28. How many positive-integer pairs $(x, y)$ are solutions to the equation $\frac{x y}{x+y}=1000$.
(ans. 49
$\left(2 a_{1}+1\right)\left(2 a_{2}+1\right) \cdots\left(2 a_{k}+1\right)$ where $1000=p_{1}^{a_{1}} p_{2}^{a_{2}} \cdots p_{k}^{a_{k}^{k}}=2^{3} 5^{3} ;$ so 49. Let $\frac{x y}{x+y}=n \Rightarrow x y-n x-n y=0 \Rightarrow(x-n)(y-n)=n^{2} \Rightarrow x>$ $n, y>n$. In the factorization $n^{2}=p_{1}^{2 a_{1}} p_{2}^{2 a_{2}} \cdots p_{k}^{2 a^{k}}$ each divisor of $n^{2}$ determines a solution, hence the answer.)