Solve the equation $x^{2}-y^{2}=1988$ in integers.

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Solve the equation in non-negative integers, and then put arbitrary signs. $(x-y)(x+y)=x^2-y^2=1988=2^2 \cdot 7 \cdot 71$. The numbers

$x-y$ and $x+y$ have the same parity, so the smaller of them, $-x-y$, is 2 or 14, and $x+y$ is 994

or 142. Hence, $(x, y)=(498,496)$ or $(78,64)$.

## Answer

$\pm(498, \pm 496), \pm(78, \pm 64)$.