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weird FE, f(x)≠f(x+h), f is semiconstant

Source: Mongolia 1999 Teachers secondary level P1

May 6, 2021
fefunctional equationalgebra

Problem Statement

Suppose that a function f:R→Rf:\mathbb R\to\mathbb R is such that for any real hh there exist at most 1999050919990509 different values of xx for which f(x)≠f(x+h)f(x)\ne f(x+h). Prove that there is a set of at most 99952569995256 real numbers such that ff is constant outside of this set.
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