Problem: For certain ordered pairs $(a, b)$ of real numbers, the system of equations $ \begin{aligned} a x+b y & =1 \\ x^2+y^2 & =50 \end{aligned} $ has at least one solution, and each solution is an ordered pair $(x, y)$ of integers. How many such ordered pairs $(a, b)$ are there? Solution: Note first that the possible integer solutions $(x,y)$ to $x^{2}+y^{2} = 50$ are $(\pm 1,\pm 7), (\pm 7, \pm 1), (\pm 5, \pm 5)$. That is, there are $12$ solutions. Now, a line can either pass through no point of the circle $x^{2}+y^{2}=50$, one point (the tangent line), or two points. Critically, for any two points whose corresponding line (passing through both points) doesn't cross the origin, we have that the line can be written $y=mx+c$ for some non-zero $c$ and hence there exist $a,b$ for which $ax+by = 1$. On the other hand, if $c=0$, no value of $a,b$ can represent the line as $ax+by=1$. As there are $12$ integral solutions to $x^{2}+y^{2}=50$, there are exactly $12$ tangent lines to the circle and none of them pass through the origin. Further, there are $\binom{12}{2} = 66$ lines between points listed above, but exactly $6$ of them pass through the origin (for each $(\pm a,\pm b)$ the line that passes through $(a,b)$ and $(-a,-b)$, along with the line that passes through $(-a,b)$ and $(a,-b)$ both pass through the origin, and as there are three of these $\pm$ points, there are $6$ origin-passing lines). Hence, the final answer is $66+12-6 = 72.$