Saudi Arabia Mathematical Competitions 2012

Note that $$\begin{array}{rl}\frac{\gcd (acn+bc,acn+ad)}{ac}& \le \gcd (an+b,cn+d)\\ & \le \gcd (acn+bc,acn+ad).\end{array}$$ Suppose $ad=bc$. We have $$\begin{array}{rl}\gcd (an+b,cn+d)& \ge \frac{\gcd (acn+bc,acn+ad)}{ac}\\ & =\frac{\gcd (acn+bc,acn+bc)}{ac}\\ & =n+\frac{b}{a}.\end{array}$$ $n+\frac{b}{a}$ is unbounded, so $\gcd (an+b,cn+d)$ is also unbounded as desired.

On the other hand, suppose $ad\ne bc$. Then $|ad-bc|\ne 0$, so therefore $$\begin{array}{rl}\gcd (an+b,cn+d)& \le \gcd (acn+bc,acn+ad)\\ & =\gcd (acn+bc,ad-bc)\\ & \le |ad-bc|.\end{array}$$ Therefore $\gcd (an+b,cn+d)$ is bounded above by a fixed number that is independent of $n$.