A number theory problem by Lucas Nascimento

Since $a$ divides $b+1$ and $b$ divides $a+1$ , we must have $a \le b+1$ and $b \le a+1$ , so that $a-1 \le b \le a+1$ . There are three cases to consider:

• If $b = a-1$ we need $a-1$ to divide $a+1$ , and hence we need $a-1$ to divide $2$ . Thus $a-1$ is either $1$ or $2$ , giving the solutions $(2,1)$ and $(3,2)$ .

• If $b=a$ we need $a$ to divide $a+1$ , and hence we need $a$ to divide $1$ . Thus $a=1$ , giving the solution $(1,1)$ .

• If $b = a+1$ we need $a$ to divide $a+2$ , and hence we need $a$ to divide $2$ . Thus $a$ is either $1$ or $2$ , giving the solutions $(1,2)$ and $(2,3)$ .

Thus there are a total of $\boxed{5}$ solutions.