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Old 2015-05-17, 11:33   #1
wildrabbitt
 
Jul 2014

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Default Will Eddington's Page

Will Eddington's page used to have a proof that

a prime divides only one prime-exponent composite Mersenne number.

As far as I can tell there's no other proof of this anywhere on the web.

His page has gone which is why I'm posting.

Am I wrong?

Last fiddled with by wildrabbitt on 2015-05-17 at 11:33
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Old 2015-05-17, 13:58   #2
PBMcL
 
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Quote:
Originally Posted by wildrabbitt View Post
As far as I can tell there's no other proof of this anywhere on the web.
This seems ... unlikely. Huge hint: for a given odd prime q, the order of 2 mod q is unique.
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Old 2015-05-17, 14:03   #3
wildrabbitt
 
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It's a well known theorem.

Did you understand what I meant properly?

e.g 23 * 89 = 2047 = (2^11) - 1

therefore 23 and 89 do not divide any other prime-exponent composite Mersenne.

I've seen the proof, checked it too.
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Old 2015-05-17, 14:17   #4
ATH
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It is discussed in this old thread: http://www.mersenneforum.org/showthread.php?t=16019

Quote:
Originally Posted by axn View Post
GCD(2^a-1,2^b-1)=2^GCD(a,b)-1. So, for prime exponent mersennes, the answer is no.
The proof for this statement is not given in the thread though.
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Old 2015-05-17, 15:09   #5
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Quote:
Originally Posted by ATH View Post

The proof for this statement is not given in the thread though.
re-quoting from above:

Quote:
GCD(2^a-1,2^b-1)=2^GCD(a,b)-1. So, for prime exponent mersennes, the answer is no.

assume for example that b=2*a; we can show that 2*(2^a-1)+1 = 2^(a+1)-1 and we end up with a general formula of 2^(n+a)-1=2^n*(2^a-1)+(2^n-1) for the 2^n-1 (the part that isn't shown as a multiple of 2^a-1) to have a common factor with (2^a-1) a being the first exponent to have that factor means n must a multiple of a but if n is a multiple of a then n+a is also a multiple of a so only if the exponent b is a multiple of a will it have that factor in common since the primes are all relatively coprime it is shown that no ratio of the exponents fit this and the proof is done.

edit: https://www.physicsforums.com/thread.../#post-3484943

Reference https://www.physicsforums.com/thread...inus-1.526047/

Last fiddled with by science_man_88 on 2015-05-17 at 15:24
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Old 2015-05-17, 21:20   #6
ewmayer
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Is Will Eddington any relation to the late great astrophysicist Arthur, who famously organized the 1919 British solar-eclipse expedition which confirmed a key cornerstone of Einstein's general relativity theory?

Yes, I *know* it's not really a double-D (he said to the buxom lass).
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