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exists f:R \to R, f(x)+f(2x)+...+f(nx) = 0 for all x \in R, f(x) = 0 iff x=0

Source: IMAR 2011 p3

September 27, 2018
functionalgebrafunctional equation

Problem Statement

Given an integer number n≥2n \ge 2, show that there exists a function f:R→Rf : R \to R such that f(x)+f(2x)+...+f(nx)=0f(x) + f(2x) + ...+ f(nx) = 0, for all x∈Rx \in R, and f(x)=0f(x) = 0 if and only if x=0x = 0.
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