Restricted equality involving prime numbers Given three real numbers such that $a + b + c = 0$, it can be proved that
\begin{align*}
\frac{a^{5} + b^{5} + c^{5}}{5} & = \frac{a^{3} + b^{3} + c^{3}}{3}\cdot \frac{a^{2} + b^{2} + c^{2}}{2}\\
\frac{a^{7} + b^{7} + c^{7}}{7} & = \frac{a^{5} + b^{5} + c^{5}}{5}\cdot \frac{a^{2} + b^{2} + c^{2}}{2}
\end{align*}
Thence I would like to ask: given three real numbers under the same restriction as above and prime numbers $p_{2} = p_{1} + 2$, for which of them does the following equation hold:
\begin{align*}
\frac{a^{p_{2}} + b^{p_{2}} + c^{p_{2}}}{p_{2}} & = \frac{a^{p_{1}} + b^{p_{1}} + c^{p_{1}}}{p_{1}}\cdot \frac{a^{2} + b^{2} + c^{2}}{2}
\end{align*}
Thank you in advance.

$p=3,5$ are the only possible solutions. 
To check this, substitute $a=2, \ b,c=-1$. 
Then 
$\displaystyle \frac{2^{p+2}-2}{p+2}=3\frac{2^p-2}{p}$. 
This equation can (after some effort) be rewritten as $2^p(p-6)=-4p-12$. 
The left hand side can only be negative if $p<6$.
