Talk:Midy's theorem

Midy Theorem And Group Theory

(Z₁₇, *) is the multiplicative group of integer modulo 17. The cyclic subgroup generated by 10 is also Z₁₇:

{1, 10, 15, 14, 4, 6, 9, 5, 16, 7, 2, 3, 13, 11, 8, 12}

The sum of these number is 8×17, divisible by 17.

Arrange these 16 numbers into a ring in the same sequence and select numbers at any intervals:

1. Select every alternative numbers, starting from the first or second number, the sum of these 8 numbers is 68, which is divisible by 17;
2. Select one out of every three. It takes 3 cycles to select all the numbers;
3. Select one number out of every four. There are four starting choices / sets. The sum of each set is 34, which is divisible by 17;
4. Select one number out of every five (or 7, 9, 11, 13 or 15). It takes 5 (or 7, 9, 11, 13, or 15) cycles to select all the numbers;
5. Select one number out of every six (or 10, 14). The selected sets are the same as item 1;
6. Select one number out of every eight. There are eight couples and the sum of each couple is 17;
7. Select one number out of every 12. the selected sets are the same as item 3;

--Ling Kah Jai (talk) 14:25, 15 October 2009 (UTC)

How to arrange positive integers not more than a prime number p in a ring such that if we select a series of numbers from the ring at a regular interval starting from any position, the sum of each selected series is divisible by p?

The solution is:

• First find the primitive root modulo p, say it s a.
• Then list the consecutive numbers generated by a in multiplication modulo p in a ring.

--Ling Kah Jai (talk) 02:48, 16 October 2009 (UTC)

Midy's theorem

I have reverted the "group theory proof" that you added to Midy's theorem. As well as being original research, your "proof" does not actually use any group theory. Please stop using Wikipedia to publish your own research. Gandalf61 (talk) 20:10, 16 October 2009 (UTC)

I am not putting up something new. I am adopting existing theory to provide a proof for Midy's theorem. If you call that original research, then perhaps there are too many original research in Wikipedia. --Ling Kah Jai (talk) 01:57, 22 October 2009 (UTC)

If you think it is not using group theory, then you can add one that truly uses group theory. --Ling Kah Jai (talk) 02:20, 22 October 2009 (UTC)

Extended Midy's theorem

The material is only supported by a link to a homepage, which is not considered proper sourcing by wiki standards. Once the paper is published in a peer-reviewed venue, it may be appropriate to mention it here. Tkuvho (talk) 05:40, 19 January 2011 (UTC)

The theorem can be also used to 1/(n), when n is a prime power

In duodecimal, ${\frac {1}{5^{2}}}$ = 0.05915343ᘔ0Ɛ62ᘔ68781Ɛ, and 05915343ᘔ0 + Ɛ62ᘔ68781Ɛ = ƐƐƐƐƐƐƐƐƐƐ. Besides, ${\frac {1}{7^{2}}}$ = 0.02Ɛ322547ᘔ05ᘔ644ᘔ9380Ɛ908996741Ɛ615771283Ɛ, and 02Ɛ322547ᘔ05ᘔ644ᘔ9380 + Ɛ908996741Ɛ615771283Ɛ = ƐƐƐƐƐƐƐƐƐƐƐƐƐƐƐƐƐƐƐƐƐ, so it can be also used in a fraction of a prime power! — Preceding unsigned comment added by 140.115.140.117 (talk • contribs) 09:39, November 9, 2014‎

We would need a source, but it is true, provided that the prime does not divide the number of parts.

Let p^t be a prime power and a/p^t be a fraction between 0 and 1. Suppose the expansion of a/p^t in base b has a period of ℓ, so

${\begin{aligned}&{\frac {a}{p^{t}}}=[0.{\overline {a_{1}a_{2}\dots a_{\ell }}}]_{b}\\[6pt]&\Rightarrow {\frac {a}{p^{t}}}b^{\ell }=[a_{1}a_{2}\dots a_{\ell }.{\overline {a_{1}a_{2}\dots a_{\ell }}}]_{b}\\[6pt]&\Rightarrow {\frac {a}{p^{t}}}b^{\ell }=N+[0.{\overline {a_{1}a_{2}\dots a_{\ell }}}]_{b}=N+{\frac {a}{p}}\\[6pt]&\Rightarrow {\frac {a}{p^{t}}}={\frac {N}{b^{\ell }-1}}\end{aligned}}$

where N is the integer whose expansion in base b is the string a₁a₂...a_ℓ.

Note that b ^ℓ − 1 is a multiple of p^t because (b ^ℓ − 1)a/p^t is an integer. Also bⁿ−1 is not a multiple of p^t for any value of n less than ℓ, because otherwise the repeating period of a/p in base b would be less than ℓ.

Now suppose that ℓ = hk. Then b ^ℓ − 1 is a multiple of b^k − 1. (To see this, substitute x for b^k; then b^ℓ = x^h and x − 1 is a factor of x^h − 1. ) Say b ^ℓ − 1 = m(b^k − 1), so

${\frac {a}{p^{t}}}={\frac {N}{m(b^{k}-1)}}.$

But b ^ℓ − 1 is a multiple of p^t; b^k − 1 is not a multiple of p^t (because k is less than ℓ )

$m=\sum _{i=0}^{h-1}x^{i},$

and

$\operatorname {gcd} (m,x-1)|m-(x-1)\left(\sum _{i=0}^{h-2}(h-1-i)x^{i}\right)=h.$

So,

1. By counting powers of p, p divides m
2. As p does not divide h, p does not divide x − 1
3. Again, by counting powers of p p^t divides m.

... and we can continue. — Arthur Rubin (talk) 16:50, 9 November 2014 (UTC)