20110303, 23:02  #1 
Jan 2010
germany
11010_{2} Posts 
Question about a mersennenumber property
I discovered a nice property about mersenne numbers.
When is not prime then this is true only for the two trivial cases: 1.) 2.) If there exist an for : the above congruence is also true. This is the case when choosing and , so that .  For the other combinations of the variable and the above congruence is NEVER true. I do not really know if I am right with this. Up til now i haven't found an counterexample. My question is: Can this conjecture be true or is this just an example for the 'laws of small numbers' ? 
20110303, 23:32  #2  
"Forget I exist"
Jul 2009
Dumbassville
3·2,801 Posts 
Quote:
Last fiddled with by science_man_88 on 20110303 at 23:41 

20110303, 23:45  #3  
"Bob Silverman"
Nov 2003
North of Boston
2^{2}×1,871 Posts 
Quote:
that they should not go near number theory until they have mastered high school mathematics. Hint: Binomial Theorem. Consider (x + y)^n mod n for arbitrary x,y \in N and n prime. 

20110303, 23:52  #4 
"Forget I exist"
Jul 2009
Dumbassville
8403_{10} Posts 
well 2^x  2^(xz) = (2^z1)*2^(xz) since I redid the math. From the fact that (2^y)/(2^a) = 2^(ya) it's impossible to add up to a multiplier (2^z1) with only one other power of 2 other than 2^(xz) except when z == 1.
Last fiddled with by science_man_88 on 20110304 at 00:52 
20110304, 00:02  #5 
Bemusing Prompter
"Danny"
Dec 2002
California
2·17·73 Posts 
Inb4miscmaththreads.

20110304, 01:03  #6 
Jan 2010
germany
2×13 Posts 

20110304, 01:24  #7 
"Forget I exist"
Jul 2009
Dumbassville
3·2,801 Posts 
according to your math you can reconvert it to 2^(xz)*(2^z1) which means unless you say the last one is a error my math holds up to yours. Last fiddled with by science_man_88 on 20110304 at 01:25 
20110304, 02:44  #8  
Jan 2010
germany
32_{8} Posts 
Quote:
I think you did not read carefully what i wrote or my english grammar was too bad. Let me show you what I did and what i meant: I played with the Lemma : if and the congruent is correct then n is prime. And shure I know that I dont may put a number to X. X is a free variable. For example: > 5 is prime But what I did was to play a little around with this Lemma. So I inserted the number 2 to the variables X and a. I know that for every Mersenne Prime n (n=(2^p)1, n is prime) this is always true: (This is because the "Kleine Fermatsche Satz".) "Kleine Fermatsche Satz" : For all and all true: If n is prime then ( is the Eulers totient function ) Therefore if is prime when for all and all the following congruation is always correct: ...And if you have : you get this: But if n is not prime, when it can be that for some x this is also true. So what I did is that I looked for numbers x where the congruation is NEVER POSSIBLE when n is NOT PRIME. So I found a pattern. And this was my conjecture. Again: Let and n is not Prime. If now and then the above congruation is ALWAYS false. This means that for these this : is NEVER true when n=2^p1 is not prime ! 

20110304, 03:40  #9 
6809 > 6502
"""""""""""""""""""
Aug 2003
101×103 Posts
2^{2}×2,687 Posts 
Sascha,
Bob Silverman is an expert. He likely did not misread what you wrote. If he suggests that you look into some area of education, do it. I think that his idea of High School math is through Math Analysis and Calculus I & II. He is often blunt, don't be scared by him though. I have not looked into your suggestion. My math skills have severely atrophied since I left school. semirandom image attached. 
20110304, 16:01  #10 
Dec 2008
you know...around...
1452_{8} Posts 

20110304, 16:08  #11  
"Bob Silverman"
Nov 2003
North of Boston
2^{2}×1,871 Posts 
Quote:
Your original statement was that x^n = x is not true when n = 2^p1 is composite. Here x = 2^a + 2^b for a!= b This is different from x^{n1} = 1 is not true when n = 2^p1 is composite. 

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