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Number theory Difficulty 8.4 Shortlist Prove it Turkey

Find all primes $p$ for which there exist an odd integer $n$ and a polynomial $Q(x)$ with integer coefficients such that the polynomial $1 + p n^2 + \prod_{i=1}^{2p-2} Q(x^i)$ has at least one integer root.

Solution

Let $P(x) = 1 + p n^2 + \prod_{i=1}^{2p-2} Q(x^i)$ . For $p = 2$ , $n = 1$ and $Q(x) = 2x + 1$ work as $P(-1) = 0$ .

We will show that for odd primes no suitable $n$ and $Q(x)$ exist.

Since all $Q(a^i)$ have the same parity for $1 \le i \le 2p-2$ for an integer $a$ ; if $P(a) = 0$ , then $p \equiv 3 \pmod{4}$ .

We also have $a^i \equiv a^{i+p-1} \pmod{p}$ for $1 \le i \le p-1$ for an integer $a$ . Therefore, $\prod_{i=1}^{2p-2} Q(a^i) \equiv (\prod_{i=1}^{p-1} Q(a^i))^2 \pmod{p}$ , and $P(a) \equiv 1 + (\prod_{i=1}^{p-1} Q(a^i))^2 \ne 0 \pmod{p}$ for $p \equiv 3 \pmod{4}$ .

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Source: MathNet, licensed CC-BY-4.0. Statement and solution reproduced as published; topic and difficulty added by this site.