^{i}< n for any odd integer n and I claimed a few things about it. Here I make another claim related to Mersenne numbers that I managed to prove.

**Claim:**If n = 2

^{i}- 1 (also known as a Mersenne number), then the order of 2 mod n is i.

**Proof:**If n is a Mersenne number then by definition n = 2

^{i}- 1 so then n - 1 = 2

^{i}and so clearly 2

^{i}= 1 modulo n.

By definition, the order of an element a modulo n is the smallest integer b such that a

^{b}= 1 mod n so it remains to show that i is the smallest integer such that 2

^{i}= 1 modulo n.

I'll show this by contradiction. Suppose there exists an integer k such that k < i and yet 2

^{k}= 1 mod n. Transforming this congruence relation to a diophantine equation yields

If 2

^{k }= 1 mod n then for some integer x, nx + (-1) = 2

^{k }so nx - 1 = 2

^{k}

Clearly, x must be greater than 1 since it is given that n - 1 = n*1 - 1 = 2

^{i}and k is not equal to i

Therefore, (nx - 1) > (n - 1) so therefore 2

^{k}> 2

^{i}so therefore k > i, which is a contradiction to the supposition that k < i .

^{.}.

Example: It is easy to calculate the order of two mod n for the first few Mersenne numbers.

The first few Mersenne numbers are 3, 7, 15, 31, 63, 127, 255 and the order of 2 mod n for each of these choices is 2,3,4,5,6,7,8 respectively.

Below is a slow calculator for finding the order of 2 mod n for any odd integer n: