20110303, 23:02  #1 
Jan 2010
germany
2·13 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
20C0_{16} Posts 
Quote:
Last fiddled with by science_man_88 on 20110303 at 23:41 

20110303, 23:45  #3  
Nov 2003
16444_{8} 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
20300_{8} 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
2×3×397 Posts 
Inb4miscmaththreads.

20110304, 01:03  #6 
Jan 2010
germany
32_{8} Posts 

20110304, 01:24  #7 
"Forget I exist"
Jul 2009
Dumbassville
20C0_{16} 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
2×13 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
2·4,783 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...
2·5^{2}·13 Posts 

20110304, 16:08  #11  
Nov 2003
2^{2}×5×373 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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