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Functional Equation with Floors

Source: 2022 Taiwan TST Round 1 Independent Study 1-A

March 16, 2022
algebrafunctional equationTaiwan

Problem Statement

Find all f:Z→Zf:\mathbb{Z}\to\mathbb{Z} such that f(⌊f(x)+f(y)2⌋)+f(x)=f(f(y))+⌊f(x)+f(y)2⌋f\left(\left\lfloor\frac{f(x)+f(y)}{2}\right\rfloor\right)+f(x)=f(f(y))+\left\lfloor\frac{f(x)+f(y)}{2}\right\rfloor holds for all x,y∈Zx,y\in\mathbb{Z}.
Proposed by usjl
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