Maths Olympiad Prep

Library / /4 of 4

, 2020

Number theory Difficulty 8.4 Shortlist Prove it Belarus

Let P(x)P(x) be a non-constant polynomial with integer coefficients such that P(0)1P(0) \neq 1. Prove that there exist infinitely many primes pp such that P(a)ap12P(a) - a^{\frac{p-1}{2}} is divisible by pp for some positive integer aa (possibly, depending on pp).

Solution

Consider the polynomial Q(x)=P(x2)1Z[x]Q(x) = P(x^2) - 1 \in \mathbb{Z}[x]. It's well-known that there exist infinitely many prime divisors of the numbers from the set M={Q(n):nN & Q(n)0}M = \{Q(n) : n \in \mathbb{N} \ \&\ Q(n) \neq 0\}. Moreover, since Q(0)=P(0)10Q(0) = P(0) - 1 \neq 0, among these prime divisors there exist infinitely many primes which doesn't divide Q(0)Q(0). Clearly, if pQ(n)p \mid Q(n) and pQ(0)p \nmid Q(0), then pnp \nmid n.

Therefore there exist infinitely primes pp and nNn \in \mathbb{N} such that pnp \nmid n and pP(n2)1p \mid P(n^2) - 1. Each such pp satisfy the problem conditions with a=n2a = n^2.

Want a route through all this instead of an archive? The track puts 2,000 problems in a working order, from AMC 10 level to the IMO shortlist.

Source: MathNet, licensed CC-BY-4.0. Statement reproduced verbatim; metadata (topic, difficulty) added by this project.