12.28 Try to find a set of positive integer solutions for the equation $x^{2}-51 y^{2}=1$.
(14th All-Russian Mathematical Olympiad, 1988)
［Solution］ Given the equation can be transformed as
$$
\begin{aligned}
x^{2} & =51 y^{2}+1 \\
& =49 y^{2}+14 y+1+2 y^{2}-14 y \\
& =(7 y+1)^{2}+2 y(y-7) .
\end{aligned}
$$

Thus, when $y-7=0$, i.e., $y=7$, the right-hand side of the above equation is a perfect square. At this point, $x=50$.
Therefore, $x=50, y=7$ is a set of positive integer solutions to the given equation.