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 sascha77 2010-01-06 21:58

Hello,

My conjecture is :

Let [TEX]2^p-1[/TEX] be an Mersenne-number, and
a is an element from [TEX]2^p-1[/TEX]

[TEX](1)\:\: a^{(p*p)} \equiv 1\: (mod\: 2^p-1)[/TEX]

[TEX](2)\:\: a^{p} \equiv 1\: (mod \: 2^p-1)[/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^p-1[/TEX] is prime, was the following:

[TEX]a^{pp} \equiv 1 [/TEX]

[TEX]a^{p} \equiv \sqrt[p]{1}[/TEX]

When [TEX]2^p-1[/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

 gd_barnes 2010-01-07 07:20

This should be posted in the GIMPS forum.

 sascha77 2010-01-07 08:06

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

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