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.
 A: 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. 
A: Start off with the fact that 
$$a^2+b^2+c^2 = (a+b+c)^2 - 2(ab+bc+ca) = - 2(ab+bc+ca)
$$
Since $a+b+c=0$ we know that ${a,b,c}$ are the three roots of some cubic missing the $x^2$ term:
$$
x^3+kx+m = 0
$$ with $ab+bc+ca = k$. And by the way, that says that 
$$a^2+b^2+c^2=-2k$$
Now start chaining upward, expressing $a^n+b^n+c^n$ in terms of $m$ and $k$.  For $n=3$ we add the cubic expression with $x=a$ to that with $x=b$ and $x=c$ to get
$$
a^3+ b^3 + c^3 +k(a+b+c)+(m+m+m) = 0 \\
a^3+ b^3 + c^3 +3m = 0 \\
a^3+ b^3 + c^3 = -3m 
$$
Now for $n=4$ we multiply each of the equations by $a$, $b$ and $c$ respectively before adding them:
$$
a^4+ b^4 + c^4 +k(a^2+b^2+c^2)+m(a+b+c) = 0 \\
a^4+ b^4 + c^4 -2k^2 = 0 \\
a^4+ b^4 + c^4 = 2k^2
$$
Now for $n=5$ we multiply each of the equations by $a^2$, $b^2$ and $c^2$ respectively before adding them:
$$
a^5+ b^5 + c^5 +k(a^3+b^3+c^3)+m(a^2+b^2+c^2) = 0 \\
a^5+ b^5 + c^5 -3mk  -2mk = 0 \\
a^5+ b^5 + c^5 = 5mk  
$$
For this problem we can afford to skip $n=6$.
$$
a^7+ b^7 + c^7 +k(a^5+b^5+c^5)+m(a^4+b^4+c^4) = 0 \\
a^7+ b^7 + c^7 +5mk^2 +2mk^2 = 0 \\
a^7+ b^7 + c^7 = -7mk^2   
$$
Then your identity reads $$
\left( -\frac{3m}{3} \right) \left( -\frac{7mk^2}{7} \right) = \left( -\frac{5mk}{5} \right)^2 \\
\left( -m \right) \left( -mk^2 \right) = (mk)^2 $$
A: $$a^2+b^2+c^2=(a+b+c)^2-2(ab+bc+ca)=-2(ab+bc+ca)$$
The key here is the identity (for all $n\in\mathbb Z_{\ge 3}$):
$$a^n+b^n+c^n=(a+b+c)\left(a^{n-1}+b^{n-1}+c^{n-1}\right)-$$
$$-(ab+bc+ca)\left(a^{n-2}+b^{n-2}+c^{n-2}\right)+abc\left(a^{n-3}+b^{n-3}+c^{n-3}\right)$$
Therefore: $$a^3+b^3+c^3=3abc\\ a^4+b^4+c^4=2(ab+bc+ca)^2\\ a^5+b^5+c^5=-5abc(ab+bc+ca)\\a^7+b^7+c^7=7abc(ab+bc+ca)^2$$
Therefore:
$$\frac{a^3+b^3+c^3}{3}\frac{a^7+b^7+c^7}{7} = \left(\frac{a^5+b^5+c^5}{5}\right)^2\\\frac{a^2+b^2+c^2}{2}\frac{a^3+b^3+c^3}{3}=\frac{a^5+b^5+c^5}{5}\\ \frac{a^2+b^2+c^2}{2}\frac{a^5+b^5+c^5}{5}=\frac{a^7+b^7+c^7}{7}\\$$
A: If we remember the well-known formulas
$(x+y)^3 - (x^3 + y^3) = 3xy(x+y)$
$(x+y)^5 - (x^5 + y^5) = 5xy(x+y)(x^2 + xy + y^2)$
$(x+y)^7 - (x^7 + y^7) = 7xy(x+y)(x^2 + xy + y^2)^2$
then your identity becomes the observation that there are equal powers of the $xy(x+y)$ and $(x^2+xy+y^2)$ factors on both sides.  (Under the change of notation $a,b,c \to -(x+y),x,y$)
A: 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.  
A: 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 $a=y-z,b=z-x,c=x-y$.
A: Let $a,b,c$ be roots of $$x^3+mx-n=0$$
From Vieta formula we get:
$$ 0 = (a+b+c)^2 = a^2+b^2+c^2+2m$$ 
so $$\boxed{a^2+b^2+c^2 =-2m}$$
Further $$x^3 = n-mx\implies a^3+b^3+c^3 = 3n-m(a+b+c) =3n $$
so $$\boxed{a^3+b^3+c^3 =3n}$$
Also $$x^5 = nx^2-mx^3 = nx^2-mn+m^2x$$ so  $$a^5+b^5+c^5 = n(a^2+b^2+c^2)-3mn = -5mn$$
so $$\boxed{a^5+b^5+c^5 =-5mn}$$
Finally $$x^7 = x(n-mx)^2 = n^2x-2mnx^2+m^2x^3 $$ $$=n^2x-2mnx^2+m^2n-m^3x$$
so $$\boxed{a^7+b^7+c^7 = 7m^2n}$$
and we are done.
