Algebraic identity involving powers of twin primes Yesterday, I verified that, if $a$,$b$ and $c$ are real numbers such that $a+b+c=0$, then
$$\frac{a^5+b^5+c^5}{5}=\frac{a^3+b^3+c^3}{3}\cdot\frac{a^2+b^2+c^2}{2}$$
and
$$\frac{a^7+b^7+c^7}{7}=\frac{a^5+b^5+c^5}{5}\cdot\frac{a^2+b^2+c^2}{2}$$
Then I asked myself: will it be true that
$$\frac{a^{p_n}+b^{p_n}+c^{p_n}}{p_n}=\frac{a^{p_{n-1}}+b^{p_{n-1}}+c^{p_{n-1}}}{p_{n-1}}\cdot\frac{a^2+b^2+c^2}{2}$$
for any pair of primes satifying $p_n=p_{n-1}+2$, i.e., for any pair of twin primes?
For the first two equalities, I wrote $a=-(b+c)$, expanded the powers $-(b+c)^x$ for $x=2,3,5,7$, multiplied the factors at the right hand side and verified that the left and right hand sides were equal. For the general case, I'm trying to do the same, but I'm having some difficulties. For the left hand side we have
$$-\frac{\sum_{i=1}^{p_n-1}{{p_n}\choose{i}}b^ic^{p_n-i}}{p_n}$$
and for the right hand side we have
$$-\frac{\sum_{i=1}^{p_{n-1}-1}{{p_{n-1}}\choose{i}}[b^{i+2}c^{p_{n-1}-i}+b^{i+1}c^{p_{n-1}+1-i}+b^ic^{p_{n-1}+2-i}]}{p_{n-1}}$$
Note that we can cancel the minus sign. For the term $bc^{p_n-1}$ we have at the left hand side
$$\frac{{p_n}\choose{1}}{p_n}bc^{p_n-1}=\frac{p_n-1}{p_n-1}bc^{p_n-1}$$
and at the right hand side
$$\frac{{p_{n-1}}\choose{1}}{p_{n-1}}bc^{p_{n-1}+2-1}=\frac{p_{n-1}-1}{p_{n-1}-1}bc^{p_n-1}$$
Note that $p_n-1=p_{n-1}+1$. So this term, as well as $b^{p_n-1}c$, is OK. But I can't handle the other terms.
My questions are, is this identity known to be true? If so, how can I manipulate the right hand side?

$\newcommand{\weird}[4]{\dfrac{{#1}^{#4}+{#2}^{#4}+{#3}^{#4}}{#4}}$
I'd say your suggestion is false, because
$$\weird {(-y-z)}yz{13}-\weird {(-y-z)}yz{11}\cdot\weird {(-y-z)}yz2=\\=-y^3z^3(y+z)^3(y^2+yz+z^2)^2\ne0$$
It's always worth checking the first hard case with a calculator.
