``` wu :: forums (http://www.ocf.berkeley.edu/~wwu/cgi-bin/yabb/YaBB.cgi) riddles >> putnam exam (pure math) >> f(n) | 2^n - 1 (Message started by: Eigenray on Mar 15th, 2008, 2:36pm) ``` Title: f(n) | 2^n - 1 Post by Eigenray on Mar 15th, 2008, 2:36pm Determine all integer polynomials f such that f(n) | 2n - 1 for all n > 0. Title: Re: f(n) | 2^n - 1 Post by Obob on Mar 15th, 2008, 4:49pm If there are infinitely many Mersenne primes, then [hide] either f(n) = 1 or f(n) = -1 for all n [/hide] .  Since we don't know if there are infinitely many Mersenne primes or not, if the problem has an answer then this is it. Title: Re: f(n) | 2^n - 1 Post by Eigenray on Mar 20th, 2008, 5:07pm Yes, that is the answer.  Hint: What do you know about f(p) when p is prime? Title: Re: f(n) | 2^n - 1 Post by Eigenray on Jul 16th, 2008, 3:01pm 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? Title: Re: f(n) | 2^n - 1 Post by Aryabhatta on Mar 14th, 2009, 1:08pm Almost a year! Title: Re: f(n) | 2^n - 1 Post by Eigenray on Mar 14th, 2009, 2:45pm Well, what do you know about the prime factors of f(p) when p is prime?Further hint: [hide]Dirichlet[/hide] Title: Re: f(n) | 2^n - 1 Post by Obob on Mar 14th, 2009, 4:15pm If q is a prime divisor of 2^p-1, 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 | q-1, so kp = q-1 for some k, and q = kp+1 for some k.  I'm not sure where this is going. Title: Re: f(n) | 2^n - 1 Post by Eigenray on Mar 14th, 2009, 6:51pm So q can't divide very many values f(p), can it? Title: Re: f(n) | 2^n - 1 Post by Obob on Mar 15th, 2009, 4:47pm 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^p-1.  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) Powered by YaBB 1 Gold - SP 1.4! Forum software copyright © 2000-2004 Yet another Bulletin Board