Given that $a+b+c=0$, show that $2(a^4+b^4+c^4)$ is a perfect square Given that $a+b+c=0$. Show that: $2(a^4+b^4+c^4)$ is a perfect square
MY ATTEMPTS:
I found that when $a+b+c=0$, $a^3+b^3+c^3=3abc$
So I did:
$(a^3+b^3+c^3)(a+b+c)$ -- $a^4+b^4+c^4=-(a^3c+a^3b+ab^3+b^3c+c^3a+c^3b)$
And then I tried to substitute $a^4+b^4+c^4$, but I found nothing that I thought relevant to the question
 A: Square $a+b+c=0$
\begin{eqnarray*}
a^2+b^2+c^2=-2(ab+bc+ca).
\end{eqnarray*}
Square this
\begin{eqnarray*}
a^4+b^4+c^4+2(a^2b^2+b^2c^2+c^2a^2)=4(a^2b^2+b^2c^2+c^2a^2)+8abc(a+b+c)
\end{eqnarray*}
The last term is zero ... rearrange 
\begin{eqnarray*}
2(a^4+b^4+c^4)=4(a^2b^2+b^2c^2+c^2a^2)=(a^2+b^2+c^2)^2.
\end{eqnarray*}
the last equality follows from the first equation.
A: From $a+b+c=0$ it follows $c=-a-b$, and then
\begin{align*}
c^4&=a^4+4a^3b+6a^2b^2+4ab^3+b^4\\
\implies \qquad 2(a^4+b^4+c^4)&=4a^4+8a^3b+12a^2b^2+8ab^3+4b^4\\
&=(2a^2+2ab+2b^2)^2
\end{align*}
A: I don't pretend that my solution is shorter, but for such questions, I try to avoid "redicovering all from scratch" ; I systematically apply the Newton-Girard identities one finds here, in particular for the present case :
$$p_4=\underbrace{e_1^4-4e_1 e_2+4e_1 e_3}_{= \ 0 \ \text{because} \ e_1= \ 0}+2e_2^2-\underbrace{4e_4}_{= \ 0}\tag{1}$$
with notations 


*

*$p_k$ for the sum of $k$-th powers and

*$e_k$ for the coefficient of $(-1)^k x^k$ in the expansion $(x-a)(x-b)(x-c)...$.
Consequence of (1): $2p_4=(2 e_2)^2$ is indeed a perfect square ; otherwise said :
$$2(a^4+b^4+c^4)=(2(ab+bc+ca))^2\tag{2}$$
Of course, the previous expression can be given different forms ; for example, variable $c$ can be eliminated by  using relationship $c=-a-b$.
