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Divisibility for the Win - Problem 3

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.

Note by Cody Johnson
3 years, 8 months ago

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We 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, 8 months ago

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Nice way to show that the set of primes which divide \( 2^n -1 \) is infinite.

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, 8 months ago

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Minor typo: this gives \(f(1)=0.\)

Sreejato Bhattacharya - 3 years, 8 months ago

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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, 8 months ago

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The general statement, I believe, would be that if \(f(m) \mid f(g(m))\) for infinitely many \(m \in \mathbb{Z}\) for some \(f, g \in \mathbb{Z}[x],\) then \(f(g(0))=0,\) although there are probably other ways generalize this.

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, 8 months ago

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