[ Formulas for abbreviated multiplication (other).] Case analysis

Solve the equation $\left(x^{2}-y^{2}\right)^{2}=16 y+1$ in integers.

#
Obviously, $y \geq 0$, and $x^{2} \neq y^{2}$. Therefore, $16 y+1=\left(y^{2}-x^{2}\right)^{2} \geq\left(y^{2}-(y-1)^{2}\right)^{2}=(2 y-1)^{2}=4 y^{2}-4 y+1$. Hence, $4 y^{2} \leq 20 y$, which means $y \leq 5$. It remains to check the values of $y$ from 0 to 5 and find the corresponding integer values of $x$.

## Answer

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