Find all functions $f: \mathbb{N} \rightarrow \mathbb{N}$ such that for all natural numbers $a$ and $b$:

$$
f(f(a)+f(b))=a+b
$$
For $b=0, f(f(a)+f(0))=a$, we then have $f$ injective (since if $f(n)=f(m), f(f(n)+f(0))=$ $f(f(m)+f(0)))$ and surjective.

Let $n$ be a natural number, we take $a=n$ and $b=1$ then $a=n+1$ and $b=0$, we have then

$$
f(f(n)+f(1))=n+1=f(f(n+1)+f(0))
$$

Thus by injectivity: $f(n+1)=f(n)+f(1)-f(0)$, we denote $f(0)=a$ and $f(1)=b$, we then have $f(n+1)=f(n)+b-a$.

By easy induction $f(n)=a+n(b-a)$, we can then conclude in two ways, either we re-inject into the initial equation to get $a=0$ and $b=1$, or, the surjectivity says that every natural number is reached, so $b-a=1$ and $a=0$. In all cases $f(n)=n$ for all $n$ and this function fits well.