A number theory problem by Ojas Singh Malhi

Relevant wiki: Quadratic Diophantine Equations - Solve by Simon's Favorite Factoring Trick

$\displaystyle \begin{aligned} y^2 + 3x^2y^2 & = 30x^2 + 517 \\ y^2 + 3x^2y^2 - 30x^2 & = 517 \\ y^2\left(3x^2 + 1\right) - 10\left(3x^2 + 1\right) & = 517 - 10 \\ \left(y^2 - 10\right)\left(3x^2 + 1\right) & = 507 \end{aligned}$

Now, $507 = 3 \times 13^2$ , so we must have $3x^2 + 1 \in \{1, 3, 13, 39, 169, 507\}$ .

$\begin{array}{c|c|c} \text{Factor} & \text{Corresponding equation} & \text{Solutions for } x^2 \\ \hline 1 & 3x^2 + 1 = 1 & 0 \\ \hline 3 & 3x^2 + 1 = 3 & \frac 23 \\ \hline 13 & 3x^2 + 1 = 13 & 4 \\ \hline 39 & 3x^2 + 1 = 39 & \frac {38}{3} \\ \hline 169 & 3x^2 + 1 = 169 & 56 \\ \hline 507 & 3x^2 + 1 = 507 & \frac{506}{3} \\ \end{array}$

We see that the only perfect squares are $0$ and $4$ . If $3x^2 + 1 = 1$ , then $y^2 - 10 = 507 \implies y^2 = 517$ , which is not a perfect square (note that $y$ must be an integer). Therefore we have $x^2 = 4$ , and it follows that $y^2 - 10 = 39 \implies y^2 = 49$ . Our answer is $3x^2y^2 = 3 \times 4 \times 49 = \boxed{588}$ .