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f(x) = 1- f(x-a_1)f(x-a_2)... f(x-a_n) for integers x>=0, f(x)=1 if integer x<0

Source: Czech And Slovak Mathematical Olympiad, Round III, Category A 2001 p6

February 11, 2020
functional equationfunctionalgebra

Problem Statement

Let be given natural numbers a1,a2,...,ana_1,a_2,...,a_n and a function f:Z→Rf : Z \to R such that f(x)=1f(x) = 1 for all integers x<0x < 0 and f(x)=1−f(x−a1)f(x−a2)...f(x−an)f(x) = 1- f(x-a_1)f(x-a_2)... f(x-a_n) for all integers x≥0x \ge 0. Prove that there exist natural numbers ss and tt such that for all integers x>sx > s it holds that f(x+t)=f(x)f(x+t) = f(x).
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