<s>[INST] Generalizing $ \frac{a^2+b^2+c^2}{2} \times \frac{a^3+b^3+c^3}{3} = \frac{a^5+b^5+c^5}{5}$ If $a+b+c=0$ as discussed in this, this, and this post, then,
$$ \frac{a^2+b^2+c^2}{2} \times \frac{a^3+b^3+c^3}{3} = \frac{a^5+b^5+c^5}{5}\tag1 $$
$$ \frac{a^3+b^3+c^3}{3} \times \frac{a^4+b^4+c^4}{2} = \frac{a^7+b^7+c^7}{7}\tag2 $$
$$ \frac{a^2+b^2+c^2}{2} \times \frac{a^5+b^5+c^5}{5} = \frac{a^7+b^7+c^7}{7}\tag3 $$
Using these basic identities, we can prove the nice squared identities here, 
$$ \frac{a^3+b^3+c^3}{3}\times \frac{a^7+b^7+c^7}{7} = \left(\frac{a^5+b^5+c^5}{5}\right)^2 $$
and here for $(a^7+b^7+c^7)^2$. Some investigation shows that,
$$\frac{(a^5+b^5+c^5)}{5\times18}\times\big(9(a^6+b^6+c^6) -(a^3+b^3+c^3)^2\big)=  \frac{a^{11}+b^{11}+c^{11}}{11}\tag4$$
$$\frac{(a^7+b^7+c^7)}{7\times18}\times\big(9(a^6+b^6+c^6) +(a^3+b^3+c^3)^2\big)=  \frac{a^{13}+b^{13}+c^{13}}{13}\tag5$$

Q: What would be the corresponding identities, as concise as possible, for $p=17$ and $p=19$?

 [/INST] Define $\;s_k := (a^k+b^k+c^k)/k.\;$ Then
 $\;s_{17}=2s_{13}s_4+3s_9s_5s_3,\; s_{19}=8s_8s_7s_4+3s_{11}s_5s_3.$
Not as nice as the ones for $\;s_{11}\;$ and $\;s_{13}\;$ but they are two terms with positive integer coefficients.
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