Integral Divisibility

Suppose wlog $a \le b$ . Then $a^2+b \ge b^2 - a$ means that $a(a+1) \ge (b-1)b$ , so clearly either $a = b$ or $a = b-1$ .

If $a = b$ we get $\frac{a^2+a}{a^2-a} = \frac{a+1}{a-1}$ is an integer. If $a \ge 4$ then the fraction is less than $2$ , but it's greater than 1, so cannot be an integer. We see that $a = b = 2,3$ are the solutions in this case.

If $a = b-1$ we get that the first fraction is $1$ and the second fraction is $\frac{a^2+3a+1}{a^2-a-1}$ . For this to be at least $2$ we must have $a^2-5a-3 \le 0$ , so $a \le 5$ . Taking $a = 1,2,3,4,5$ we see that $a=1,2$ are the only solutions.

So we've found six solutions in all: $(2,2), (3,3), (1,2), (2,3), (2,1), (3,2)$ . The answer is thus $\fbox{29}$ .