Let $f\colon \mathbb{N_0}\to \mathbb{N_0}$
$f(n+1)\gt f(n)$ , with $n \in \mathbb{N_0}$
$f(n+f(m))= f(n)+m+1$, with $n,m \in \mathbb{N_0}$
$f(2017)=?$
I tried with a equation system, but i can't figure it out.
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First we have to determine what $f(0)$ is. Suppose $f(0)=0$. Then $f(0)=f(0+f(0))=f(0)+1$, a contradiction. Now suppose $f(0)=k$ for some $k>1$. Then $f(k)=f(0+f(0))=f(0)+1=k+1$. But note that by the first property, we have $$k=f(0)<f(1)<\cdots<f(k)=k+1.$$ Since $0<1<k$, there is no valid value for $f(1)$, a contradiction.
We deduce that $f(0)=1$. This and the second condition suggests that perhaps $f(n)=n+1$ holds for all $n$. Indeed, the function $n\mapsto n+1$ satisfies all of the above conditions, so is a candidate for the solution. We would like to prove that this is the only such function.
Let's prove this by induction. Suppose $f(m)=m+1$ holds for all $0\leqslant m<n$. In particular, $n=f(n-1)$. Thus, $$f(n)=f(0+f(n-1))=f(0)+n-1+1=f(0)+n=n+1.$$
Therefore, by induction, $f(n)=n+1$ holds for all $n$. In particular, $f(2017)=2018$.
JonCC
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