}\right),c=\left(2k+d+{\frac {2k^{2}}{d}}\right)$

Corresponds to $2k=a+b-c,\left(d,{\frac {2k^{2}}{d}}\right)=(c-b,c-a)$

Which corresponds to, using the standard formulation $2k=2KN(M-N),\left(d,{\frac {2k^{2}}{d}}\right)=\left(K(M-N)^{2},2KN^{2}\right)$

or $k=KN(M-N),d=K(M-N)^{2},2KN^{2}$

Now, the Becker-Sievert formula above reads

$A=a+b;B=\pm \left({\frac {a^{2}+ab}{a-b}}-a\right),C=\pm \left({\frac {a^{2}+ab}{a-b}}-b\right)$ or $A=a+b={\frac {\left|a^{2}-b^{2}\right|}{|a-b|}},B={\frac {2ab}{|a-b|}},C={\frac {a^{2}+b^{2}}{|a-b|}}$ which can be solved by $a=KM(M-N),b=K(M-N)N$ or $a=KN(M+N),b=KN(M-N)$ (with A and B reversed)

So both formulas have the advantage over the "standard" formula that no explicit scale factor is required, but the disadvantage of there being an odd divisibility condition. — Arthur Rubin (talk) 11:05, 20 November 2014 (UTC)

After going through all that, the equations are related (for a > b) by

b=k,a=k+d

. :) — Arthur Rubin (talk) 11:26, 20 November 2014 (UTC)

THE THREE SIDES OF A PPT.

The odd leg a is of the form 2p + 1, p ε Z⁺. Each number of the form 2p + 1 can serve as as the odd leg in a PPT. If a = p₁^A•p₂^B•••••p_k^X with p_i's odd primes and A, B,....,X ε Z⁺ than 2^k-1 different PPT's can be constructed with a as odd leg.

The even leg b is not of the form 4p + 2, p ε Z⁺. Each number of the form 4p, p ε Z⁺ can serve as the even leg in a PPT. If b = 2^A•p₁^B•••••p_k^X with p_i's odd primes and A ≥ 2, B, ....,X ε Z⁺ than 2^k different PPT's can be constructed with b as even leg.

The hypothenuse is of the form 4p + 1. The hypothenuse c = m² + n² (Euclid's formula) is irreducible and this complicates things. Not each number of the form 4p + 1 can serve as hypothenuse (example c = 9). If c is of the form 4p + 1 and prime it can serve as hypothenuse. But there are also numbers of the form 4p + 1 and not prime that can serve as hypothenuse. For example c = 65 can serve as hypothenuse in two different PPT's.C.W. Vugs (talk) 17:18, 14 February 2013 (UTC)

Yes, but these numbers are the products of two or more such 4k + 1 primes. In the case of 65, it is the product of 5 and 13, both of which are 4k + 1 primes. — Preceding unsigned comment added by 79.113.241.126 (talk) 09:16, 16 February 2013 (UTC)

I suppose the rule is : if c = p₁^A•p₂^B•••••p_k^X with p_i's primes of the form 4n + 1 than 2^k-1 different PPT's can be constructed with c as hypothenuse.I can't prove it, especially not the property written in this article that all prime factors of c are primes of the form 4n + 1. This seems to be original but I doubt that the writer can prove it.C.W. Vugs (talk) 10:13, 18 March 2013 (UTC)

The fact that all 4k + 1 primes can be written as the sum of two squares has been first discovered by Fermat and then proven by Euler [2]. And I didn't say "all factors", but merely two or more. — 79.113.221.97 (talk) 01:32, 20 March 2013 (UTC)

Your exact text : "All prime factors of c are primes of the form 4n + 1". Here your hypothenuse c is no longer a prime and Fermat (Euler) can not be applied. Nevertheless I think that your statement is correct, but I don't have the proof. Directed at the writer, I would like to ask if I can correct the mistakes as indicated in my remarks above. Your article is frequently visited because of the long list of properties that students can use in writing a project. It is a pity that there are mistakes.C.W. Vugs (talk) 10:38, 20 March 2013 (UTC)

I never said or wrote that... either here, or on your Talk page. (Perhaps you are confusing me with someone else, who wrote or said something similar ?) — 79.113.214.121 (talk) 16:45, 20 March 2013 (UTC)

Another identity discovered by Euler, which I forgot to mention, was that the product of numbers who are each the sum of two squares is itself the sum of two squares: (aa + bb)(cc + dd) = (ac + bd)² + (ad - bc)². — 79.113.214.121 (talk) 07:48, 21 March 2013 (UTC)

Sorry 79.113.214.121. My comment "Your exact text ..... mistakes" was directed at the writer of the article in question, perhaps 79.113.221.97. Point 19 of the list of properties in his/her article states : "All prime factors of c are primes of the form 4n + 1". This without references. I am interested in the proof of his/her statement and, like you, I think that somewhere we have to use the identity of Brahmagupta-Fibonacci.C.W. Vugs (talk) 09:00, 21 March 2013 (UTC)

Well, I guess it's kinda easy to understand why (though understanding and demonstrating are two different things): if you'd add any other kind of factors to the hypothenuse, then -by their very nature- they would not contribute anything to its Pythagorean property. So then you'd have to have the same kind of factors present in the two cathetes as well. Which would mean the triple is Pythagorean, but not prime. Something of the sort. — 79.113.245.81 (talk) 22:04, 21 March 2013 (UTC)

Starting with philosophical effusions me too like to add some slogans. A good proof demonstrates the "Why" of a theorem. In this article the definition of a PPT and the demonstration of Euclid's formula are sloppy (see for this my earlier remarks). The "Why"of Euclid's formula is to reduce an equation with three unknowns (a,b,c) to an equation with two unknowns (m,n) and to solve the diophantine equation. In the part "Some relationships" he ends with "If two numbers of a triple are known, the third can be found by using the Pythagorean theorem". This knows a child. He should have stated : if one side of a PPT is known, the two others can be found using Euclid's formula. For instance : find a PPT with a leg of 32, and find two PPT's with a leg of 33.C.W. Vugs (talk) 11:29, 22 March 2013 (UTC)

(33, 56, 65). — 79.113.231.88 (talk) 13:36, 22 March 2013 (UTC)

This and the other solution can be found by a correct use of Euclid's formula. Maths has to be applied and is not a list of uncontrolled learned slogans.C.W. Vugs (talk) 12:12, 23 March 2013 (UTC)

(33, 544, 545). But there aren't any primitive solutions for 32. — 79.113.210.71 (talk) 04:13, 24 March 2013 (UTC)

Correct, but there is a solution for 32. There is no solution for 34. Try to prove this as well. Today I wanted to send my usual financial donation to wikipedia, but the writer of the article deleted all my mathematical contributions to his article. Perhaps he is paid for this job. So I like to wait a bit.C.W. Vugs (talk) 19:40, 24 March 2013 (UTC)

LOL! Same thing happened to me a while back, and I told that guy the same thing! :-) 32 cannot be of the form mm - nn, because then mm + nn would be even as well, and the other way around, thus making all three sides even. And it cannot be of the form 2mn either, since it is a power of two, meaning that both m and n are themselves even, being also powers of two, thus making all three sides even as well. But if all three sides are even, the triple is no longer primitive. — 79.113.235.6 (talk) 22:30, 24 March 2013 (UTC)

UNLESS we're dealing with the ONLY power of two (or of ANY number for that matter), which is NOT divisible through the number itself: 1. Then m = 16 and n = 1, giving (32, 255, 257). — 79.113.235.6 (talk) 22:33, 24 March 2013 (UTC)

P.S.: Could you help me out in finding a similar formula for generating all numbers which can be written as the sum of two cubes in two (or more) different ways ? I'm stuck. — 79.113.235.6 (talk) 22:44, 24 March 2013 (UTC)

This came up recently on the maths reference desk and I put the citations and solution from them into Euler's sum of powers conjecture. Dmcq (talk) 23:40, 24 March 2013 (UTC)

Then perhaps if someone could help me understand how the Swiss genius arrived at that particular conclusion... :-) — 79.113.235.6 (talk) 00:17, 25 March 2013 (UTC)

One of the citations there derives a solution, and here is Euler's work, it is in Latin but it can be followed fairly easily. Dmcq (talk) 16:07, 25 March 2013 (UTC)

LOL, INDERDAAD. (255,32,257) is a sound PPT. I am struggling with the same problem.C.W. Vugs (talk) 07:26, 25 March 2013 (UTC)

PROPERTY ONE

Your first property reads : (c-a)(c-b)/2 is always a perfect square. This is particularly useful in checking if a given triple is a Pythagorean triple, but is only a necessary ccndition, not a sufficient one. The triple {6, 12, 18} passes the test that (c-a)(c-b)/2 is a perfect square but is not a Pythagorean triple. When a triple of numbers a, b, c forms a PPT, then (c-b) and (c-a)/2 are both squares; however this is not a sufficient condition, as the triple {1, 8, 9} is a counterexample since 1² + 8² ≠ 9².

This is an easy property, that can be proved by the m,n - substitution in one line. I don't like the counterexamples : {6, 12, 18} is even not a triangle and {1, 8, 9} not a right triangle. We should not start fishing outside the set of PT's. Then the condition in statement 1 is sufficient and we don't need a counterexample. This is not the case in statement 2 where (12, 16, 20) could be a counterexample. A definition of a PT could be : (a, b, c) a PT <—> there exist k, m and n ε Ζ⁺, m > n, m and n impair, gcd(m,n) = 1 and (a,b,c) = k(m² - n², 2mn, m² + n²). I was always a teacher and I can't resist to add an exercise. Find k, m, and n of the two PT's (2925, 3536, 4589) and (2697, 3904, 4745).C.W. Vugs (talk) 14:37, 25 March 2013 (UTC)

PROPERTY NINETEEN

Property 19 : All prime factors of c are primes of the form 4n + 1. There were many reactions when I started to talk about this property. Other people seem to struggle with that as well. A good site to make a start with this property is : 19. The Fermat-Euler Prime Number Theorem. It is good maths. I am still not paying my donation.C.W. Vugs (talk) 11:19, 3 April 2013 (UTC)

Um. Is there a point to all of your postings here? Wikipedia articles are supposed to reflect the published literature, and Wikipedia talk pages are supposed to be discussions about how to improve the articles. This all appears to be neither. Can you give me a reason why I shouldn't just remove it all as irrelevant and off-topic? —David Eppstein (talk) 16:22,

When I say that the article's definition of a PPT claims too many conditions and prove my statement is that irrelevant and off-topic? When I say that the theorem around Euclid's formula is no well formulated and the proof is not complete and I prove that statement is that irrelevant and off-topic? When I say that diophantin equations has to be included in an article on PPT's and that there is not a single one in this article (besides the properties) and I prove that statement is that irrelevant and off-topic? I can go on in this way. When you allow me to make changes I will do it and restart my donations. But up to now all my contributions in the article itself were removed.C.W. Vugs (talk) 15:39, 4 April 2013 (UTC)

That's because your contributions have been some combination of the following: difficult to decipher, unencyclopaedic, original research, poorly formatted, unreferenced, and not worked in to the surrounding text. For example, you have raised in several places and at great length the question of whether the definition of primitive Pythagorean triple requires setwise or pairwise coprimality. It is indeed true that in the case of PPTs setwise coprimality implies pairwise coprimality and so the weaker condition would suffice. The normal way to handle this situation would be to write two or three brief sentences on the talk page pointing this out, and editing the article to change mentions of pairwise coprimality to just coprimality. Instead, you've posted essay-length stuff on this subject on the talk page while making edits to the article that make it less clear without actually making the relevant change! Speaking for myself but probably also for some large number of other observers, this behavior is really baffling, and I think it largely explains why your edits keep being reverted. I don't really know what to suggest; perhaps you should try to restrict the volume of your "demonstrations" and instead make shorter, to-the-point comments with suggestions of particular revisions? Then if no one objects after a day or two you can try to make the edit and see what happens. --JBL (talk) 18:34, 4 April 2013 (UTC)

Suggestions of particular revisions 1

JBL's text. A Pythagorean triple consists .......

Proposed revision : A Pythagorean triple (PT) consists .......

JBL's text. A primitive Pythagorean triple (PPT) is one in which a, b, c are pairwise coprime.

Proposed revision. A primitive Pythagorean triple (PPT) is one with gcd (a, b, c) = 1.C.W. Vugs (talk) 12:30, 5 April 2013 (UTC)

No. This sort of formalism just makes things more difficult for the reader. Better to use clear English prose than formulas whenever possible. Of course, "pairwise coprime" and "fcd(a,b,c)=1" mean different things mathematically but I don't see why the difference is important. —David Eppstein (talk) 14:57, 5 April 2013 (UTC)

David, it is clear, you don't like the gcd. But coprimality is based on the gcd and this on Euclid's algorithm. Anybody knows that 11 and 20 are coprime but what about 12769 and 14803?C.W. Vugs (talk) 17:13, 5 April 2013 (UTC)

I don't have any idea why you're attributing text to me; perhaps you mean the current article? I agree with David Eppstein that the jargony abbreviation and the formula with GCD are unnecessary here. (Perhaps you're missing the fact that the word "coprime" is a link to an article all about coprimality, and anyone who wants to read about that can click the link? This article is not meant to be a self-contained mathematical treatise, it's an encyclopedia article.) On the other hand, I've done the obvious change of removing the unnecessary technical term "pairwise" in two places. One more statement about pairwise coprimality exists in the article, about Heronian rather than Pythagorean triples, and I haven't bothered to check whether it, too, could be removed. --JBL (talk) 19:54, 5 April 2013 (UTC)

Joel, you changed your definition. It reads now : a primitive Pythagorean triple is one in which a, b and c are coprime.

The term coprime is, like the gcd, used for two numbers. The numbers a and b are called coprime if gcd(a,b) = 1 and gcd(a,b) can be calculated by the 2300 year old algorithm of Euclid. What is the meaning of "a, b, c are coprime" ? The meaning of gcd(a,b.c) = gcd(gcd(a,b),c). C.W. Vugs (talk) 21:00, 5 April 2013 (UTC)

The term coprime applies both to pairs and to larger sets of numbers; a set of integers is coprime if the only positive integer dividing all the elements of the set is 1. Obviously this is equivalent to the set having GCD 1. You could learn this by e.g. clicking on the word "coprime" (either in this comment or in the article) and reading the linked article (notably, the section "generalizations").

Incidentally, while it's certainly true that GCD(a, b, c) = GCD(GCD(a, b), c), this would be an unusual definition of the GCD (though it is perhaps a good way to think about things for purposes of computing the GCD) -- a more natural definition is that the GCD of a set of integers is the largest positive integer that divides all of them. This definition doesn't tell you how to compute a GCD, but the name "GCD" is not justified without this defining property. --JBL (talk) 03:04, 6 April 2013 (UTC)

Also, I repeat: please stop attributing the definition in the article to me. It is confusing and unhelpful. I changed the definition of PPT in the article; I did not change "[my] definition." --JBL (talk) 03:05, 6 April 2013 (UTC)

The defining properties of the gcd are part of Euclid's algorithm. Like Euclid's formula the algorithm should be embedded in a theorem "For each a, b ε Z⁺ there exists a gcd(a,b) ε Z⁺. Gcd(a,b) can be calculated by Euclid's algorithm. The algorithm shows (i) gcd(a,b) is a common divisor of a and b (ii) each common divisor of a and b is a divisor of gcd(a,b)(this together with (i) means gcd(a,b) is the greatest common divisor of a and b. (iii) the diophantin equation ax - by = gcd(a,b) has a solution in Z⁺". — Preceding unsigned comment added by C.W. Vugs (talk • contribs) 08:27, 6 April 2013 (UTC)

I am satisfied. You changed 'pairwise coprime" in "coprime" and here you tell me that a, b, c coprime is the same as gcd(a, b, c) = 1. I will bank my usual donation and start with suggestions of particular revisions 2.C.W. Vugs (talk) 09:47, 6 April 2013 (UTC)

Suggestions of particular revisions 2

(i) Text of the article : If both m and n are odd, then a, b and c will be even, and so .......

Proposed revision : If both m and n are odd or if both are even, then a, b and c will be even, and so .......

(ii) Text of the article : ....., however, dividing a, b and c by 2 will yield a primitive triple if m and n are coprime.

Proposed revision : ....., moreover it can be proved that if a, b and c are coprime, then m and n are coprime as well.

(iii) Text of the article : The relation of a and b to m and n .......

Proposed revision : The relation of a, b and c to m and n .......

(iv) Text of the article : Many formulas for generating triples have been developed since the time of Euclid.

Proposed revision : Many formulas for generating triples have been developed before and after Euclid.

Additional comment : (i), (iii) and (iv) are slips of the pen. (ii) is a serious mistake.C.W. Vugs (talk) 12:25, 6 April 2013 (UTC)

If somebody is interested I can prove fact 1 : gcd(m² - n², 2mn, m² + n²) = 1 ----> gcd(m, n) = 1.It will be a proof by contradiction. If somebody has a proof without contradiction, please show it here.C.W. Vugs (talk) 12:39, 7 April 2013 (UTC)

A comment on your wording: "moreover it can be proved that" is completely superfluous. It adds no useful meaning to the sentence it is part of. (There are mathematical articles where the distinction between truth and provability is important, but this isn't one of them.) —David Eppstein (talk) 14:42, 7 April 2013 (UTC)

David is "...., however, dividing a, b and c by 2 ....." a truth or a part of a proof? It is not true and not provable, it is an impossibility, because (a, b, c) is a PPT. — Preceding unsigned comment added by C.W. Vugs (talk • contribs) 16:41, 7 April 2013 (UTC)

Example: m = 5, n = 3, both are coprime, and lead to the triple 16, 30, 34 which is certainly not primitive (nor a PPT if you insist on calling things by idiosyncratic initialisms). However, dividing by two gives the triple 8, 15, 17 which is primitive. The text of the article looks correct and your proposed replacement (ii) looks overly restrictive, to me. —David Eppstein (talk) 17:58, 7 April 2013 (UTC)

David, m = 5, n = 3 is a good example (not a proof) of the statement : (a, b, c) = (m² - n², 2mn, m² + n²) and m and n impair ----> (a, b, c) is not a PPT.

Here you have to prove the statement : (a, b, c) = (m² - n², 2mn, m² + n²) a PPT ----> Gcd( m, n) = 1.C.W. Vugs (talk) 19:12, 7 April 2013 (UTC)

If that's what you think it says now, then try reading it again. You have the implication backwards. What the part of the text you propose to replace in (ii) actually says is:

(\gcd(m,n)=1\wedge mn=1{\bmod {2}})\Rightarrow (a/2,b/2,c/2)

is primitive.

I don't see why you think this is a mistake. —David Eppstein (talk) 20:40, 7 April 2013 (UTC)

In this part of the article the writer is summing up the properties of m and n. He told us already that m and n ε Z⁺ and m > n. He is now saying that m - n is odd. He clarifies this by telling us : if both m and n are odd (or both are even), then (a, b, c) is not a PPT. Correct. He likes still to add an other property, namely gcd(m, n) = 1. To make this clear for us he states : if gcd(m, n) = 1, then (a, b, c) is not a PPT. He should say or prove : if (a, b, c) is a PPT, then gcd(m, n) = 1. In his statement he uses as given what he has to prove. I think this is not allowed. — Preceding unsigned comment added by C.W. Vugs (talk • contribs) 06:44, 8 April 2013 (UTC)

You are still misreading this. The point of this section is to describe a way to generate primitive triples from pairs (m,n). One very obvious necessary condition for the triple to be primitive is that gcd(m,n)=1; that is why it is the first assumption. But it is not a sufficient condition for primitivity, because of the (m odd,n odd) case. The section describes this issue and then shows how to modify the construction to get primitive triples always, for every (m,n) with gcd(m,n)=1: by dividing by two when both are odd. Because you don't seem to understand the flow of this reasoning, your proposed replacements don't make sense in context. —David Eppstein (talk) 08:34, 8 April 2013 (UTC)

Your translation of writer's text was correct : gcd(m, n) = 1 and mn = 1(mod 2) ----> (a/2, b/2, c/2) is a PPT. You clearly can see that this is a mistake. m, n ε Z⁺, m > n, mn = 1(mod 2) and gcd(m, n)= 1 ----> (m² - n², 2mn, m² + n²) = (a, b, c) is a PPT. This is the necessity part of Euclid's formula, which he is going to prove in next section.

I am going to edit the proposed revisions and, as the writer told me, see what happens.C.W. Vugs (talk) 10:25, 8 April 2013 (UTC)

I'm sorry to say that you seem very confused; David Eppstein is absolutely correct, and I'm not sure how to more clearly explain it, but here's a try. The sentences in question read, in full, " The triple generated by Euclid's formula is primitive if and only if m and n are coprime and m − n is odd. If both m and n are odd, then a, b, and c will be even, and so the triple will not be primitive; however, dividing a, b, and c by 2 will yield a primitive triple if m and n are coprime." These sentences are absolutely correct as written. The first sentence asserts that if m and n are coprime and one of them is even while the other is odd then the associated triple (a, b, c) will be primitive, and conversely that every primitive triple can be achieved only by such m, n. The second sentence provides some additional information about what happens if m and n are coprime but both are odd. As the first sentence explains, in this case the resulting triple is not primitive, and indeed a, b, c will all be even in this case. The second sentence elaborates by noting that this common factor of 2 is the full extent of the imprimitivity (to coin a word) of the triple; that is, when m and n are relatively prime and both odd, the associated triple (a, b, c) is not primitive but the triple (a/2, b/2, c/2) is primitive (so the former triple is "almost" primitive, with the only obstruction being the common factor of 2). All of this is correct. --JBL (talk) 21:02, 8 April 2013 (UTC)

In this part the writer is busy to clarify the properties of the numbers m and n. He told us already that in a PT m and n ε Ζ⁺ and m > n. He now is going to the property m + n is odd. He gives a proof by contradiction. m + n not odd -----> (m² - n², 2mn, m² + n²) = (a, b, c) is not a PPT. Example : m = 4, n = 2 -----> (12, 16, 20) is not a PPT. Correct. Finally he comes to the property gcd (m, n) = 1. I would say : gcd (m, n) = g > 1 -----> (m² - n², 2mn, m² + n²) = (a, b, c) is not a PPT. This is because a, b and c are in that case divisible by g. Example : m = 9, n = 3 -----> (72, 54, 90) is not a PPT . Correct 72, 54 and 90 are divisible by 3. The writer actually says : gcd (m, n) = 1 -----> (m² - n², 2mn, m² + n²) = (a, b, c) is not a PPT. This is because because a, b and c are divisible by 2. Example m = 2, n = 1 -----> (3, 4, 5) is not a PPT. Hopeless incorrect.C.W. Vugs (talk) 06:01, 10 April 2013 (UTC)

Writers text in full : the triple generated by Euclid's formula is primitive if and only if m and n are coprime and m - n is odd.

My comment. This is more or less the theorem of Euclid's formula.

Proposed revision : Theorem of Euclid's formula : (a, b, c) a PPT <-----> there are m, n ε Z⁺, m > n, m and n impair,gcd (m, n) = 1 with (m² - n², 2mn, m² + n²) = (a, b, c).

My comment : This theorem will be proved in next section and may certainly not be used, but the numbers in the right hand site can be clarified.

Writers's text in full : if both m and n are odd then a, b, c will be even and so the triple will not be primitive.

My comment : correct but not complete.

Proposed revision : if both m and are odd or both m and are even -----> (a, b, c) is not a PPT. Because a, b and c are even. Examples : see above or m = 3 and n = 1 -----> (8, 6, 10) is not a PPT.

Writer's text in full : however, dividing a, b, and c by 2 will yield a primitive triple if m and n are coprime.

My comment : Not correct. Counterexample : 2 and 1 coprime -----> (3, 4, 5) a PPT. In further "explanations" the writer is using the theorem and especially the "only if" part and that is forbidden.

Proposed revision : moreover gcd(m, n) = g > 1 -----> gcd(a, b, c) ≥ g.

My comment : of course the last statement has still to be proved in next section.C.W. Vugs (talk) 10:28, 10 April 2013 (UTC)

C. W. Vugs, I'm sorry to say that I don't much see the point of continuing this discussion. If you can't understand what David Eppstein and I have written, I am skeptical that you will be able to make constructive edits to the article. Of course you are (like everyone else) welcome to try. So long as your edits remain in the same style and spirit as what's happened so far, in my opinion they are likely to be reverted. --JBL (talk) 13:18, 10 April 2013 (UTC)

Joel, before answering your concern, I like to talk with the writer of this article. Please, writer, bring back all these nice properties written in the "a is even convention". There were a lot of nice ones. You can, as you know, simply interchange a and b and delete the heading of the properties list a < b < c.C.W. Vugs (talk) 20:06, 10 April 2013 (UTC)

Joel, I added something in the text. Let us see what the writer is going to do. The writer didn't bring back the properties I asked for. He should delete reference [2]. 85.26 is talking about PT's not about PPT's. The guy intends to advertise his book.C.W. Vugs (talk) 10:06, 12 April 2013 (UTC)

Suggestions of particular revisions 3

Text of the article : That satisfaction ..... all integers, and from the fact that a² + b² = (m² - n²)² + (2mn)² = (m² + n²)² = c².

Proposed revision : That satisfaction ..... all integers, the fact that (the same text as above) and from fact 2 : gcd(m,n) = 1 -----> gcd(m² - n², 2mn, m² + n²) = gcd(a, b, c) = 1.

Note that fact 2 is the reverse of fact 1. Both can be proved by contradiction.C.W. Vugs (talk) 13:46, 8 April 2013 (UTC)C.W. Vugs (talk) 13:48, 8 April 2013 (UTC)

A bit further I did read : b and c have opposite parities. I was not convinced and tried to find a counterexample. I made a long list of PPT's according to writer's claim a < b < c on top of his list of properties. I was lucky and found soon a counterexample, (8, 15, 17) and a bit further (20, 21, 29). To my surprise the writer has the same example in the same list. I am confused. Where did it go wrong? Is his statement : b and c have opposite parities not correct? Is the claim a < b < c not correct? Are our calculations not correct? Please, writer of the article, David, JBL or somebody else can you help me? I am stuck. — Preceding unsigned comment added by C.W. Vugs (talk C.W. Vugs (talk) 19:26, 8 April 2013 (UTC) • contribs) 19:05, 8 April 2013 (UTC)

The correspondence between the symbols a, b and the expressions 2mn,

m^{2}-n^{2}

is inconsistent in the article; various sections have different conventions, and in some of them it is announced which convention is being used. If there are particular spots where it is not made clear at the beginning of a section and it matters, that should be changed. Which section do you believe has this problem? --JBL (talk) 21:04, 8 April 2013 (UTC)

Somebody deleted all properties written in the "a is even convention". I found still one in the section "Geometry of Euclid's formula". He deleted even the whole section "Some relations". It is a pity, because some properties were very interesting, like the property "if j and k are odd, then there exist exactly one PPT with a + j² = c = b + 2^k. He could have interchanged a and b. Specially "exactly one" is surprising in the midst of "infinitely many ones".

Going through the properties I found : "exactly one of a, b is odd, c is odd". This is of course evident when one can use the m,n-substitution. But now we are proving the correctness of this sustitution. In my opinion the proof of "one odd, one even" together with the "b is even" convention should come in the very first section. What do you think ? — Preceding unsigned comment added by C.W. Vugs (talk •C.W. Vugs (talk) 09:19, 9 April 2013 (UTC) contribs) 09:08, 9 April 2013 (UTC)

Your "fact 2" is false. It is possible for gcd(m,n)=1 and yet for gcd(m^2+n^2,2mn,m^2-n^2)≠1. The case m=5 n=3 is an example. —David Eppstein (talk) 17:21, 10 April 2013 (UTC)

David things are becoming complicated. Firstly my wording of the theorem of euclid's formula : (a, b, c) a PPT <-----> there exist m, n ε Z⁺, m > n, m and n impair, gcd(m, n) = 1 with (m² - n², 2mn, m² + n²) = (a, b, c). Do you agree with this wording or do you prefer another expression? If this is the case then write it on this talk page.84.24.10.61 (talk) 19:16, 10 April 2013 (UTC)C.W. Vugs (talk) 19:52, 10 April 2013 (UTC)

Again your predictions came out, Joel. He deleted all my "improvements", but only after two days. There is progress. I am going over to new tactics. I am going to ask questions. I think he should give an answer, doesn't he?C.W. Vugs (talk) 18:45, 13 April 2013 (UTC)

~~== Questions about possible revisions 4 ==

I found something in the text related to his irrevocable deleting of properties written in the "a is even convention". It reads :

All such triples can be written as (a, b, c) where a² + b² = c² and a, b an c are coprime and b and c have opposite parities ...

Why is he choosing b and c and not for instance a and b? Is he going to prove that? It continues :

If c has the same parity as both legs then if all were even the parameters would not be coprime and if all were odd then aa + bb = cc would equate an even to an odd.

He pays attention to two possibilities, but how many possibilities are there if c has not the same parity as both legs, for instance odd + odd = even? Are those possibilities trivial?C.W. Vugs (talk) 19:16, 13 April 2013 (UTC)

Yes, they are trivial. If a and b are both odd then a^2+b^2 = 2 mod 4, and cannot be a square. —David Eppstein (talk) 06:23, 16 April 2013 (UTC)

Thank you. I understand what you are saying. But all together there are 2³ possibilities. What about the 5 other possibilities?C.W. Vugs (talk) 08:27, 16 April 2013 (UTC)

David, point 5 of the list of properties reads : "Exactly one of a, b is odd; c is odd". If this is true, then 2 conventions are acceptable : the 'a is even' or the 'b is even' convention. He can simply added that he is going to stay with the last one. But a proof or a reference for point 5 and his choice of the 'b is even' convention has to come before talking about the m,n-substitution, because there he is using it.It can come in just after the definition of a PPT.C.W. Vugs (talk) 11:33, 19 April 2013 (UTC)

Suggestions of particular revisions 4

Definition : (a, b, c) a PPT <----> a, b and c ε Ζ⁺, a² + b² = c² and GCD (a, b, c) = 1.

Theorem : (a, b, c) a PPT <----> there are m, n ε Z⁺, m > n, m and n impair, GCD (m, n) = 1 with (m² - n², 2mn, m² + n²) = (a, b, c).

Writer's text : That satisfaction of Euclid's formula by a, b and c is sufficient for the triangle to be Pythagorean is apparent from the fact that for integers m and n, the a, b and c , given by the formula are all integers, and from the fact that a² + b² = (m² - n²)² + (2mn)² = (m² + n²)² = c².

Writer's text : for integers m and n

Proposed revision : for m and n ε Ζ⁺, m > n

Writer's text : a, b and c, given by the formula are all integers

Proposed revision : a, b and c, given by the formula are ε Ζ⁺ C.W. Vugs (talk) 19:32, 27 April 2013 (UTC)

In the article I added : 2 x the word positive and m > n.C.W. Vugs (talk) 13:09, 30 April 2013 (UTC)

Suggestions of particular revisions 5

Writer's text now : That satisfaction of Euclid's formula by a, b and c is sufficient for the triangle to be Pythagorean is apparent from the fact that for positive integers m and n, m > n, the a, b and c given by the formula are all positieve integers and from the fact that a² + b² = (m² - n²)² + (2mn)² = (m² + n²)² = c².

Comment : Here is proved that (a, b, c) is a PT. You still have to prove that (a, b, c) is a PPT. In other words :

{There are m and n ε Z⁺, m > n, m and n impair, GCD (m, n) = 1 with (m² - n², 2mn, m² - n²) = (a, b, c)} ==== > GCD (a, b, c) = 1. C.W. Vugs (talk) 09:56, 3 May 2013 (UTC)

This subsection does not contain suggestions for a particular revision, so I'm not sure why you have so-labelled it. It is true that the current proof does not show that a, b, c are coprime. --JBL (talk) 14:08, 3 May 2013 (UTC)

If you like I can give a proof or here or in the article. It is a proof by contradiction and contains quite a lot of sup's and /sup's.C.W. Vugs (talk) 14:51, 3 May 2013 (UTC)

In the proof that follows I use continuously the fact that a common divisor of a and b is also a divisor of their sum and difference, the theorem of Euclid's algorithm, Euclid's lemma and the fundamental theorem of arithmetic. Here comes the text to be added.

Still to prove GCD (a, b, c) = 1

(i) GCD (a, b, c) ≠ even, because m and n are impair and therefore are both m² - n² and m² + n² odd. Thus GCD (a, b, c) = odd.

Hi Joel. Again your expectation was correct. He removed everything. His speciality seems to be to remove. Perhaps I was too early, I'm going to wait three days and then trying again. Perhaps he or somebody else has a better proof. I'm interested in that, because I also don't like a proof by contradiction.C.W. Vugs (talk) 18:13, 5 May 2013 (UTC)

Please familiarize yourself with the workings of Wikipedia: the person who removed your most recent addition is David Eppstein, and he left an explanatory comment in his edit summary, which you can see by clicking the "view history" tab at the top of the article. (NB: if you click "view history" at the top of this page, you will see the history for this talk page, not the article itself.) --JBL (talk) 02:57, 6 May 2013 (UTC)

I got the message of David Eppstein, unfortunedly not the article. I will try to translate the proof in clear English. It will be a bit long-winded.

Still to prove GCD (a, b, c) = 1

(i) GCD (a, b, c) ≠ even, because m and n are impair and therefore both m² - n² and m² + n² both odd. GCD (a, b) = odd, 1, 3, 5, ......

(ii) Assume GCD (a, b, c) = g ≥ 3. The factorization of g contains at least one odd prime number, say p. This p is a divisor of g and therefore a common divisor of m² - n² and m² + n². But then is p also a divisor of their difference 2n² and their sum 2m². It follows that p a common divisor is of m² and n², because p is odd. Applying Euclid's lemma (p is prime) we have p is a common divisor of m and n. But then GCD (m, n) ≥ p, while it was given GCD (m, n) = 1. Contradiction. Conclusion GCD (a, b, c) = 1. C.W. Vugs (talk) 07:04, 6 May 2013 (UTC)

This last part, approved by David Eppstein, was copied to the article and removed 9 May. C.W. Vugs (talk) 06:40, 10 May 2013 (UTC)

Suggestions of particular revisions 6

Writer's text : (if c had the same parity as both legs then if all were even the parameters would not be coprime, and if all were odd then a² + b² = c² would equate an even to an odd.)

Comment : The sentence should be removed. You have already started with the m,n-substitution and chosen for the b is even convention by writing b = 2mn.

To show that your choice is acceptable you can prove point 5 of your properties list, Exactly one of a, b is odd, c is odd. After that you can point out that a and b are interchangable in both a² + b² = c² and GCD (a, b, c} = 1 and that you take b = 2mn. However this should not be done in the middle of this proof of Euclid's formula, but before you started with the m,n-substitution. C.W. Vugs (talk) 04:00, 8 May 2013 (UTC)

Proposal : to delete this sentence and to add some text just before the m,n-substitution. For instance.

Exactly one of a, b is odd, c is odd.

(i) Assume both a and b are even then c is also even and GCD (a, b, c) ≥ 2.

(ii) Assume both a and b are odd then c² = (2s + 1)² + (2t + 1)² = 4(s² + t² + s + t) + 2 but a square can not be written in this form.

(iii) Conclusion : exactly one of a, b is odd. It follows that c is odd.

(iv) The numbers a and b are interchangable in both a² + b² = c² and GCD (a, b, c) = 1. In this article we will take a is odd and b is even. C.W. Vugs (talk) 05:01, 9 May 2013 (UTC)

THIS CONTRIBUTION WAS COPIED TO THE ARTICLE 12 MAY. C.W. Vugs (talk) 12:34, 12 May 2013 (UTC)

THE CONTRIBUTION WAS REVERTED THE SAME DAY. Reasons given : proof in the wrong section, badly formatted and might require a source. 84.24.10.61 (talk) 18:16, 12 May 2013 (UTC)

I got this message from Arthur Rubin. One of his most famous contributions to high maths can be found in the talk page of "greatest common divisor". It reads :

Yes, it does follow from the fundamental theorem of arithmetic, but α, β, γ and δ are products of 0 or more primes, not primes themselves. C.W. Vugs (talk) 07:19, 14 May 2013 (UTC)

Suggestions of particular revisions 7

Writer's text : Since (c + a)/b is rational, we set it equal to m/n in lowest terms.

Question : What is the meaning of this ? (i) Is (c + a)/b already reduced and you give the rational number another name m/n ? or (ii) You reduce first (c + a)/b by dividing numerator and denominator by their common factor and then give it the name m/n ? C.W. Vugs (talk) 05:29, 13 May 2013 (UTC)

(ii) — Arthur Rubin (talk) 11:41, 13 May 2013 (UTC)

Correct : (c + a)/b and (c - a)/b are certainly not fully reduced because of the "b is even" convention ( b is even and a is odd, c is odd and odd + odd = even). Now comes the blackboard part and after that : Since c/b and a/b are fully reduced by assumption. My question : by what assumption? This proof doesn't run nicely. C.W. Vugs (talk) 15:35, 15 May 2013 (UTC)

That was the difference between "relatively prime" (GCD(a, b, c) = 1) and pairwise relatively prime (GCD(a, b) = GCD (a, c) = GCD (b, c) = 1). The latter implies c/b and a/b are fully reduced; the former is what is now the definition, and the proof that the former implies the latter is non-trivial, obvious. and not presently in the article. — Arthur Rubin (talk) 18:19, 15 May 2013 (UTC)

Exactly, I fully agree. This is non-trivial and not in the article. The choice for the "b is even" convention too is non-trivial and not in the article. Both should be there, just below the definition of a PPT. Without these this proof becomes unreadable. C.W. Vugs (talk) 08:49, 16 May 2013 (UTC)

My proposal : to add in the article, just before the m,n-substitution the following facts :

Fact 1 : (a, b, c) a PPT ----> GCD (a, b) = GCD (b, c) = GCD (c, a) = 1

Assume GCD (c, a) = g > 1. Then the factorization of g in prime factors gives us at least one prime, say p. This p is a common divisor of c and a, and therefore also a divisor of c², a² and their difference c² - a² = b². Thus p is a divisor of b (Euclid's lemma, p prime). But then GCD (a, b, c) ≥ p. Contradiction. Thus GCD (c, a) = 1 and GCD (a, b) = GCD (b, c) = 1 (analogues cases).

Fact 2 :Every a ε Z⁺ can be written as 4m, 4m + 1, 4m + 2 or 4m + 3. But every a² is of the form 4n

or 4n + 1, because (4p)² = 4(4p²), (4p + 1)² = 4(4p² + 2p) + 1, (4p + 2)² = 4(4p² + 4p + 1) and (4p + 3)² = 4(4p² + 6p + 2) + 1 C.W. Vugs (talk) 12:51, 21 May 2013 (UTC)

Fact 3 : Exactly one of a, b is odd, c is odd.

(i) Assume both a and b are even then c is also even and GCD (a, b, c) ≥ 2.

(ii) Assume both a and b are odd then c² = (2s + 1)² + (2t + 1)² = 4(s² + t² + s + t) + 2 but a square can not be written in this form.

(iii) Conclusion : exactly one of a, b is odd. It follows that c is odd.

(iv) The numbers a and b are interchangable in both a² + b² = c² and GCD (a, b, c) = 1. In this article we will take a is odd and b is even. C.W. Vugs (talk) 12:59, 21 May 2013 (UTC)

These three facts were copied and pasted to the article on 23 of May and reverted by Arthur Rubin on 24 May, His notification : Reverted good faith edits by C.W. Vugs. Remove badly written proof − when we want proofs, they must be either elegant, important or sourced. — Preceding unsigned comment added by C.W. Vugs (talk • contribs) 17:26, 25 May 2013 (UTC)

That's badly written; I can recognize it as the outline of a proof, but no one who doesn't know the real proof would so recognize it. Furthermore, this uses some mathematical shorthand which is inappropriate for Wikipedia. — Arthur Rubin (talk) 18:32, 25 May 2013 (UTC)

Please can you give the real proof. Me too dislike proofs by contradiction. C.W. Vugs (talk) 13:34, 26 May 2013 (UTC)

Please, I repeat, can you give the real proof. C.W. Vugs (talk) 08:37, 28 May 2013 (UTC)

All assuming

a^{2}+b^{2}=c^{2}

Lemma 1

The following are all equivalent:

GCD(a, b, c) = 1
GCD(a, b) = 1
GCD(b, c) = 1
GCD(c, a) = 1

Proof

2, 3 and 4 obviously imply 1.

To show that 1 implies 2, assume that 2 fails, and h = GCD(a, b) ≠ 1, then:

h divides a and b.

h² divides a² and b².

h² divides c² = a² + b².

h divides c.

h divides GCD(a, b, c).

QED

Careful study shows the we have proved that GCD(a, b, c) = GCD(a, b) = GCD(b, c) = GCD(c, a).

Lemma 2

The square of an even number is congruent to 0 modulo 4. The square of an odd number is congruent to 1 modulo 4.

Proof: $(2n)^{2}=4n^{2}.$; $(2n+1)^{2}=4n^{2}+4n+1=4(n^{2}+n)+1.$; QED.
Theorem: If (a,b,c) is a PPT then a and b are of opposite parity and c is odd.
Proof: If a and b are both even, then 2 | GCD(a, b) , so (a,b,c) is not a PPT.; If a and b are both odd, then a² + b² is congruent to 2 modulo 4, a contradiction.; If a and b are of opposite parity, then a² + b² is congruent to 1 modulo 4, and c is odd.; QED

— Arthur Rubin (talk) 10:11, 28 May 2013 (UTC)

Lemma 1 : h divides a and b or a = k₁•h and b = k₂•h. Further c² = a² + b² = (k₁² + k₂²)•h² or c² = k₃•h² and c = k₃^½•h. Question is k₃^½ an integer?

Lemma 2: Excellent.

Theorem : Excellent. Add a note on the "b is even" convention. C.W. Vugs (talk) 11:35, 28 May 2013 (UTC)

Your comment in regard Lemma 1 is flawed. I was assuming (2), GCD(a, b) = h > 1.

And I'm not proposing adding the proof, as it's not notable. Lemma 1 is important, so should be stated, with a reference, not with a proof.

Okay. State lemma 1 in the article with a reference and go for my proof above or for Sierpinsky's proof. They are the same. My comment is not flawed. You have to use Euclid's lemma : if h is a divisor of c² and h is prime then h is a divisor of c. Please add also some notes on the b is even convention. 84.24.10.61 (talk) 16:26, 1 June 2013 (UTC)

Suggestions of particular revisions 8.

Writer's text at the end of the necessity proof : (And m and n must have opposite parity : if both were odd then dividing through (m² + n²)/2mn by 2 would give the ratio of two odd numbers, equating this ratio to c/b, which is a ratio of two numbers with opposite parities, would give conflicting parities when the equation is cross–multiplied.)

Comment : Arthur is this an elegant proof? Cross-multiplied with what? Most probably with a/b. If this is so, where are the conflicting parities? This looks to be a dead end !! C.W. Vugs (talk) 18:24, 25 May 2013 (UTC)

Changed to 2-adic order; it's not exactly elementary, but it's formally correct, and can be interpreted in an elementary manner. — Arthur Rubin (talk) 19:44, 25 May 2013 (UTC)

Perhaps the following is a road to a successful proof that m and n are impair. Cross-multiply c/b and a/b to get a/c = (m² – n²)/(m² + n²). If the later fraction is fully reduced we can state that a = m² - n² and c = m² + n² and calculate b = 2mn. Task : Given GCD (m, n) = 1 and (m² - n²)/ (m² + n²) = a/c = odd/odd -- > m and n of opposite parity C.W. Vugs (talk) 13:49, 26 May 2013 (UTC)

Your change to 2-adic order, not exactly elementary, formally correct, interpretable in an elementary manner, already edited in the article, is a porridge of learned slogans not understandable by normal thinking people. Please can you explain a bit. C.W. Vugs (talk) 08:50, 28 May 2013 (UTC)

"Cross-multiply" is just wrong. And your "proof" makes little sense. "Impair"? However, if m and n are both odd, then, using the elementary definition of ν₂ given in the p-adic order article,

\nu _{2}(2mn)=\nu _{2}(m^{2}+n^{2})=1,

so

\nu _{2}{\big (}{\frac {m^{2}+n^{2}}{2mn}}{\big )}=0.

If b and c are of opposite parity, then

\nu _{2}(b)\neq \nu _{2}(c),

so

\nu _{2}{\big (}{\frac {c}{b}}{\big )}\neq 0.

But

{\frac {m^{2}+n^{2}}{2mn}}={\frac {c}{b}},

a contradiction. — Arthur Rubin (talk) 09:53, 28 May 2013 (UTC)

Your proof is correct. You refer to the p-adic order article, better is additive valuation. The ranges of the function ν₂ are well defined in additive valuation; ν₂ : Z --- > N υ {∞} and ν₂ : Q --- > Z υ {∞}. Note that ∞ is not an element of the domains Z and Q. The article gives also some rules for working with ∞ in Z υ {∞} : for all a ε Z we have a ≤ ∞, a + ∞ = ∞ + a = ∞, a - ∞ = ∞ - a = ∞ and a•∞ = ∞•a = ∞. 84.24.10.61 (talk) 11:07, 31 May 2013 (UTC)

My proof too is correct. (i) Both m an n can not be even, because then GCD (m, n) ≥ 2. (ii) Both m and n can not be odd because then (m² - n²)/(m² + n²) = (4t)/(4s + 2) = (2t)/(2s + 1) = even/odd. Contradiction. C.W. Vugs (talk) 12:24, 31 May 2013 (UTC)

Suggestions of particular revisions 9

The necessity part of the proof of Euclid's formula is herewith completed. The sufficient part of the proof misses still : if m, n ε Ζ⁺, m > n, m and n of opposite parity, and GCD (m, n) = 1, then GCD (m² – n², 2mn, m² + n²) = 1.

See for a proof : suggestions of particular revisions 5. C.W. Vugs (talk) 11:24, 4 June 2013 (UTC)

At the end of these proofs the reader need a summary, because we lost in the long story the properties of the numbers a, b, c, m and n. For instance :

Definition : (a, b, c) a PPT ←→ a, b and c ε Z⁺ , a² + b² = c² and GCD (a, b, c) = 1.

Theorem of Euclid's formula : (a, b, c) a PPT ←→ There are m and n ε Ζ⁺, m > n, m and n of opposite parity, GCD (m, n) = 1 with (m² - n², 2mn, m² + n²) = (a, b, c). C.W. Vugs (talk) 16:51, 6 June 2013 (UTC)

This improvement was deleted 6 minutes later by David Epstein. Reason given to me : learn to write English ! — Preceding unsigned comment added by C.W. Vugs (talk • contribs) 18:26, 7 June 2013 (UTC)

Yes, and I stand by that reason. Stuff like "a PPT" and "←→" may be acceptable on this talk page as shorthand but need to be spelled out in the actual article. Write in sentences, not in formulas. See for example Wikipedia:Make technical articles understandable. —David Eppstein (talk) 19:11, 7 June 2013 (UTC)

Use of parameters in Euclid's formula

After this more or less correct definition of a PPT and the more or less correctly proved Theorem of Euclid's formula, I am interested in how to use this formula in concrete applications, as I am teaching my students. To give an example.

Find 4 PPT's with a leg of 315 ! — Preceding unsigned comment added by C.W. Vugs (talk • contribs) 17:53, 7 June 2013 (UTC)

We know b = m² - n² = (m - n)(m + n) = 315 = 3²•5•7, m and n impair, GCD (m, n) = 1, m > n.

(i) m - n = 1 and m + n = 315 or 2m = 316 and 2n = 314 or m = 158 and n = 157. Thus PPT = (315, 49612, 49613)

(ii) m - n = 3 and m + n = 105 or 2m = 108 and 2n = 102 or m = 54 and n = 51, but GCD (54, 51) = 3

(iii) m - n = 5 and m + n = 63 or 2m = 68 and 2n = 58 or m = 34 and n = 29. Thus PPT = (315, 1972, 1997)

(iv) m - n = 7 and m + n = 45 or 2m = 52 and 2n = 38 or m = 26 and n = 19. Thus PPT = (315, 988, 1037)

(v) m - n = 9 and m + n = 35 or 2m = 44 and 2n = 26 or m = 22 and n = 13. Thus PPT = (315, 572, 653)

(vi) m - n = 15 and m + n = 21 or 2m = 36 and 2n = 6 or m = 18 and n = 3, but GCD (18, 3) = 3.

(vii) Further factorizations are impossible because then m - n > m + n C.W. Vugs (talk) 16:28, 17 June 2013 (UTC)

Points (ii) and (vi) can be eliminated on structural grounds, resulting in the theorem that the number of PPTs with a specified (odd) leg is always a power of 2. A similar result applies for even legs, which are then always a multiple of 4. Still, there's no reason why anything like this (other than the theorem, if a source is provided) should be in the article. — Arthur Rubin (talk) 20:06, 17 June 2013 (UTC)

YOUR STATEMENT : "For each natural number n, there exist at least n different PPT's with the same leg a, where a is some natural number". I think the following is a sharper result.

Step 1 : Each odd number 2p + 1, p ε Z⁺, can act as the odd leg(a) of a PPT. Take m = p + 1 and n = p. Then m, n ε Z⁺, m > n, m and n of opposite parity and gcd(m,n) = 1, because m and n are consecutive numbers. By Euclid's formula is (m² - n², 2mn, m² + n²) = (2p + 1, 2p² + 2p, 2p² + 2p + 1) a PPT.

Step 2 : A number of the form p^A, p prime and A ε Z⁺ can act only once as an odd leg, because p^A = m² - n² = (m - n)(m + n) and the only allowed factorization is m + n = p^A and m - n = 1. All other factorizations will lead to gcd(m - n,m + n) ≥ p and this together with the imparity of m and n to gcd(m,n) ≠ 1.

Step 3 : If a = p₁^A·p₂^B· · · · · p_k^T than a can 2^k-1< times act as odd leg in different PPT's. We prove this by mathematical induction. The statement is true for k = 1 (see step 2). Assume the statement is true for k and consider p₁^A·p₂^B ·····p_k^T·p_k+1^S. Without the factor p_k+1^S we have 2^k-1 factorizations in two factors. Each of the two factors in the 2^k-1 factorizations can be multiplied by p_k+1^S to give a total of 2·2^k-1 = 2^k factorizations.

Step 4 : By a given natural t one can take a = p₁·p₂·····p_t, for instance the first t odd prime numbers. One has 2^t-1 different PPT's, all with the same a and 2^t-1 ≥ t for all t ε Z⁺.

Note : The even leg b can be investigated in a similar way. The hypotenuse is much more difficult.

No sources.C.W. Vugs (talk) 15:38, 8 November 2012 (UTC) C.W. Vugs (talk) 12:18, 9 November 2012 (UTC) — Preceding unsigned comment added by C.W. Vugs (talk • contribs)

I agree with C.W.Vugs : to find the numbers m and n and then a, b and c of the four PPT's with a leg of 315 is a meaningful application of m,n-substitution and the article misses a worked out example of this kind. Xoagus (talk) 12:18, 20 June 2013 (UTC)

Of course you agree. You're the same editor. Similar style, CWV's last edit was on June 18, X's first edit was on June 20. See WP:DUCK. — Arthur Rubin (talk) 10:37, 28 June 2013 (UTC)

Elementary properties of primitive Pythagorean triples 1

Heading : the properties of a primitive Pythagorean triple (a, b, c) with a < b < c (without specifying which of a or b is even and which is odd) include :

Proposed change : we follow the b is even convention (b = 2mn) in which leg a is odd, leg b is even and hypotenuse c is odd. Xoagus (talk) 18:42, 26 June 2013 (UTC)

No reason for change, loses information (no PPT). The convention might be appropriate elsewhere in the article, but not in that section, and not replacing any text. But I don't think it is, and I don't think you would be able to find it. — Arthur Rubin (talk) 10:35, 28 June 2013 (UTC)

Elementary properties of primitive Pythagorean triples 2

Point 2 of properties states the following. At most one of a, b, c is a square. See infinite descent#Non-solvability of r² + s⁴ = t⁴ for a proof. The proof there is ridiculous. A good proof can be found on this talk page and also in the article mathematical proof by Fermat. Xoagus (talk) 10:59, 10 July 2013 (UTC)

I am not undoing another editor's work. I try to delete my own contributions. In fact I like to delete them all because they are not appreciated. I mean contributions of C.W.Vugs and Xoagus.Xoagus (talk) 11:37, 12 July 2013 (UTC)

I'm sure you're deleting your own work, but, if it's been replied to, it removes context. — Arthur Rubin (talk) 15:07, 12 July 2013 (UTC)

You think all my contributions are worthless. I think all your replies are nonsense. Can we not remove both. Not only in Pythagorean triple, but also in Euclidean algorithm, in extended Euclidean algorithm and in greatest common divisor. I like to end this fruitless discussion. Xoagus (talk) 17:27, 12 July 2013 (UTC)

Mr. Eppstein, 3 April 2013 you wrote on this talk page : Um, is there a point to all your postings here? Wikipedia are supposed to reflect the published literature, and Wikipedia talk pages are supposed to be discussions about how to improve the articles. This all appears to be neither. Can you give me a reason why I shouldn't just remove it all as irrelevant and off-topic? Here is a reason why you should remove it all : the conductors of the articles on Euclidean number theory (D.Lazard, Glrx, Dmcq, Arthur Rubin, David Eppstein and JBL) see my contributions as worthless and I see their replies as nonsense. There is original work in my contributions (as far as I know) and it should be removed. That is Wikipedia policy.Xoagus (talk) 09:05, 15 July 2013 (UTC)

Any inappropriate (your term) comments that have been replied to cannot be removed without those replies being removed, and you have been removing parts of your comments making the rest of your comments (the part replied to) mean something completely different. Your comment about "original research" is incorrect; only articles are subject to the restriction against original research. Talk pages can discuss original research, in order to find where it might be published, and whether it would be suitable even if it were published. — Arthur Rubin (talk) 17:55, 15 July 2013 (UTC)

Is there any formula to find a and b in Pythagorean triple a^2+b^2=c^2 in a way that a and b are consecutive?

In Pythagorean triple a, b, c where a^2+b^2=c^2, I found a way to get some pairs in which a and b are consecutive.

Let An = a1, a2, a3, ......., an, ........

    Bn = b1, b2, b3, ......., bn, .......
    Cn = c1, c2, c3, ......., cn, .......
                                                  where an^2 + bn^2 = cn^2 and an + 1 = bn.

I found that 3(an) + 2(cn) + 1 = an+1 and

                 4(an) + 3(cn) + 2 = cn+1.

Let's explain. I'm not a professional mathematician but an amateur, so I'm sorry if there's a mistake in my presentation.

Apparently, the first pair is 3^2 + 4^2 = 5^2.

So 3(3) + 2(5) + 1 = 20 and

           4(3) + 3(5) + 2 = 29.          i.e.      20^2 + 21^2 = 29^2

And 3(20) + 2(29) + 1 = 119 and

            4(20) + 3(29) +2 = 169      i.e.     119^2 + 120^2 = 169^2

and so on.

May my presentation be useful to a limit to our maths world.

presented by drkathant@gmail.com203.81.93.79 (talk) 14:18, 13 September 2013 (UTC)

Yes, there's a formula for these. See Special right triangles#Almost-isosceles Pythagorean triples. —David Eppstein (talk) 15:40, 13 September 2013 (UTC)

Thanks a lot, Professor. — Preceding unsigned comment added by 203.81.93.79 (talk) 16:09, 13 September 2013 (UTC)

For clarity, I've added more detail (in terms of a Pell equation) to the link given by Eppstein.Titus III (talk) 04:33, 14 September 2013 (UTC)

Infinite series of Pythagorean Triple

I would like to add / contribute the following, but since I am not a professional in this field (hobbyist) and have worked long and hard enough I hope you would see the relevance and indulge in proper presentation as a favour.

See here for my Pythagorean Triple formula

a = s(2r+1), b = s(2r(r+1)), c = s((2r(r+1))+1)

Where r and s are elements of the natural numbers

-- Yes, I know, this is bad representation here --

See this link ^[2] (of my discussion on a related topic) for an explanation and demonstration. Pcproblemdoctor (talk) 11:13, 24 September 2013 (UTC)

The first is clearly an example of the main formula with

k = s, m = r+1, n = r

— Arthur Rubin (talk) 01:40, 25 September 2013 (UTC)

and does not generate all Pythagorean triples (see Formulas for generating Pythagorean triples), it only generates an infinite subset. Frank M Jackson (talk) 10:48, 25 September 2013 (UTC)

Thanks for the response. I see now that it only generates a subset. It was not obvious for me to be an example of the main formula. Comment accepted. Pcproblemdoctor (talk) 01:02, 26 September 2013 (UTC)

Algorithm to generate all Pythagorean triples

The algorithm creates all triples not only the primitive ones. It is based on the following relation using the differences between the legs and the hypotenuse:

2({\tfrac {a+b-c}{2}})^{2}=(c-a)(c-b)

In principal one constructs all numbers of the form 2n^2 and evaluates all possibilities to write these numbers as a product of two positive integers which are then (c-a) and (c-b).

{\begin{array}{|r|r|r|r|r|r|r|}n={\tfrac {a+b-c}{2}}&2n^{2}&(c-b)&(c-a)&a=2n+(c-b)&b=2n+(c-a)&c=2n+(c-a)+(c-b)\\\hline 1&2&1&2&3&4&5\\\hline 2&8&1&8&5&12&13\\\hline &&2&4&6&8&10\\\hline 3&18&1&18&7&24&25\\\hline &&2&9&8&15&17\\\hline &&3&6&9&12&15\\\hline 4&32&1&32&9&40&41\\\hline &&2&16&10&24&26\\\hline &&4&8&12&16&20\\\hline 5&50&1&50&11&60&61\\\hline &&2&25&12&35&37\\\hline &&5&10&15&20&25\end{array}}

— Preceding unsigned comment added by Samuel Fux (talk • contribs) 14:29, 16 October 2013 (UTC)

I'm afraid we would need a source. But it's not "one parameter". It's one parameter (n) and a factor of 2 n². — Arthur Rubin (talk) 16:21, 16 October 2013 (UTC)

This procedure is already known and results in Euler's generating equations. Taking the equation

r^{2}=2(c-a)(c-b)

and substituting for each factor some integers

k,p,q

with

p,q

coprime such that

(c-a)=2kp^{2}

and

(c-b)=kq^{2}

then

r=a+b-c=2kpq.

This resolves to

c=k((p+q)^{2}+p^{2}),

b=2kp(p+q),

a=k((p+q)^{2}-p^{2}).

Now by substituting

m

for

(p+q)

and

n

for

p

it gives Euler's generating equation. Frank M Jackson (talk) 22:33, 16 October 2013 (UTC)

Thank you very much for your comments. @Arthur Rubin: I only have access to public math literature and will therefore most likely not be able to provide the necessary references to add something to this wikipedia article about Pythagorean triples. I therefore decided to put my ideas in my personal talk page on wikipedia instead of here. @Frank M Jackson: Thank you for providing information about the algorithm. Do you by chance have a reference for the equations you put in your comment ? I would be interested in reading more about it. Hopefully the literature about it is public. — Preceding unsigned comment added by Samuel Fux (talk • contribs) 11:30, 17 October 2013 (UTC)

It is deep within http://www.mathpages.com/home/kmath367.htm - (the second half) and choose p=2 for PPT's. Frank M Jackson (talk) 21:56, 17 October 2013 (UTC)

References

^ M. G. Teigen and D. W. Hadwin, The American Mathematical Monthly, Vol. 78, No. 4 (Apr., 1971), pp. 378-379
^ http://sites.google.com/site/pcproblemdoctor/fermat-s-little-law

Elementary properties 6

is this right "Exactly one of a, b is divisible by 3.[9]"

the triple 9, 12, 15 seems to dispute it

174.97.134.52 (talk) 01:29, 1 March 2014 (UTC)JP

It doesn't. It is not primitive. Georgia guy (talk) 01:31, 1 March 2014 (UTC)