steppewolf wrote: Find all functions $f: \mathbb{N} \rightarrow \mathbb{Z}$ such that $|f(k)| \leq k$ for all positive integers $k$ and there is a prime number $p>2024$ which satisfies both of the following conditions: $1)$ For all $a \in \mathbb{N}$ we have $af(a+p) = af(a)+pf(a),$ $2)$ For all $a \in \mathbb{N}$ we have $p|a^{\frac{p+1}{2}}-f(a).$ $p|f(p)$ and so $f(p)=\epsilon p$ where $\epsilon\in\{-1,+1\}$ A simple induction implies then $f(kp)=\epsilon kp$ Considering then $x\not\equiv 0\pmod p$ : We have $\frac{f(x+p)}{x+p}=\frac{f(x)}x$ and so $f(kp+r)=(kp+r)\frac{f(r)}r$ where $r\in\{1,2,...,p-1\}$ This implies $r|f(r)$ and so $f(x)=\epsilon(x)x$ where $\epsilon(x)\in\{-1,+1\}$ And so $p|x(x^{\frac{p-1}2}-\epsilon(x))$ And solution : $f(kp+r)=\epsilon(r)(kp+r)$ where $\epsilon (0)=\pm 1$ (both choices fit) For $r\ne 0$ : If $r^{\frac{p-1}2}\equiv 1\pmod p$, then $\epsilon(r)=1$ If $r^{\frac{p-1}2}\equiv p-1\pmod p$, then $\epsilon(r)=-1$