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Baltic Way Shortlisted Problems, 2015

Algebra Difficulty 7.5 National Olympiad, round 2 Prove it Baltic Way

Let $P$ be a real polynomial of degree $2015$ and $Q$ a real quadratic polynomial. Could it be that the polynomial $P(Q(x))$ has precisely the roots
$-2014, -2013, \dots, -2, -1, 1, 2, \dots, 2014, 2015, 2016?$

Solution

The values of $Q$ at the $4030$ points indicated in the problem need be a subset of the zeroes of $P$ . But these are at most $2015$ in number, and $Q$ , being quadratic, assumes any given value at most twice. Therefore, the $4030$ numbers can be split into $2015$ pairs $(p_i, q_i)$ , for which $Q(p_i) = Q(q_i)$ runs through all the $2015$ zeroes of $P$ as $i = 1, \dots, 2015$ .

Put $Q(x) = a x^2 + b x + c$ . By Vieta's formulae, $p_i + q_i = -\frac{b}{a}$ for such a pair, so the $2015$ pairs need have equal sums. Since the numbers are integers, this is possible only if their sum is a multiple of $2015$ . But it is not, in fact the sum is
$-1007 \cdot 2015 + 1008 \cdot 2017,$
which is not even a multiple of $5$ . $\square$

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