10. Solve the equation $\left(x_{5}\right)^{2}+\left(y^{4}\right)_{5}=2 x y^{2}+51$, where $n_{5}$ denotes the closest multiple of 5 to the integer $n$, and $x, y$ are integers.
(51st Czech and Slovak Mathematical Olympiad (Open Question))
Solution: Since $\left(x_{5}\right)^{2}+\left(y^{4}\right)_{5}$ can be divisible by 5, and $2 x y^{2}$ has a remainder of 4 when divided by 5, i.e., $51\left(2 x y^{2}-4\right)$, then $y=5 k \pm 1$ or $5 k \pm 2$, at this time $y_{5}=5 k$.

If $y=5 k \pm 1$, by $51\left(y^{2}-1\right)$, we get $51(2 x-4)$, i.e., $51(x-2)$. Let $x=5 n+2$, we get $x_{5}=5 n$. Since $51\left(y^{2}-1\right)$, so, $51\left(y^{4}-1\right),\left(y^{4}\right)_{5}=y^{4}-1$. Thus, the original equation becomes
$$
\begin{array}{l}
(5 n)^{2}+\left(y^{4}-1\right)=2(5 n+2) y^{2}+51, \\
\left(y^{2}-5 n\right)^{2}-4 y^{2}=52, \\
\left(y^{2}-5 n-2 y\right)\left(y^{2}-5 n+2 y\right)=52 .
\end{array}
$$

Since the difference between the two expressions on the left side is $4 y$, 52 can only be factored as 2 and 26 or -2 and -26. Hence $y= \pm 6$. If $6^{2}-5 n-12=2$, there is no solution; if $6^{2}-5 n-12=-26$, then $n=10$. At this time, $x=52$, $y= \pm 6$.

If $y=5 k \pm 2$, by $51\left(y^{2}+1\right)$, we get $51(-2 x-4)$, i.e., $51(x+2)$. Let $x=5 n-2$, we get $x_{5}=5 n$. Since $51\left(y^{2}+1\right)$, so, $51\left(y^{4}-1\right),\left(y^{4}\right)_{5}=y^{4}-1$. Thus, the original equation becomes
$(5 n)^{2}+\left(y^{4}-1\right)=2(5 n-2) y^{2}+51$,
$\left(y^{2}-5 n\right)^{2}+4 y^{2}=52$.
Since $4 y^{2} \leqslant 52$, and $y=5 k \pm 2$, then $y= \pm 2$ or $\pm 3$.
If $y= \pm 2$, then $n=2, x=8$;
If $y= \pm 3$, then $n=1, x=3$.
In summary, there are 6 solutions: $(52,6)$, $(52,-6)$, $(8,2)$, $(8,-2)$, $(3,3)$, $(3,-3)$.