smc

$(n+i{)}^4=n^4+4i{n}^3-6n^2-4in+1$, and this has to be an integer, so the sum of the imaginary parts must be $0$. $$4i{n}^3-4in=0$$ $$4i{n}^3=4in$$ $$n^3=n$$ Since $n^3=n$, there are $\boxed{3}$ solutions for $n$: $0$ and $\pm 1$.