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Let be $\quad x+y+z=0$

show this:

$$\frac{x^2+y^2+z^2}{2}\times\frac{x^5+y^5+z^5}{5}=\frac{x^7+y^7+z^7}{7}$$

I solved ,but Im interesting what are you thinking about this,how can we arrive to solution quickly?

Tito Piezas III
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Micheal Brain Hurts
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  • I'm almost sure this is a duplicate – mercio Jun 04 '16 at 20:27
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    This one is not exactly the same, but similar: http://math.stackexchange.com/questions/851985/big-dfracx7y7z77-big2-big-dfracx5y5z55-big2-cdot-big – Hans Lundmark Jun 04 '16 at 20:31
  • @mercio and Hans Lundmark thank you,

    I am like that found.

    $3(x^{2n+1}+y^{2n+1}+z^{2n+1})=(-1)^{n+1}.(2n+1)(x.y.z)[x^{n-1}.y^{n-1}+x^{n-1}.z^{n-1}+y^{n-1}.z^{n-1}]$

     – Micheal Brain Hurts Jun 04 '16 at 20:45
  • "Show this" or "solve this"? those are different questions. I'm not sure that equation is always true so I think you mean "solve this". – fleablood Jun 04 '16 at 22:21
  • for $x+y+z=0$ , $x+y=-z$, $x+z=-y$, $y+z=-x$,

    $(x+y)^3=-z^3=x^3+y^3+3xy(x+y)$ , $x^3+y^3+z^3=3xyz$ , and $x^5+y^5+z^5=-5x^2y^2z$ , $\vdots$ $\vdots$

    $n\in\mathbb{N^{+}}$ ....................................................... for $x+y=-z$ , $x^{2n+1}+y^{2n+1}+z^{2n+1}=(-1)^{n+1}.(2n+1)(x^n.y^n.z)$ ........................................... for $x+z=-y$ , $x^{2n+1}n+y^{2n+1}+z^{2n+1}=(-1)^{n+1}.(2n+1)(x^n.y.z^n)$

    ............................................ for $y+z=-x$ , $x^{2n+1}+y^{2n+1}+z^{2n+1}=(-1)^{n+1}.(2n+1)(x.y^n.z^n)$

     – Micheal Brain Hurts Jun 05 '16 at 14:39
  • and let's make a together form;

    $\forall x,y,z$ : $x+y+z=0$

    $3(x^{2n+1}+y^{2n+1}+z^{2n+1})=(-1)^{n+1}.(2n+1)[x^n.y^n.z+x^n.y.z^n+x.y^n.z^n]$

    and

    $3(x^{2n+1}+y^{2n+1}+z^{2n+1})=(-1)^{n+1}.(2n+1)(x.y.z)[x^{n-1}.y^{n-1}+x^{n-1}.z^{n-1}+y^{n-1}.z^{n-1}]$ $\quad\Box$

     – Micheal Brain Hurts Jun 05 '16 at 14:39

1 Answers1

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Your polynomials are invariant by the action of $S_3$ on $(x,y,z)$.

A standard result says that the graded ring $\Bbb Q[x,y,z]^{S_3}$ is $\Bbb Q[x+y+z;xy+yz+xz;xyz]$.

Furthermore here, quotienting by $(x+y+z)$ we can work in the ring $(\Bbb Q[x,y,z]/(x+y+z))^{S_3} \cong \Bbb Q[xy+yz+xz;xyz]$.

Now in this graded ring, the pieces of degree $2,5,7$ are all of dimension $1$ (there is only one decomposition of $2,5,7$ as a sum of $2$s and $3$s).

Picking $x,y,z = 1,1,-2$ shows that $x^k+y^k+z^k$ are nonzero in that ring for $k > 1$, which proves that $(x^2+y^2+z^2)(x^5+y^5+z^5) = q (x^7+y^7+z^7) \pmod {x+y+z}$ for a rational number $q$ (and also that $(x^2+y^2+z^2)(x^3+y^3+z^3) = q' (x^5+y^5+z^5)$ but we don't particularly care).

Then you can easily check that when replacing $z$ with $-x-y$ in your expressions, the dominant coefficient in your polynomials is $1$ after doing the divisions (note that for an odd prime $p$, by Fermat's little theorem, $(x^p+y^p+(-x-y)^p)/p$ even has integral coefficients, ... and the case $p=2$ just works), and you only have to check $1 \times 1 = 1$ to finish the proof.

You can also evaluate $(x^2+y^2+z^2)(x^5+y^5+z^5) = q (x^7+y^7+z^7)$ at $x,y,z = 1,1,-2$ to get $6\times (-30) = q \times (-126)$ hence $q=10/7$.

mercio
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