## Task A-1.5.

Determine all pairs $(m, n)$ of integers for which $m n + 5 m + 2 n = 121$.
## Solution.

By rearranging we have:

$$
\begin{aligned}
m n+5 m+2 n & =121 \\
m(n+5)+2(n+5) & -10=121 \\
(m+2)(n+5) & =131
\end{aligned}
$$

Notice that 131 is a prime number.

Therefore, it must be that $m+2 \in\{ \pm 1, \pm 131\}$ and $n+5=\frac{131}{m+2}$.

By listing all possibilities, we get:

- For $m+2=1$, we have $n+5=131$, so $m=-1, n=126$.
- For $m+2=-1$, we have $n+5=-131$, so $m=-3, n=-136$.
- For $m+2=131$, we have $n+5=1$, so $m=129, n=-4$.
- For $m+2=-131$, we have $n+5=-1$, so $m=-133, n=-6$.

Therefore, the solutions are

$$
(m, n) \in\{(-1,126),(-3,-136),(129,-4),(-133,-6)\}
$$