Find all pairs ( $a, b$ ) of strictly positive integers such that $a b^{2}+b+7$ divides $a^{2} b+a+b$.

## - Exercise Solutions -

Translate the above text into English, please keep the original text's line breaks and format, and output the translation result directly.
If $a b^{2}+b+7$ divides $a^{2} b+a+b$, then it also divides $a\left(a b^{2}+b+7\right)-b\left(a^{2} b+a+b\right)=7 a-b^{2}$. Several possibilities present themselves to us.

- If $7 \mathrm{a}b^{2}$, then $7 a-b^{2}>0$ so $a^{2}<a^{2}+b+7 \leqslant 7 a-b^{2}<7 a$ and $b \leqslant 2$:
- if $b=1$, then $a+8$ must divide $7 a-1$, so $7 a-1-7(a+8)=-57=-3 \times 19$ as well: we have $a+8=19$ or $a+8=57$, and conversely the pairs $(11,1)$ and $(49,1)$ are both solutions to the problem
- if $b=2$, then $4 a+9$ must divide $7 a-4$, so $4(7 a-4)-7(4 a+9)=-79$ as well: since 79 is prime, we must have $4 a+9=79$, which has no integer solution.

The solutions to the problem are the pairs $(11,1),(49,1)$ and $\left(7 \mathrm{k}^{2}, 7 \mathrm{k}\right)$ for $\mathrm{k} \geqslant 1$.

## 3 Arithmetic Course/TD

## - Exercises -