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Old 2010-01-06, 21:58   #1
sascha77
 
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Jan 2010
germany

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Default conjecture about mersenne numbers

Hello,

My conjecture is :

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

(1)\:\: a^{(p*p)} \equiv 1\: (mod\: 2^p-1)


(2)\:\: a^{p} \equiv 1\: (mod \: 2^p-1)

(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 2^p-1 is prime, was the following:


a^{pp}  \equiv 1

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

When 2^p-1 is prime, when the Elements with the form
2^x are the only ones, that have order of p

and therefore:
a^p   \equiv \sqrt[p]{1} \equiv 2^x

a \equiv \sqrt[p]{2^x}

- But the only solution to this is 1:
-->
a \equiv \sqrt[p]{2^x}\equiv 1

a \equiv 1

-> So  pp 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
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Old 2010-01-07, 07:20   #2
gd_barnes
 
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This should be posted in the GIMPS forum.
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Old 2010-01-07, 08:06   #3
sascha77
 
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Jan 2010
germany

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