Find all $f: \mathbb{N}^{*} \rightarrow \mathbb{N}^{*}$ such that $\forall n \geq 1$,

$$
\begin{gathered}
\text { (i) } f(n)+f(n+1)=f(n+2) f(n+3)-11 \\
\text { (ii) } f(n) \geq 2
\end{gathered}
$$

## Solutions
Let's show that for all $n \geq 1, f(n+2)=f(n)$. Indeed, by writing the functional equation at rank $n$ and rank $n+1$, we obtain by taking the difference of the two equations: $f(n+2)-f(n)=f(n+3)(f(n+4)-f(n+2))$. Let $g(n)=\|f(n+2)-f(n)\|$: $g(n)=f(n+3) f(n+3+2) \ldots f(n+3+2(k-1)) g(n+2 k)$. If the $g(n+2 k)$ are non-zero, then since $f(i) \geq 2, g(n)$ is not defined (tends to infinity in fact), which is impossible. Therefore, $g(n)=0$ so $f(n+2)=f(n)$. We set $f(1)=a, f(2)=b$, so $f(3)=f(1)=a$, and we have: $a+b=ab-11$. Hence $(a-1)(b-1)=12$ so $(a, b) \in\{(13,2),(7,3),(5,4),(2,13),(3,7),(4,5)\}$.

## 3 Exercises of the day

## 1 Statements

## Thursday August 19