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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.

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  • $\begingroup$ Where did you get these two identities from? $\endgroup$ – user21820 Jun 16 '16 at 3:37
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    $\begingroup$ I got them from the book entitled "Solving Problems in Algebra and Trigonometry" by V. Litvinenko and A. Mordkovich, Mir Publishers Moscow. $\endgroup$ – APCorreia Jun 16 '16 at 3:39
  • $\begingroup$ They're surprising. Thanks for the reference! $\endgroup$ – user21820 Jun 16 '16 at 3:49
  • $\begingroup$ See math.stackexchange.com/questions/851985/… $\endgroup$ – lab bhattacharjee Jun 16 '16 at 4:56
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$p=3,5$ 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 $p<6$.

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