Determine the integers $a, b \geqslant 3$ such that $a b^{2}+b+7 \mid a^{2} b+a+b$
On a :

$$
\begin{gathered}
a b^{2}+b+7 \mid a^{2} b^{2}+a b+b^{2} \\
a b^{2}+b+7 \mid a^{2} b^{2}+a b+7 a
\end{gathered}
$$

It follows that: $a b^{2}+b+7 \mid b^{2}-7 a$. Since $a, b \geqslant 3: b^{2}-7 a < b^{2} < a b^{2}+b+7$ and $7 a - b^{2} < 7 a < a b^{2}+b+7$. Therefore, $b^{2}-7 a=0$. It follows that $b=7 k$ and $a=7 k^{2}$ with $k \in \mathbb{N}^{*}$.

Let's now verify the solutions of the form $(a, b)=\left(7 k^{2}, 7 k\right)$: we have $a^{2} b+a+b=7^{3} k^{5}+7 k^{2}+7 k=k\left(7 k^{2} \cdot(7 k)^{2}+7 k+7\right)=k\left(a b^{2}+b+7\right)$.

Thus, the solutions are exactly of the form $(a, b)=\left(7 k^{2}, 7 k\right)$ with $k \in \mathbb{N}^{*}$.