Find all functions $f: \mathbb{N}^{*} \rightarrow \mathbb{N}^{*}$ that satisfy for all $n$

$$
h(h(n))+h(n+1)=n+2
$$
Let's start by calculating $h(1)$ and $h(2)$. Taking $n=1$, we have $h(h(1)) + h(2) = 3$, and since $h$ takes values in $\mathbb{N}^{*}$, $h(2) \leq 2$.

First case: $h(2) = 1$. Then we have $h(h(1)) = 2$. Taking $n=2$, we get $h(h(2)) + h(3) = h(1) + h(3) = 4$, so $h(1) \leq 3$. If $h(1) = 1$, then $h(h(1)) = 1$; if $h(1) = 2$, then $h(h(1)) = 1$, which does not work. Finally, if $h(1) = 3$, then $h(h(1)) = h(3) = 4 - h(1) = 1$. Therefore, this case has no solution.

Second case: $h(2) = 2$. Then we have $h(h(1)) = 1$. Setting $n=2$, we find $h(3) = 2$, then $h(4) = 3$, $h(5) = 4$, $h(6) = 4$. It is easy to show that if $n \geq 2$, then $h(n) \geq 2$, and thus $h(1) = 1$. We then see that if there is a solution, it is unique. It is sufficient to find a solution that works, and it will be the only one. The function in question is then

$$
h(n) = [n \alpha] + 1, \text{ with } \alpha = \frac{-1 + \sqrt{5}}{2} \approx 0.618
$$

The proof that this function works uses the fact that $\alpha^2 + \alpha = 1$ (but it is not so easy).