2. Given positive integers $x, y$. Then $\frac{10}{x^{2}}-\frac{1}{y}=\frac{1}{5}$ has $(\quad)$ solutions $(x, y)$.
(A) 0
(B) 1
(C) 2
(D) More than 2, but finite
2. C.

From $\frac{10}{x^{2}}-\frac{1}{y}=\frac{1}{5} \Rightarrow 50 y=x^{2}(y+5)$
$\Rightarrow x^{2}=\frac{50 y}{y+5}=50-\frac{250}{y+5}$.
If $(y, 5)=1$, then
$(5, y+5)=1 \Rightarrow(y+5) \mid 2$.
Also, $y+5>2$, which leads to a contradiction. Therefore, 5 । $y$.
In equation (1), from $y+5>10$, we have $25 \leqslant x^{2}<50$.
Thus, $x$ can only be one of the numbers 5, 6, 7.
Upon verification, $(x, y)$ has only two positive integer solutions: $(5,5),(7,245)$.