A polynomial \(f(x)\in\mathbb{Z}[x]\) has the property that \(2^{n-1}-1\mid f\left(2^{n}-1\right)\) for all integers \(n>1\). Show that \(f(x)=0\) has an integer root.

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TopNewestWe have the condition \( A\mid f(2A+1)\) where \(A\) is a power of two minus one. Assuming that \(n\) is big enough, let \(p\) be the biggest prime number dividing \(A\): we have \(p\mid f(2A+1)\), or just \(p\mid f(1)\). Since the set of prime numbers dividing a number of the form \(2^m-1\) is infinite (*), we have that \(f(1)\) is divisible by an infinite number of primes - this gives \(f(1)=0\).

(*) Assuming that every \(2^n-1\) can be factored in terms of \(p_1,\ldots,p_k\), we have a contradiction, since: \[ \forall i\in[1,k],\qquad 2^{1+\prod_{j=1}^k(p_j-1)}-1\equiv 2-1\equiv 1\not\equiv 0\pmod{p_i}.\] – Jack D'Aurizio · 3 years, 1 month ago

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Note that \(p\) doesn't need to be the biggest prime number that divides \(A\). (doesn't impact your proof) – Calvin Lin Staff · 3 years, 1 month ago

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– Sreejato Bhattacharya · 3 years, 1 month ago

Minor typo: this gives \(f(1)=0.\)Log in to reply

Is there a more general statement that can be made here?

On a related note, show that if \( n \mid f(n) \) for infinitely many integers \(n\), then \( f(0) = 0 \). – Calvin Lin Staff · 3 years, 1 month ago

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Note that \(f(n) \equiv f(0) \pmod{n},\) so \(n \mid f(0)\) for infinitely many \(n \in \mathbb{Z},\) which is possible if and only if \(f(0)=0,\) since \(0\) is the only integer with infinitely many divisors. – Sreejato Bhattacharya · 3 years, 1 month ago

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