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.

Although I find solution given by GNU Supporter definitely more straightforward and elementary than mine, I will place what I found after asking the question, because it might give another perspective and some idea about what one can do when can't recognize correct formula.
Let $l_i$ denote elementary symmetric polynomial of degree $i$ and $s_i$ power symmetric polynomial of degree i. We need to show that $l_1=s_1=0$ implies that $32s_4$ is a perfect square. By Newton's Identity (but in case of three variables it can be easily checked by hand) $s_4=s_3l_1-s_2l_2+s_1l_3$ and $s_2=s_1^2-2l_2$, which by assumption takes form $s_4=-s_2l_2=-(s_1^2-2l_2)l_2=(-l_2)(-2l_2)=2l_2^2$ and from this $32s_4=64l_2^2=(8l_2)^2$.
