13. Determine the number of pairs $(a, b)$ of integers with $1 \leq b<a \leq 200$ such that the sum $(a+b)+(a-b)+a b+a / b$ is a square of a number.
13. Ans: 112
Since the number $(a+b)+(a-b)+a b+a / b=(a / b)(b+1)^{2}$ is a perfect square with $b$ and $b+1$ relatively prime, the number $a / b$ must be a perfect square. Let $a / b=n^{2}$. As $a>b$, the number $n \geq 2$ so that $a=b n^{2} \leq 200$. From this, $a$ can be determined once $b$ and $n$ are chosen. Hence, it suffices to count the number of pairs of $(b, n)$ satisfying $b n^{2} \leq 200$ with $b \geq 1$ and $n \geq 2$. Hence the answer is $\left\lfloor 200 / 2^{2}\right\rfloor+\cdots+\left\lfloor 200 / 14^{2}\right\rfloor=112$.