1. Given a strictly increasing function $f: \mathbb{N}_{0} \rightarrow \mathbb{N}_{0}$ (where $\mathbb{N}_{0}$ is the set of non-negative integers), which satisfies the relation $f(n+f(m))=f(n)+m+1$ for any $m, n \in \mathbb{N}_{0}$. Find all possible values that $f(2023)$ can take.

(T.A. Garmanova)
Solution. 1) Substituting $m=0, n=0$, we get $f(f(0))=f(0)+1$.

If $f(0)=0$, then we get $f(0)=f(0)+1$, which is impossible.

2) Let $f(0)=a$, then $a \in \mathbb{N}$. From the first point, we get that $f(a)=a+1$. If we substitute $m=0, n=a$, then we get that $f(2a)=f(a)+1=a+2$. Therefore, the values of the function at the ends of the segment $[a ; 2a]$ are two consecutive natural numbers. According to the condition, the function $f: \mathbb{N}_{0} \rightarrow \mathbb{N}_{0}$ is strictly increasing, which means there should be no other integer points on the segment $[a ; 2a]$ except $a$ and $2a$, because otherwise, the values at these points would coincide with $a+1$ or $a+2$, which would contradict the strict increase. $2a-a=1$, i.e., $a=1$.

Substituting $m=0$ into the original relation and considering the equality $f(0)=1$, we get $f(n+1)=$ $f(n)+1$. Thus, $f(n)=n+1$, therefore, $f(2023)=2024$.