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Topic: f(n)  2^n  1 (Read 1402 times) 

Eigenray
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f(n)  2^n  1
« on: Mar 15^{th}, 2008, 2:36pm » 
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Determine all integer polynomials f such that f(n)  2^{n}  1 for all n > 0.


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Obob
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Re: f(n)  2^n  1
« Reply #1 on: Mar 15^{th}, 2008, 4:49pm » 
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If there are infinitely many Mersenne primes, then either f(n) = 1 or f(n) = 1 for all n . Since we don't know if there are infinitely many Mersenne primes or not, if the problem has an answer then this is it.


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Eigenray
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Re: f(n)  2^n  1
« Reply #2 on: Mar 20^{th}, 2008, 5:07pm » 
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Yes, that is the answer. Hint: What do you know about f(p) when p is prime?


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Eigenray
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Re: f(n)  2^n  1
« Reply #3 on: Jul 16^{th}, 2008, 3:01pm » 
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Looks like I forgot about this one. I guess that was a bad hint. I should have said: what do you know about the prime factors of f(p) when p is prime?


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Aryabhatta
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Re: f(n)  2^n  1
« Reply #4 on: Mar 14^{th}, 2009, 1:08pm » 
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Almost a year!


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Eigenray
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Re: f(n)  2^n  1
« Reply #5 on: Mar 14^{th}, 2009, 2:45pm » 
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Well, what do you know about the prime factors of f(p) when p is prime? Further hint: Dirichlet


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Obob
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Re: f(n)  2^n  1
« Reply #6 on: Mar 14^{th}, 2009, 4:15pm » 
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If q is a prime divisor of 2^p1, for p prime, then 2^p = 1 (mod q). Hence p is a multiple of the order of 2 in (Z/qZ)*, and since p is prime actually p is the order of 2 in (Z/qZ)*. This implies p  q1, so kp = q1 for some k, and q = kp+1 for some k. I'm not sure where this is going.


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Eigenray
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Re: f(n)  2^n  1
« Reply #7 on: Mar 14^{th}, 2009, 6:51pm » 
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So q can't divide very many values f(p), can it?


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Obob
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Re: f(n)  2^n  1
« Reply #8 on: Mar 15^{th}, 2009, 4:47pm » 
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Ah. Suppose the prime q other than p divides f(p) for some prime p, so that f(p) = 0 (mod q). In Z/qZ, we have the identity f(p+kq) = f(p); see this at the level of monomials by applying the binomial theorem. Letting k vary, the expression u = p+kq is prime infinitely often by Dirichlet's theorem since p != q, so q divides f(u) for infinitely many primes u. But q > u for all such primes u by my previous post, a contradiction. This forces the only prime factor of f(p) to be p for each prime p. But again by my previous post p never divides 2^p1. Thus in fact f(p) = + 1 for each p, from which we easily conclude that either f(n) = 1 for all n or f(n)= 1 for all n. That was a nice problem. Number theory isn't really my thing, but this was fun to think about. (I do algebraic geometry, preferably in characteristic 0)

« Last Edit: Mar 15^{th}, 2009, 4:50pm by Obob » 
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