Find all integers $m, n \in \mathbb{N}$ such that:

$$
1+(m+n) m \mid (n+1)(m+n)-1
$$

## 2 Solution
If $1+m(m+n)|(m+n)(n+1)-1, 1+m(m+n)|(m+n)(n+1)-1+1+$ $m(m+n)$ and thus $1+m(m+n) \mid (m+n)(m+n+1)$. Now, $(1+m(m+n))-m(m+n)=1$, so $m+n$ and $1+m(m+n)$ are coprime. The Gauss lemma gives $1+m(m+n) \mid m+n+1$ and thus $1+m(m+n) \leq m+n+1$, from which $(m-1)(m+n) \leq 0$, so $m=n=0$ or $m \leq 1$, so $m \leq 1$. We verify that the property holds whenever $m=0$. For $m=1$, it is a matter of finding all $n$ such that $n+2 \mid (n+1)^2-1$. But any $n$ works since $(n+1)^2-1=n(n+2)$. The solutions are therefore $m=0, n \in \mathbb{N}$ and $m=1, n \in \mathbb{N}$.

## 3 Group C

## 1 Statement