62nd Ukrainian National Mathematical Olympiad

Consider the given equation modulo $3$. The right-hand side, $2023^k=1(\mathrm{mod}3)$. Among any three consecutive natural numbers $n$, $n+1$, $n+2$, one is divisible by $3$, one leaves a remainder of $1$ when divided by $3$, and one leaves a remainder of $2$.

If $x$ is divisible by $3$, then $x^x$ is also divisible by $3$; if $x$ has a remainder $1$, then $x^x$ also has a remainder $1$. If $x$ has a remainder $2$, then $x^x$ has remainder $1$ or $2$. Then, $n^n+(n+1{)}^{n+1}+(n+2{)}^{n+2}$ has a remainder $2$ or $0$ modulo $3$. This contradiction completes the proof.