By Fermat's Last Theorem, we know $n<3$. Suppose $n \leq-3$. Then $a^{n}+b^{n}=c^{n} \Longrightarrow(b c)^{-n}+$ $(a c)^{-n}=(a b)^{-n}$, but since $-n \geq 3$, this is also impossible by Fermat's Last Theorem. As a result, $|n|<3$. Furthermore, $n \neq 0$, as $a^{0}+b^{0}=c^{0} \Longrightarrow 1+1=1$, which is false. We now just need to find constructions for $n=-2,-1,1,2$. When $n=1,(a, b, c)=(1,2,3)$ suffices, and when $n=2,(a, b, c)=$ $(3,4,5)$ works nicely. When $n=-1,(a, b, c)=(6,3,2)$ works, and when $n=-2,(a, b, c)=(20,15,12)$ is one example. Therefore, the working values are $n= \pm 1, \pm 2$.