Find all integer solutions $(a, b)$ of the equation $3 a^{2} b^{2}+b^{2}=517+30 a^{2}$.
With the factorization of the rectangle, the equation rewrites as $\left(3 a^{2}+1\right)\left(b^{2}-10\right)=507=3 \cdot 13^{2}$. Let's discuss the cases by noting that $3 a^{2}+1$ is coprime with 3.

$-3 a^{2}+1=1$ thus $a=0$ and $b^{2}-10=507$: impossible.

$-3 a^{2}+1=13$ thus $a= \pm 2$ and $b^{2}-10=39$, i.e., $b= \pm 7$.

$-3 a^{2}+1=13^{2}$ thus $a^{2}=56$: impossible.

The only solutions are $(2,7),(-2,7),(2,-7)$ and $(-2,-7)$.