Mersenne Conjecture

sascha77 · 2010-01-07, 08:13

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

maxal · 2010-01-07, 18:28

(1) implies (2) for all $a$ if and only if $p^2$ does not divide $M_p-1=2^p-2$ .
In other words, you need to prove either of the two statements:
* if $p$ is Wieferich prime then $M_p$ is not prime;
* if $M_p$ is prime then $p$ is not a Wieferich prime.
It may be hard to prove.

sascha77 · 2010-01-08, 08:44

Thanks maxal,

Now I see the connection to the Wieferich primes and
I feel confident now that this is hard to solve.

blob100 · 2010-02-09, 04:58

Do you have a proof to this conjecture (without inclding the 2^p-1 prime number option)?

retina · 2010-02-09, 05:29

Quote:

Originally Posted by blob100

Do you have a proof to this conjecture (without inclding the 2^p-1 prime number option)?

If the OP had proof then it wouldn't be a conjecture.

blob100 · 2010-02-09, 19:17

What do you mean by element ( I tought element is a member in a set).
Do you have an example for an element of a merssene number?

R.D. Silverman · 2010-02-09, 20:31

Quote:

Originally Posted by blob100

What do you mean by element ( I tought element is a member in a set).
Do you have an example for an element of a merssene number?

It is bad nomenclature. He means that a is an element of the group of
units of Z/(2^p-1)Z

, i.e. an element of its maximal cyclic subgroup.

blob100 · 2010-02-09, 20:47

Silverman,
So he just mean a factor of 2^p-1.
BTW: I started reading "solved and unsloved problems in mathematics",
I really enjoy it and it is really readable for my level.

henryzz · 2010-02-10, 07:34

Quote:

Originally Posted by blob100

Silverman,
So he just mean a factor of 2^p-1.
BTW: I started reading "solved and unsloved problems in mathematics",
I really enjoy it and it is really readable for my level.

Do you mean D. Shanks "Solved and Unsolved Problems in Number Theory"?
Are you learning from it or just enjoying it?
Might get it myself

blob100 · 2010-02-10, 16:30

Henry,
I do read it, it isn't easy and it takes me time to understand.
When I say "enjoy", I mean that it is really fun for me to learn more mathematics, I love mathematics, so why won't I enjoy reading a math book?
I'm not going to read a book if I'm going to be bored of it..

sascha77 · 2010-04-25, 23:31

Hello ,

I wrote an PDF-document to show why I am interested
that "my" conjecture is true.
Sorry that it took so long time ;-(

In this PDF there might be some "new" Mersenne Properties, but
I am not really shure if they are not well known.
So I post this document here, in hope that the information might
be usefull for someone.
If I did mathematical mistakes in the document I would really be glad if you can say what I did false.
( I know there are plenty of english-grammar-mistakes ;-) Please excuse my bad english)

kind regards,

Sascha

2010-01-07, 08:13	#1
sascha77 Jan 2010 germany 2·13 Posts	Mersenne Conjecture 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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2010-01-07, 18:28	#2
maxal Feb 2005 2²·5·13 Posts	(1) implies (2) for all $a$ if and only if $p^2$ does not divide $M_p-1=2^p-2$ . In other words, you need to prove either of the two statements: * if $p$ is Wieferich prime then $M_p$ is not prime; * if $M_p$ is prime then $p$ is not a Wieferich prime. It may be hard to prove.

2010-01-08, 08:44	#3
sascha77 Jan 2010 germany 11010₂ Posts	Thanks maxal, Now I see the connection to the Wieferich primes and I feel confident now that this is hard to solve.

2010-02-09, 04:58	#4
blob100 Jan 2010 379 Posts	Do you have a proof to this conjecture (without inclding the 2^p-1 prime number option)?

2010-02-09, 19:17	#6
blob100 Jan 2010 573₈ Posts	What do you mean by element ( I tought element is a member in a set). Do you have an example for an element of a merssene number?

2010-02-09, 20:47	#8
blob100 Jan 2010 379 Posts	Silverman, So he just mean a factor of 2^p-1. BTW: I started reading "solved and unsloved problems in mathematics", I really enjoy it and it is really readable for my level.

2010-02-10, 16:30	#10
blob100 Jan 2010 379 Posts	Henry, I do read it, it isn't easy and it takes me time to understand. When I say "enjoy", I mean that it is really fun for me to learn more mathematics, I love mathematics, so why won't I enjoy reading a math book? I'm not going to read a book if I'm going to be bored of it..