, 2016

Solution:
First consider the case $a, b > 0$. We have $720 = 2^4 \cdot 3^2 \cdot 5$, so the number of divisors of $720$ is $5 \times 3 \times 2 = 30$. We consider the number of ways to select an ordered pair $(a, b)$ such that $a, b, ab$ all divide $720$. Using the balls and urns method on each of the prime factors, we find the number of ways to distribute the factors of $2$ across $a$ and $b$ is $\binom{6}{2}$, the factors of $3$ is $\binom{4}{2}$, the factors of $5$ is $\binom{3}{2}$. So the total number of ways to select $(a, b)$ with $a, b, ab$ all dividing $720$ is $15 \times 6 \times 3 = 270$. The number of ways to select any $(a, b)$ with $a$ and $b$ dividing $720$ is $30 \times 30 = 900$, so there are $900 - 270 = 630$ ways to select $a$ and $b$ such that $a, b$ divide $720$ but $ab$ doesn't.

Now, each $a, b > 0$ corresponds to four solutions ($\pm a, \pm b$) giving the final answer of $2520$. (Note that $ab \neq 0$.)