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.

  • $\begingroup$ Guessing the form this square (I have no idea how it can look) and expressing $(a+b+c)^4$ in elementary symmetric polynomials (this form makes cancelation easier). I posted it also because I would prefer to see some more systematic approach (even if I will find out it by myself, it will just a guess). I know that from that condition I can obtain many identities easily, but I don't know which I should use here. Perhaps at first it will be better to just see some hint. $\endgroup$ – Shingle Feb 25 '17 at 23:15
  • $\begingroup$ +1 for your solution, you should consider posting it as an answer. $\endgroup$ – dxiv Feb 26 '17 at 1:08
  • $\begingroup$ Thank you, I will do as you suggest. $\endgroup$ – Shingle Feb 26 '17 at 1:21

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}

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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 $s_2=s_1^2-2l_1$, which by assumption takes form $s_4=-s_2l_2=-(s_1^2-2l_1)l_2=(-l_2)(-2l_2)=2l_2^2$ and from this $32s_4=64l_2^2=(8l_2)^2$.

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  • $\begingroup$ +1 for exploiting the symmetry to the end. in case of three variables it can be easily checked by hand I took the liberty to post a variation on that idea as a separate answer, since it may be instructive on its own for those less familiar with Newton's identities. $\endgroup$ – dxiv Feb 26 '17 at 1:51

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

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  • 1
    $\begingroup$ +1, this shows directly how symmetric polynomial identities (which to me seem a little bit abstract) give "exact" formula results (I didn't expected that this translation can be done in so nice way) $\endgroup$ – Shingle Feb 26 '17 at 2:11

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