Prove $\frac{a^3+b^3+c^3}{3}\frac{a^7+b^7+c^7}{7} = \left(\frac{a^5+b^5+c^5}{5}\right)^2$ if $a+b+c=0$ Found this lovely identity the other day, and thought it was fun enough to share as a problem:

If $a+b+c=0$ then show $$\frac{a^3+b^3+c^3}{3}\frac{a^7+b^7+c^7}{7} = \left(\frac{a^5+b^5+c^5}{5}\right)^2.$$

There are, of course, brute force techniques for showing this, but I'm hoping for something elegant.

Here's the symmetric polynomial approach.
$a^n+b^n+c^n$ is a symmetric homogeneous polynomial of degree $n$. So it can be expresses as a linear combination of polynomials $s_1^is_2^js_3^k$ where $i+2j+3k=n$, and $s_1=a+b+c,$ $s_2=ab+bc+ac,$ $s_3=abc$ are the elementary symmetric polynomials.
Now,  $s_1=a+b+c=0$ implies that $a^n+b^n+c^n$ can be written as a linear combination of $s_2^js_3^k$ with $2j+3k=n$.
It the case where $n=2,3,4,5,7$, there is only one pair $j,k$ such that $2j+3k=n$, so $a^n+b^n+c^n$ becomes a monomial in $s_2,s_3$.
When $n=2$, this means $a^2+b^2+c^2=k_2s_2$ for some constant $k_2$. Setting $(a,b,c)=(2,-1,-1)$, we see $k_2=-2$.
When $n=3$, this means $a^3+b^3+c^3=k_3s_3$ for some constant $k_3$. Setting $a=2,b=c=-1$, we get $k_3=3$.
Similarly $a^5+b^5+c^5=k_5s_2s_3$, and again we use $(a,b,c)=(2,-1,-1)$ to get that $k_5=-5$.
Likewise, $a^7+b^7+c^7=7s_2^2s_3$.
Similar formula under the same condition:
$$\frac{a^2+b^2+c^2}{2}\frac{a^3+b^3+c^3}{3} = \frac{a^5+b^5+c^5}{5}\\
\frac{a^2+b^2+c^2}{2}\frac{a^5+b^5+c^5}{5} = \frac{a^7+b^7+c^7}{7}
$$
More generally, if $a,b,c\in\mathbb Z$ with $a+b+c=0$ and $n$ is relatively prime to $6$ then 
$a^n+b^n+c^n$ is divisible by $s_2s_3=(ab+ac+bc)abc$.

There's actually an expression for $a^n+b^n+c^n$ with explicit coefficients:
$$a^n+b^n+c^n = \sum_{i+2j+3k=n} (-1)^j\frac{n}{i+j+k}\binom{i+j+k}{i,j,k} s_1^is_2^js_3^k\tag{1}$$
I confess I found that formula when looking at Fermat's Last many many years ago. It occurred to me at the time that Fermat can be expresses, for odd $n$, as 

Given odd positive integer $n>1$. Then for integers $a,b,c$, $a^n+b^n+c^n=0$ if and only if $a+b+c=0$ and $abc=0$.

which seemed to imply it is a question about symmetric polynomials. 
