Solution June 03, 2007

Problem

i) Find all infinite arithmetic progressions of positive integers $(d_n)_{n\in\mathbb{N}}$ such that $d p$ is prime for all sufficiently large primes $p$ .

ii) Find all polynomials $f(X) \in \mathbb{Z}[X]$ such that $\left|f(p)\right|$ is prime for all sufficiently large primes $p$ .

Solution

i) is a special case of ii) so we will only solve ii) below.

Let $f(x)\,$ satisfy the condition in ii).

$f(x) = x\,$ and $f(x) = -x\,$ are obviously solutions.

Assume that $f(x)\neq \pm x$ . Then there is a prime $p\,$ with $q:=|f(p)|\neq p\,$ . Then $\gcd(p,q)=1\,$ so the sequence $a_n:=n\cdot q+p$ contains infinitely many primes by Dirichlet's theorem (http://en.wikipedia.org/wiki/Dirichlet's_theorem_on_arithmetic_progressions).

Let $a_n\,$ be prime. Because $f\,$ is a polynomial with integer coefficients, $f(a_n)-f(p)\,$ is divisible by $a_n-p\,$ , which in turn is divisible by $q\,$ by the definition of $a_n\,$ . We also have $f(p)\equiv 0 \pmod q$ so $f(a_n)\equiv 0\pmod q$ . But $f(a_n)\,$ must be prime so that $|f(a_n)| = q\,$ . So $f(a_n)\,$ assumes one of the values $-q\,$ or $+q\,$ infinitely often which means that it is a constant polynomial.

$f(x) = \pm q$ for prime numbers $q\,$ are indeed solutions of ii). There are no other solutions.

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