Case 1: $n$ is odd Then $m=1^{2k+1}+(2^{2k+1}+n^{2k+1})+(3^{2k+1}+(n-1)^{2k+1})+... \equiv 1 \pmod {n+2}$ so $n+2 \not | m$ Case 2: $n=2t$ is even Then $m=m=1^{2k+1}+(2^{2k+1}+n^{2k+1})+(3^{2k+1}+(n-1)^{2k+1})+...\equiv 1+(\frac{n+2}{2})^{2k+1}=1+(t+1)^{2k+1} \pmod {2t+2}$ So $n+2 \not | m$