[Number Theory]
Problem: Compute the number of ordered pairs $(x, y)$ of positive integers satisfying $x^{2}-8 x+y^{2}+4 y=5$.

Solution: Completing the square twice in $x$ and $y$, we obtain the equivalent equation $(x-4)^{2}+(y+2)^{2}=$ 25 , which describes a circle centered at $(4,-2)$ with radius 5 . The lattice points on this circle are points 5 units up, down, left, or right of the center, or points 3 units away on one axis and 4 units away on the other. Because the center is below the $x$-axis, we know that $y$ must increase by at least 2 units; $x$ cannot decrease by 4 or more units if it is to remain positive. Thus, we have:

$$
\begin{aligned}
& (x, y)=(4,-2)+(-3,4)=(1,2) \\
& (x, y)=(4,-2)+(0,5)=(4,3) \\
& (x, y)=(4,-2)+(3,4)=(7,2) \\
& (x, y)=(4,-2)+(4,3)=(8,1) .
\end{aligned}
$$

There are $\mathbf{4}$ such ordered pairs.

Final answer: 4