16. Given positive integers $x, y$, then $\frac{10}{x^{2}}-\frac{1}{y}=\frac{1}{5}$ has ( ) solutions of the form $(x, 1)$.
(A) 0
(B) 1
(C) 2
(D) More than 2, but finite
(E) Infinite

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16. (C).
$$
\text { Given } \frac{10}{x^{2}}-\frac{1}{y}=\frac{1}{5} \Rightarrow 50 y=x^{2}(y+5) \Rightarrow x^{2}=\frac{50 y}{y+5} \text {. }
$$

If $(y, 5)=1$, then $(y, y+5)=1$, so $(y+5) \mid 50=5^{2} \times 2$. Since $y+5>2$, then $(y+5) \mid 5^{2}$, which contradicts $(y+5,5)=1$. Therefore, $5 \mid y$.
Let $y=5 y_{0}$, then $x^{2}=\frac{50 y_{0}}{y_{0}+1}=\frac{5^{2} \cdot 2 y_{0}}{y_{0}+1} \geqslant 5^{2}$.
In (I), from $y+5>y$ we have $x^{2}<50$.
Therefore, $x$ can only be 5, 6, or 7. Upon verification, $(x, y)$ has only two positive integer solutions: $(5,5),(7,245)$.