[ Limitedness, monotonicity]

Given a strictly increasing function \$f \colon \mathbb{\mathrm{N}}_0 \to \mathbb{\mathrm{N}}_0\$ (where $\mathbb{\mathrm{N}}_0$ is the set of non-negative integers), which satisfies the relation $\mathrm{f}(\mathrm{n}+\mathrm{f}(\mathrm{m}))=\mathrm{f}(\mathrm{n})+\mathrm{m}+1$ for any $m,n \in \mathbb{\mathrm{N}}_0$. Find all possible values that \$f(2023)\$ can take.
1) Substituting $\$ \mathrm{~m}=0 \$, \$ n=0 \$$, we get $\$ \mathrm{f}(\mathrm{f}(0))=\mathrm{f}(0)+1 \$$. If $\$ \mathrm{f}(0)=0 \$$, then we get $\$ \mathrm{f}(0)=\mathrm{f}(0)+1 \$$, which is impossible.

![](https://cdn.mathpix.com/cropped/2024_05_06_80a25f3745320796e6c6g-29.jpg?height=60&width=1946&top_left_y=2062&top_left_x=-1)
$\$ \mathrm{n}=\mathrm{a} \$$, then we get that $\$ \mathrm{f}(2 \mathrm{a})=\mathrm{f}(\mathrm{a})+1=\mathrm{a}+2$ \$ Therefore, the values of the function at the endpoints of the interval $\$[\mathrm{a} ; 2 \mathrm{a}]$ \$ are two consecutive natural numbers. By the condition, the function \$flcolon \mathbb $\{\mathrm{N}\}$ _0lto \mathbb\{N\}_0\$ is strictly increasing, so on the interval \$ otherwise the values at these points would coincide with $\$ a+1 \$$ or $\$ a+2 \$$, which would contradict the strict increase. Therefore, $\$ 2 \mathrm{a}-\mathrm{a}=1, \$$ i.e., $\$ \mathrm{a}=1 \$$.

Substituting into the original relation $\$ \mathrm{~m}=0 \$$ and considering the equality $\$ \mathrm{f}(0)=1 \$$, we get $\$ \mathrm{f}(\mathrm{n}+1)=\mathrm{f}(\mathrm{n})+1 \$$. Thus, \$f(n)=n+1\$, hence, \$f(2023)=2024\$.

## Answer

$\$(2023)=2024 \$$.

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