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IMC 2010, Problem 5, Day 2

Source:

July 27, 2010
functionalgebrapolynomialvectoranalytic geometryreal analysisnumber theory

Problem Statement

Suppose that for a function f:R→Rf: \mathbb{R}\to \mathbb{R} and real numbers a<ba<b one has f(x)=0f(x)=0 for all x∈(a,b).x\in (a,b). Prove that f(x)=0f(x)=0 for all x∈Rx\in \mathbb{R} if ∑k=0p−1f(y+kp)=0\sum^{p-1}_{k=0}f\left(y+\frac{k}{p}\right)=0 for every prime number pp and every real number y.y.
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