2023 OIM Problems/Problem 2

Problem

Let $\mathbb{Z}$ be the set of integers. Find all functions $f: \mathbb{Z} \to \mathbb{Z}$ such that:

$2023f(f(x))+2022x^2=2022f(x)+2023[f(x)]^2+1$

for each integer $x$ .

Solution

Consider the equation mod $2023$ : \begin{align*} 0f(f(x))+(-1)x^2&\equiv(-1)f(x)+0[f(x)]^2+1\pmod{2023}\\ -x^2 &\equiv -f(x)+1 \pmod{2023}\\ f(x) &\equiv x^2+1 \pmod{2023}\\ \end{align*} Thus by definition, for every integer $x$ , there exists some integer $k_x$ such that $f(x)=x^2+1+2023k_x$ If we substitute this into the initial functional equation, we get: $2023((x^2+1+2023k_x)^2+1+2023k_x)+2022x^2=2022(x^2+1+2023k_x)+2023(x^2+1+2023k_x)^2+1$ $\Rightarrow 2023(1+2023k_x)+2022x^2=2022(x^2+1+2023k_x)+1$ $\Rightarrow2023(1+2023k_x)=2022(2023k_x)+2023$ $\Rightarrow2023(2023k_x)=2022(2023k_x)$ $\Rightarrow k_x=0$ Thus the only functional equation possible is $\boxed{f(x)\equiv x^2+1}$ , which works upon substitution.

~ eevee9406

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2023 OIM Problems/Problem 2

Problem

Solution

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