Divisibilité d’un nombre pair

Suppose that $n-1$ divides $2n$. This means that there exists an integer $k$ such that $2n=k(n-1)=kn-k$. We obtain that $k$ is divisible by $n$ : there exists an integer $l$ such that $k=nl$. Plugging this back in the equation above, we get that $2n=nl(n-1)$, and because $n$ is non-zero, we obtain $2=l(n-1)$.
$n-1$ is therefore a divisor of $2$, which means it is either $1$ or $2$, and so $n=2$ or $n=3$. We now just have to manually check if these cases work : for $n=2$, $4$ is indeed divisible by 1 ; and for $n=3$, $6$ is indeed divisible by $2$.

So the only solutions are $n=2$ and $n=3$.

Soit $n \ge 2$, supposons que $n-1|2n$, alors : $\exists k \in \mathbb{N}/k=\frac{2n}{n-1} \Rightarrow \exists k \in \mathbb{N} / (k-2=\frac{2}{n-1} )$.
$$\text {Or, }k\in \mathbb{N} \Rightarrow k-2 \in \mathbb{Z}\Rightarrow (n-1)|2 \Rightarrow n-1\le 2 \Rightarrow n \in \left\{2,3\right\}.$$Réciproquement, ces 2 solutions fonctionnent. Au final, l’ensemble S des solutions est S={2,3}.