Show that $a+b+c=0$ implies that $32(a^4+b^4+c^4)$ is a perfect square. There are given integers $a, b, c$ satysfaying $a+b+c=0$. Show that $32(a^4+b^4+c^4)$ is a perfect square.
EDIT: I found solution by symmetric polynomials, which is posted below.


EDIT: I found solution by symmetric polynomials (in variables a, b, c)

The following more or less transcribes OP's solution in direct calculations, without explicitly using Newton's relations. From the assumption that $a+b+c=0\,$:
$$
0 = (a+b+c)^2 = a^2+b^2+c^2 + 2(ab+bc+ca)
$$
$$
 \implies 2(ab+bc+ca)=-(a^2+b^2+c^2) \tag{1}
$$
$$
\require{cancel}
(ab+bc+ca)^2 = a^2b^2+b^2c^2+c^2a^2 + \cancel{2abc(a+b+c) } \tag{2}
$$
$$
\begin{align}
a^4+b^4+c^4 & = (a^2+b^2+c^2)^2-2(a^2b^2+b^2c^2+c^2a^2) \\[5px]
 & \overset{(1),(2)}{=} 4(ab+bc+ca)^2 - 2(ab+bc+ca)^2 \\
 & = 2 (ab+bc+ca)^2
\end{align}
$$
The latter gives $32(a^4+b^4+c^4)=\big(8 (ab+bc+ca)\big)^2\,$.
