User talk:Gwaihir

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Redwolf24 9 July 2005 07:19 (UTC)

P.S. I like messages :-P

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Your clue was the final piece of the puzzle I needed to nail this problem. In fact I had hit upon the idea of $d\mid ab$ as a criteria for generating the triples for the original equation. But I was unable to prove that multiples of basic Pythagorean triples where $a'd'\leq 100$ were all that's needed. Essentially, my mental block was on the fact that either the numbers in a Pythagorean triple are all coprime with each other or all three share the same common factor. I kept on trying to find the case where two number out of the three had a common factor that was not divisible by the third, turned out this is impossible, because for a Pythagorean triple $(a,b,c)$:

Assume that $c$ has no common factor with $a$ or $b$, but $a$ and $b$ have a greatest common factor, $d$, that is greater than 1. Then

$c^{2}=a^{2}+b^{2}=(a'd)^{2}+(b'd)^{2}=a'^{2}d^{2}+b'^{2}d^{2}=d^{2}(a'^{2}+b'^{2})$

Clearly, $c$ is also divisible by $d'$, which is a contradiction with the original assumption that $c$ has no common factor with $a$ or $b$.

Assume that $a$ has no common factor with $b$ or $c$, but $b$ and $c$ have a greatest common factor, $d$, that is greater than 1. Then

$c^{2}=a^{2}+b^{2}\Rightarrow a^{2}=c^{2}-b^{2}=(c'd)^{2}+(b'd)^{2}=c'^{2}d^{2}+b'^{2}d^{2}=d^{2}(c'^{2}+b'^{2})$

Clearly, $a$ is also divisible by $d'$, which is a contradiction with the original assumption that $a$ has no common factor with $b$ or $c$.

Symmetrically, it could be shown that it is impossible for $a$ and $c$ to share a greater than 1 common factor that is not divisible by $b$.

Hence, for Pythagorean triples, either all three numbers have a greater-than-one common factor or they are pairwise coprime, that means $gcd(a,b)=1$ and $gcd(b,c)=1$ and $gcd(c,a)=1$ all at the same time! There are no other situations possible in terms of common factors.

With this proof, al[[Image:ong with gwaihir's clue, it is not hard to figure out the following possible values for $a$:

15, 30, 45, 60, 75, 90	for the (3, 4, 5) case
65	for the (5, 12, 13) case

Thus, the answer to the original question (the sum of all such $a$) is: 350.

Hurrrrrrrrrrrah]], thank you so much for your help! I award you this barnstar:

129.97.252.63 05:02, 21 February 2006 (UTC)

Thank you, I'm glad I could help you. (But see my comments on WP:RD/Math.)--gwaihir 10:42, 21 February 2006 (UTC)

Waaaaaaaaaaaaaah, I'm stupid, I'm stupid, yeah (4, 3, 5) should be counted, whereas (12, 5, 13) would pop you over 100. 129.97.252.63 21:07, 21 February 2006 (UTC)

I added up the numbers again and it's still 350. That is assuming my incomplete table of $a$ values. 129.97.252.63 21:07, 21 February 2006 (UTC)

No, the negative b's and c's need not to be counted, since all I care about here are $a$'s such that ${\frac {1}{a^{2}}}+{\frac {1}{b^{2}}}={\frac {1}{c^{2}}}$ has integer/diophantine solutions. so the overlap of a=60 also is included only once. 129.97.252.63 21:07, 21 February 2006 (UTC)

thanks for the translations

I appended them to the German headers. --Jtir 20:51, 9 October 2006 (UTC)