(given to Juliette Fournier). Find all strictly positive integers $a$ and $b$ such that $a b^{2}+b+7$ divides $a^{2} b+a+b$.


Let $q=a b^{2}+b+7$, and $n=a^{2} b+a+b$. Since $q$ divides, $q$ divides $n b$ as well as $n b-a q=b^{2}-7 a$. And $-7 a-q$. If $b \geq 3$, the quotient $n / q$, if it is an integer, can only be 0. Therefore, there are three cases to study: $b=1, b=2$ and $b^{2}-7 a=0$.

If $b=1, q=a+8$ must divide $n=a^{2}+a+1=(a+8)(a-7)+57$. Therefore, $a+8$ must divide 57: either $a=11$, or $a=49$, which provides two solutions: $b=1$ and $a=11$, $q=19$ and $n=133=19 \times 7$, and $b=1, a=49, q=57, n=2451=57 \times 43$.

If $b=2, q=4 a+9$ must divide $2 a^{2}+a+2=\frac{1}{8}(4 a+9)(4 a-7)+\frac{79}{8}$. Therefore, $4 a+9$ must divide 79, which is clearly not possible (79 is prime and $79-9=70$ is not divisible by 4).

But let's not forget the third case: $b^{2}-7 a=0$. If $7 a$ is a perfect square divisible by $7, a=7 k^{2}, b=7 k$, and we notice that for any value of $k$, we have a solution, because $q=343 k^{4}+7 k+7$ divides $n=343 k^{5}+7 k^{2}+7 k=k q$. The solutions are therefore $(11,1),(49,1)$ and $\left(7 k^{2}, 7 k\right)$ for any value of $k$.