3A. Determine all pairs of integers \(x\) and \(y\) that are solutions to the equation

\[
(x+y+1993)^{2}=x^{2}+y^{2}+1993^{2}
\]
Solution. We have

$$
\begin{aligned}
& (x+y+1993)^{2}=x^{2}+y^{2}+1993^{2} \\
& x^{2}+y^{2}+1993^{2}+2 x y+2 \cdot 1993 x+2 \cdot 1993 y=x^{2}+y^{2}+1993^{2} \\
& x y+1993 x+1993 y=0 \\
& (x+1993)(y+1993)=1993^{2}
\end{aligned}
$$

Given that 1993 is a prime number, and $x$ and $y$ are integers, the last equation yields the following six systems of linear equations:

$$
\begin{aligned}
& \left\{\begin{array}{l}
x+1993=1 \\
y+1993=1993^{2}
\end{array}\right. \\
& \left\{\begin{array}{l}
x+1993=1993^{2} \\
y+1993=1
\end{array}\right. \\
& \left\{\begin{array}{l}
x+1993=1993 \\
y+1993=1993
\end{array}\right. \\
& \left\{\begin{array}{l}
x+1993=-1 \\
y+1993=-1993^{2}
\end{array}\right. \\
& \left\{\begin{array}{l}
x+1993=-1993^{2} \\
y+1993=-1
\end{array}\right. \\
& \left\{\begin{array}{l}
x+1993=-1993 \\
y+1993=-1993
\end{array}\right.
\end{aligned}
$$

whose solutions are the pairs:

$$
\begin{aligned}
& (-1992,3970056),(3970056,-1992),(0,0), \\
& (1994,-3974042),(-3974042,-1994),(-3986,3986)
\end{aligned}
$$