1. Find all integer pairs $x$ and $y$ such that the difference of their squares is 203.
Solution. If the pair $(x, y)$ is a solution to the equation

$$
x^{2}-y^{2}=203
$$

then the pairs $(-x, y),(x,-y),(-x,-y)$ are also solutions to the equation (1). Therefore, it is sufficient to find only the positive solutions of the equation.

From $203=1 \cdot 203=7 \cdot 29$ and from

$x^{2}-y^{2}=(x-y)(x+y)$,

we get

$$
\left\{\begin{array}{l}
x-y=1 \\
x+y=203
\end{array}\right.
$$

or

$$
\left\{\begin{array}{l}
x-y=7 \\
x+y=29
\end{array}\right.
$$

The solution to system (2) is $x=102, y=101$, and the solution to system (3) is $x=18$, $y=11$. Therefore, the solutions to equation (1) are the pairs

$$
\begin{array}{llll}
(102,101), & (-102,101), & (102,-101), & (-102,-101) \\
(18,11), & (-18,11), & (18,-11), & (-18,-11)
\end{array}
$$