12.114 Find all prime numbers $x$ and $y$ that satisfy the equation $x^{2}-2 y^{2}=1$.
(Kyiv Mathematical Olympiad, 1966)
［Solution］ From $x^{2}=2 y^{2}+1$, we know that $x^{2}$ is an odd number, hence $x$ is odd.
Let $x=2 n+1,(n \in N)$. Then the original equation becomes
$$
\begin{array}{c}
(2 n+1)^{2}=2 y^{2}+1, \\
y^{2}=2\left(n^{2}+n\right) .
\end{array}
$$

Thus, $y^{2}$ is even, so $y$ is even.
Since $y$ is a prime number, then $y=2$.
Substituting $y=2$ into the original equation gives $x=3$.
Therefore, the original equation has only one prime solution $x=3, y=2$.