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.

It suffices to show that $2(a^4+b^4+c^4)$ is a perfect square.
\begin{align}
& 2(a^4+b^4+c^4) \\
=& 2((b+c)^4+b^4+c^4) \\
=& 2(2b^4+4b^3c+6b^2c^2+4bc^3+2c^4) \\
=& 4(b^4+\bbox[2px, border:1px solid]{2b^3c}+\bbox[2px, border:1px dashed]{3b^2c^2}+2bc^3+c^4) \\
=& 4((b^4+\bbox[2px, border:1px solid]{b^3c}+\bbox[2px, border:1px dashed]{b^2c^2})+(\bbox[2px, border:1px solid]{b^3c}+\bbox[2px, border:1px dashed]{b^2c^2}+bc^3)+(\bbox[2px, border:1px dashed]{b^2c^2}+bc^3+c^4)) \\
=& 2^2 (b^2(b^2+bc+c^2)+bc(b^2+bc+c^2)+c^2(b^2+bc+c^2))\\
=& 2^2(b^2+bc+c^2)^2
\end{align}
