Competition problem — Algebra — Maths Olympiad Prep

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Algebra Difficulty 5.5 AIME, harder Prove it Russia

Let $P(x)$ be a quadratic polynomial with a unit leading coefficient. Given that the polynomials $P(x)$ and $P(P(P(x)))$ have a common root, prove that $P(0)P(1) = 0$ . (A. Khrabrov)

Квадратный трёхчлен $P(x)$ с единичным старшим коэффициентом таков, что многочлены $P(x)$ и $P(P(P(x)))$ имеют общий корень. Докажите, что $P(0) \cdot P(1) = 0$ . (А. Храбров)

Solution

Пусть $t$ — общий корень данных многочленов. Тогда $0 = P(P(P(t))) = P(P(0))$ . Пусть $P(x) = x^2 + a x + b$ ; тогда $P(0) = b$ , $P(1) = a + b + 1$ , а значит, $0 = P(P(0)) = P(b) = a b + b^2 + b = b(a + b + 1) = P(0) \cdot P(1)$ , что и требовалось доказать.

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