22. How many pairs of integers $(x, y)$ are there such that $x \geq y$ and $\frac{1}{x}+\frac{1}{y}=\frac{1}{211}$ ?
22. Answer: 3
Note that
$$
\frac{1}{x}+\frac{1}{y}=\frac{1}{211} \Rightarrow x y-211 x-211 y=0 \Rightarrow(x-211)(y-211)=211^{2} .
$$

Since 211 is a prime number, the factors of $211^{2}$ are $1,211,211^{2},-1,-211,-211^{2}$. Thus the pairs of integers $(x, y)$ satisfying the last equation are given by:
$$
\begin{aligned}
(x-211, y-211)= & \left(1,211^{2}\right),(211,211),\left(211^{2}, 1\right),\left(-1,-211^{2}\right), \\
& (-211,-211),\left(-211^{2},-1\right) .
\end{aligned}
$$

Equivalently, $(x, y)$ are given by
$$
\begin{array}{l}
\left(212,211+211^{2}\right),(422,422),\left(211+211^{2}, 212\right),\left(210,211-211^{2}\right), \\
(0,0),\left(211-211^{2}, 210\right) .
\end{array}
$$

Note that $(0,0)$ does not satisfy the first equation. Among the remaining 5 pairs which satisfy the first equation, three of them satisfy the inequality $x \geq y$, and they are given by $(x, y)=(422,422),\left(211+211^{2}, 212\right),\left(210,211-211^{2}\right)$.