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 $2023$ : $\begin{aligned} 0 f (f (x)) + (- 1) x^{2} & \equiv (- 1) f (x) + 0 [f (x)]^{2} + 1 (\mod 2023) \\ - x^{2} & \equiv - f (x) + 1 (\mod 2023) \\ f (x) & \equiv x^{2} + 1 (\mod 2023) \end{aligned}$ 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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