Sonkin $M$.

Solve the equation $\left(x^{2}-y^{2}\right)^{2}=1+16 y$ in integers.
Clearly, $y \geq 0$, so the left side is not less than $(2 y-1)^{2}$, since the absolute difference between $y^{2}$ and any perfect square (if $y \geq 0$ and the squares are different) is not less than $|2 y-1|$. Therefore, $(2 y-1)^{2} \leq 1+16 y$, from which $y \leq 5$. Thus, the right side can take the values $1, 17, 33, 49, 65, 81$, of which only $1, 49, 81$ are squares.

Consider the three cases.

1) $y=0, (x^{2})^{2}=1$, hence $x= \pm 1$.
2) $y=3, (x^{2}-9)^{2}=49$, hence $x^{2}=16$, that is, $x= \pm 4$.
3) $y=5, (x^{2}-25)^{2}=81, x^{2}=34$ or 16, that is, $x= \pm 4$.

## Answer

$(\pm 1,0), (\pm 4,3), (\pm 4,5)$.