jmc

By the Pythagorean theorem, the distance from $(2,n)$ to $(-1,1)$ is $\sqrt{(2-(-1){)}^2+(n-1{)}^2}$. Setting this equal to $5$, we find $$\begin{array}{rl}9+(n-1{)}^2& =25⟹\\ (n-1{)}^2& =16⟹\\ n-1=4& \text{or}n-1=-4⟹\\ n=5& \text{or}n=-3.\end{array}$$ Both of these solutions are integers, and their product is $\boxed{-15}$.