Integral solutions 1

$\begin{aligned} y^2(5x^2+1) & = 25(2x^2+13) \\ \implies y^2 & = 5^2 \cdot \frac {2x^2+13}{5x^2+1} \end{aligned}$

Since the LHS is a perfect square, the RHS must also be a perfect square and hence $\dfrac {2x^2+13}{5x^2+1}$ is also a perfect square and let it be $m^2$ , where $m$ is an integer.

$\begin{aligned} \frac {2x^2+13}{5x^2+1} & = m^2 & \small \color{#3D99F6} \text{Rearranging} \\ \implies x^2 & = \frac {13-m^2}{5m^2-2} \end{aligned}$

Considering the non-negative $m$ , we note that the RHS $< 0$ for $m \ge 4$ and hence no solution for $x$ because the LHS $\ge 0$ . For $m = 0, 1, 2, 3$ , there is only a solution for $m=1$ when $x^2 = 4$ $\implies x = \pm 2$ . Putting this back in the following equation:

$\begin{aligned} y^2 & = 5^2 \cdot \frac {2(4)+13}{5(4)+1} = 5^2 \\ \implies y & = \pm 5 \end{aligned}$

Therefore, there are $\boxed{4}$ integer pairs $(-2,-5), (-2,5), (2,-5), (2, 5)$ satisfying the equation.