conjecture about mersenne numbers
Hello,
My conjecture is : Let [TEX]2^p1[/TEX] be an Mersennenumber, and a is an element from [TEX]2^p1[/TEX] [TEX](1)\:\: a^{(p*p)} \equiv 1\: (mod\: 2^p1)[/TEX] [TEX](2)\:\: a^{p} \equiv 1\: (mod \: 2^p1)[/TEX] (1) > (2) This means, that if (1) is true, than is (2) also true. It is easy to show that ? My idea to do this with [TEX]2^p1[/TEX] is prime, was the following: [TEX]a^{pp} \equiv 1 [/TEX] [TEX]a^{p} \equiv \sqrt[p]{1}[/TEX] When [TEX]2^p1[/TEX] is prime, when the Elements with the form [TEX]2^x [/TEX] are the only ones, that have order of [TEX]p[/TEX] and therefore: [TEX]a^p \equiv \sqrt[p]{1} \equiv 2^x[/TEX] [TEX]a \equiv \sqrt[p]{2^x}[/TEX]  But the only solution to this is 1: > [TEX]a \equiv \sqrt[p]{2^x}\equiv 1[/TEX] [TEX]a \equiv 1[/TEX] > So [TEX] pp[/TEX] can not be the order of a, because a is 1. I have searched with google many sites, but could not find the answer to this "problem". I hope that anybody can help me with the conjecture. kind regards, sascha 
This should be posted in the GIMPS forum.

ok. thanks.
I will post this in the gimps>math Forum 
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