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 $a+b+c=0$

Let $T_{m}$ be $a^m+b^m+c^m$.

Let $k=-ab-bc-ca$, and $l=abc$.

Note that this implies $a,b,c$ are solutions to $x^3=kx+l$.

Using Newton's Identity, note the fact that $T_{m+3}=kT_{m+1}+lT_{m}$(which can be proved by summing $x^3+kx+l$)

It is not to difficult to see that $T_{2}=2k$, $T_3=3l$, from $a+b+c=0$.

From here, note that $T_{4}=2k^2$ using the identity above.

In the same method, note that $T_{5}=5kl$.

From here, note $T_{7}=5k^2l+2k^2l=7k^2l$ from $T_{m+3}=kT_{m+1}+lT_{m}$ . Therefore, the equation simplifies to showing that $k^2l \times l=(kl)^2$, which is true.

Useful identities:

$(y-z)^{3}+(z-x)^{3}+(x-y)^{3}= 3(y-z)(z-x)(x-y)$

$(y-z)^{5}+(z-x)^{5}+(x-y)^{5}= 5(y-z)(z-x)(x-y)(x^{2}+y^{2}+z^{2}-yz-zx-xy)$

$(y-z)^{7}+(z-x)^{7}+(x-y)^{7}= 7(y-z)(z-x)(x-y)(x^{2}+y^{2}+z^{2}-yz-zx-xy)^{2}$

By letting $a=y-z,b=z-x,c=x-y$.

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.