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Team Selection Test for IMO, 2009

Number theory Difficulty 8.4 Shortlist Prove it Turkey

Find all primes pp for which there exist an odd integer nn and a polynomial Q(x)Q(x) with integer coefficients such that the polynomial 1+pn2+∏i=12p−2Q(xi)1 + p n^2 + \prod_{i=1}^{2p-2} Q(x^i) has at least one integer root.

Solution

Let P(x)=1+pn2+∏i=12p−2Q(xi)P(x) = 1 + p n^2 + \prod_{i=1}^{2p-2} Q(x^i). For p=2p = 2, n=1n = 1 and Q(x)=2x+1Q(x) = 2x + 1 work as P(−1)=0P(-1) = 0.

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

Since all Q(ai)Q(a^i) have the same parity for 1≤i≤2p−21 \le i \le 2p-2 for an integer aa; if P(a)=0P(a) = 0, then p≡3(mod4)p \equiv 3 \pmod{4}.

We also have ai≡ai+p−1(modp)a^i \equiv a^{i+p-1} \pmod{p} for 1≤i≤p−11 \le i \le p-1 for an integer aa. Therefore, ∏i=12p−2Q(ai)≡(∏i=1p−1Q(ai))2(modp)\prod_{i=1}^{2p-2} Q(a^i) \equiv (\prod_{i=1}^{p-1} Q(a^i))^2 \pmod{p}, and P(a)≡1+(∏i=1p−1Q(ai))2≠0(modp)P(a) \equiv 1 + (\prod_{i=1}^{p-1} Q(a^i))^2 \ne 0 \pmod{p} for p≡3(mod4)p \equiv 3 \pmod{4}.

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