Standard Fractions!

Simplifying the expression, we have $B=\frac{A^2+1}{A-1}=\frac{\big((A-1)+1\big)^2+1}{A-1}=(A-1)+2+\frac{2}{A-1}$ Since $B$ is an integer, $A-1$ must divide $2$ . Also, since $A$ is a positive integer, there are only two possibilities $A-1=1$ or $A-1=2$ , i.e., $A=2$ or $A=3$ . Both cases yields $B=5$ . Hence only two ordered pairs of positive solutions are possible.

We can write the given equation as $A^2-BA+B+1=0$ , and we can solve for $A$ to find $A=\frac{B\pm{\sqrt{(B-2)^2-8}}}{2}$ . The only way that $(B-2)^2-8$ can be a perfect square is $(B-2)^2=9$ and $B=5$ . Then $A=\frac{5\pm{1}}{2}$ , giving us $\boxed{2}$ solutions.

$\begin{aligned} \frac{1}{A} + \frac{A}{B} + \frac{1}{AB} & = 1 \\ \Rightarrow \frac{B + A^{2} + 1}{AB} & = 1 \\ \Rightarrow B + A^{2} + 1 - AB & = 0 \\ \Rightarrow (B - AB) + (A^{2} - 1) + 2 & = 0 \\ \Rightarrow (A - B + 1)(A - 1) = -2 \end{aligned}$ Now, if $x, y \in \mathbb{Z}$ , $xy = -2$ , then $(x, y) = (\pm{1}, \pm{2}).$ \begin{equation*}A - B + 1 = 1, A - 1 = -2 \Rightarrow (A, B) = (-1, -1) \\ A - B + 1 = -2, A - 1 = 1 \Rightarrow (A, B) = (2, 5) \\ A - B + 1 = -1, A - 1 = 2 \Rightarrow (A, B) = (3, 5) \\ A - B + 1 = 2, A - 1 = -1 \Rightarrow (A, B) = (0, -1) \end{equation*} Only $(2, 5)$ and $(3, 5)$ are positive integers. Hence, $\boxed{2}$ is the answer.