Is it possible to either prove or disprove the following?(Nerd)

If m is a positive odd integer such that 2^{m}= 2 (mod. m(m-1)) then m is a prime.

Examples:

2^{7}= 2 (mod. 42 = 7x6). 7 is prime.

2^{43}= 2 (mod. 1806 = 43x42). 43 is prime.

Printable View

- Sep 6th 2012, 02:34 AMStanPossible Primes.
Is it possible to either prove or disprove the following?(Nerd)

If m is a positive odd integer such that 2^{m}= 2 (mod. m(m-1)) then m is a prime.

Examples:

2^{7}= 2 (mod. 42 = 7x6). 7 is prime.

2^{43}= 2 (mod. 1806 = 43x42). 43 is prime. - Sep 7th 2012, 08:13 AMStanRe: Possible Primes.
Is the following proof valid?

Theorem.

If m is a positive odd integer such that 2m ≡ 2 (mod. m(m-1)) then m ≡ 3 (mod. 4) and m is a prime.

Proof.

Let 2m ≡ 2 (mod. m(m-1)) then 2m - 2 =km(m-1) for some positive integer k.

This gives 2m-1 – 1 = km(m-1)/2 and since 2m-1 – 1 is odd, (m-1)/2 is odd implying m ≡ 3 (mod. 4).

Let p be the least valued prime dividing (m-1)/2 . Then p is odd and 2m-1 ≡ 1 (mod. p).

Let d be the order of 2 (mod. p).

Then d is a positive integer < p and 2d ≡ 1 (mod. p).

So d divides (m-1) and d divides (p-1).

This implies there is a prime dividing d, less than p, which divides (m-1)/2 .

This contrdicts p is the least valued prime dividing (m-1)/2 .

Hence, m is a prime. - Sep 8th 2012, 02:51 AMStanRe: Possible Primes.
The proof is invalid. However, there is an updated proof of the theorem in the post ' Primes Conjecture' which may or may not be valid.