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}$

Question

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 $p=19$?

Answer 1

When denote $$ s_k = \dfrac{a^k+b^k+c^k}{k}, $$

then some expressions for $s_{17}$ and $s_{19}$:

$$ s_5 \left(6s_{12}-7s_5s_7-\frac{1}{2}{s_3^4}\right) = s_{17}, $$

$$ s_7 \left(6s_{12}-5s_5s_7+\frac{3}{2}{s_3^4}\right) = s_{19}. $$

Or $$ s_5 \left(s_{12}+3s_6^2+4s_4s_8\right) = s_{17}, $$

$$ s_7 \left(3s_{12}+9s_6^2-4s_4s_8\right) = s_{19}. $$

Answer 2

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.