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}


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

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

  • 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.