Note that $a^{p+2}+(p-a)^{p+2} \equiv 2a^{p+1} \cdot p \pmod{p^2}$ (binomial expansion) and that $\sum a^2 \equiv 0 \pmod{p}$ (using $a^{p-1} \equiv 0 \pmod{p}$).

p is odd p+2 is also odd we can combine $k^{p+2}+(p-k)^{p+2}$ is divisible by p with binomial expansion then we are done