HMMT November

Let $\operatorname{gcd}(a, b)=d, a=d a^{\prime}, b=d b^{\prime}$. Then, $d^{2}\left(a^{\prime}+b^{\prime 2}\right)=100^{2}$. Consider each divisor $d$ of 100. Then, we need to find the number of solutions in coprime integers to $a^{\prime}+b^{\prime 2}=\frac{100^{2}}{d^{2}}$. Note that every $b^{\prime}<100 / d$ coprime to $\frac{100^{2}}{d^{2}}$ satisfies this equation, which is equivalent to being coprime to $\frac{100}{d}$, so then there are $\varphi\left(\frac{100}{d}\right)$ choices for each $d$, except for $d=100$, which would count the solution $(0,100)$. Then we just need $\sum_{d \mid n} \varphi\left(\frac{100}{d}\right)-1=100-1=99$.