IMO 2004 SL A4

Origin: KOR | Category: Algebra

Problem

Find all polynomials P(x) with real coefficients that satisfy the equality P(a −b) + P(b −c) + P(c −a) = 2P(a + b + c) for all triples a, b, c of real numbers such that ab + bc + ca = 0.

Solution

Let P(x) = a0 + a1x + \cdot \cdot \cdot + anxn. For every x \inR the triple (a, b, c) = (6x, 3x, −2x) satisfies the condition ab + bc + ca = 0. Then the condition on P gives us P(3x) + P(5x) + P(−8x) = 2P(7x) for all x, implying that for all i = 0, 1, 2, . . ., n the following equality holds:
3i + 5i + (−8)i −2 \cdot 7i ai = 0. Suppose that ai ̸= 0. Then K(i) = 3i + 5i + (−8)i −2 \cdot 7i = 0. But K(i) is negative for i odd and positive for i = 0 or i \geq6 even. Only for i = 2 and i = 4 do we have K(i) = 0. It follows that P(x) = a2x2 + a4x4 for some real numbers a2, a4. It is easily verified that all such P(x) satisfy the required condition.