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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

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3 Answers3

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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.

Donald Splutterwit
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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*}

Ángel Mario Gallegos
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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 $c=-a-b$.

Jean Marie
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