If $$abc=xyz=mnp$$ $a^4+b^4+c^4-2a^2b^2-2b^2c^2-2c^2a^2=x^4+y^4+z^4-2x^2y^2-2y^2z^2-2z^2x^2=a^4+b^4+c^4-2a^2b^2-2b^2c^2-2c^2a^2=m^4+n^4+p^4-2m^2n^2-2n^2p^2-2p^2m^2$ $$\frac{x}{m}=\frac{n}{b}=\frac{c}{z}$$ Then $a+b+c=x+y+z=m+n+p$? $a,b,c$ positive and are lenghts of triangles, $x,y,z$ same, $m,n,p$ same.

I edited it.


closed as off-topic by Crostul, Martin R, Aqua, José Carlos Santos, mrtaurho Mar 30 at 16:21

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  • $\begingroup$ What have you tried? $\endgroup$ – Jacob Jones Mar 28 at 19:11
  • $\begingroup$ This is false: Consider $a,b=-1,c,x,y,z=1$. $\endgroup$ – Don Thousand Mar 28 at 19:13
  • $\begingroup$ I edited the question a bit. Sorry for making this an unclear question. $\endgroup$ – furfur Mar 28 at 19:15
  • $\begingroup$ What are $m$ and $n$? $\endgroup$ – Aqua Mar 28 at 19:31
  • $\begingroup$ This is the full thing now. $\endgroup$ – furfur Mar 28 at 19:31

Hint: Note that $$ a^4+b^4+c^4-2a^2b^2-2b^2c^2-2c^2a^2=(a + b + c)(a + b - c)(a - b + c)(a - b - c). $$

  • $\begingroup$ I don't think you need to do that much work. $a,b=-1,c,x,y,z=1$ is a simple counterexample. $\endgroup$ – Don Thousand Mar 28 at 19:13
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    $\begingroup$ Wuau!!!! How can one note that? $\endgroup$ – Aqua Mar 28 at 19:14
  • $\begingroup$ I echo @MariaMazur's sentiment: That's quite stunning, and I'm quite sure I would not have come up with that unless I had a really long time $\endgroup$ – Siddharth Bhat Mar 28 at 19:17
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    $\begingroup$ @MariaMazur that is Heron's formula for the area of a triangle, at least it is the integer polynomial part. Something the kids would memorize $\endgroup$ – Will Jagy Mar 28 at 19:24
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    $\begingroup$ en.wikipedia.org/wiki/Heron%27s_formula#Formulation $\endgroup$ – Will Jagy Mar 28 at 19:59

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