, 2022

Solution:
Let $\operatorname{gcd}(a, b) = d$, $a = d a'$, $b = d b'$. Then,
$$d^2\left(a' + b'^2\right) = 100^2.$$
Consider each divisor $d$ of $100$. Then, we need to find the number of solutions in coprime integers to $a' + b'^2 = \frac{100^2}{d^2}$. Note that every $b' < 100 / d$ 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 $\sum_{d \mid n} \varphi\left(\frac{100}{d}\right) - 1 = 100 - 1 = 99$.

Solution 2:
Note that $b$ must be at most $99$ in order for $a$ to be positive. Now we claim that each choice of $b \in \{1, 2, \ldots, 99\}$ corresponds to exactly one value of $a$ that satisfies the equation.
To see why this is true, we rewrite the formula as
$$\operatorname{gcd}(a, b) \cdot a = (100 - b)(100 + b).$$
For any prime $p$, let $v_p(n)$ be the largest integer $k$ such that $p^k \mid n$. We wish to show that once we fix $b$, the value $v_p(a)$ is uniquely determined for all $p$ (which will give us a unique solution for $a$). Applying $v_p$ to both sides of the equation gives us
$$\min \left(v_p(b), v_p(a)\right) + v_p(a) = v_p(100 - b) + v_p(100 + b)$$
We immediately see that there is at most one solution for $v_p(a)$ since the left-hand side increases with $v_p(a)$. Further, as we increment $v_p(a)$, the left-hand side takes on all even numbers up to $2 v_p(b)$, and then all integers larger than $2 v_p(b)$. So to show that a solution exists, we need to prove that the right-hand side is always either even or at least $2 v_p(b)$.
If $v_p(b) \neq v_p(100)$, then $v_p(100 - b) = v_p(100 + b) = \min \left(v_p(100), v_p(b)\right)$. In this case, $v_p(100 - b) + v_p(100 + b)$ is clearly even. Otherwise, assume $v_p(b) = v_p(100)$. Then $v_p(100 - b) \geq v_p(b)$ and $v_p(100 + b) \geq v_p(b)$, so $v_p(100 - b) + v_p(100 + b) \geq 2 v_p(b)$ as desired. Thus, there is always a unique solution for $v_p(a)$. Once we fix $1 \leq b \leq 99$, the value of $a$ is uniquely determined, so the answer is $99$.