* With n = 4k+1 or n = 4k+3 \(\left(k\in N\right)\)

\(\Rightarrow1^n+2^n+3^n+4^n\equiv1^n+2^n+\left(-2\right)^n+\left(-1\right)^n\equiv0\left(mod5\right)\)

So \(1^n+2^n+3^n+4^n⋮5\)

* With n = 4k+2

\(\Rightarrow1^n+2^n+3^n+4^n\equiv1+2^{4k+2}+3^{4k+2}+4^{4k+2}\equiv1+4\cdot16^k+9\cdot81^k+16\cdot256^k\equiv1+4+9+16\equiv30\equiv0\left(mod5\right)\)

So \(1^n+2^n+3^n+4^n⋮5\)

* With n = 4k

\(\Rightarrow1^n+2^n+3^n+4^n\equiv1+2^{4k}+3^{4k}+4^{4k}\equiv1+16^k+81^k+256^k\equiv1+1+1+1\equiv4\left(mod5\right)\)

So \(1^n+2^n+3^n+4^n⋮̸5\)

==> With \(n\ne4k\Leftrightarrow1^n+2^n+3^n+4^n⋮5\)