Show that $a+b+c=0$ implies that $32(a^4+b^4+c^4)$ is a perfect square.

Question

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.

Answer 1

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}

Answer 2

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 $32s_4=64l_2^2=(8l_2)^2$.

Answer 3

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 $32(a^4+b^4+c^4)=\big(8 (ab+bc+ca)\big)^2\,$.