# x^3 + y^3 + z^3 = 3xyz

A.N.W.Hone A.N.W.Hone at kent.ac.uk
Tue Apr 25 12:14:31 CEST 2006

```To prove that x^2+y^2+z^2-xy-yz-zx>=0 for all real values, note that
it is a quadratic form which can be written as

1/2 * X^T M X

for vector X=(x y z)^T and matrix

M= (2 -1 -1)
(-1 2 -1)
(-1 -1 2)

(which as it happens is the Cartan matrix of the affine root system A_2^(1)).

The matrix M is symmetric with eigenvalues 3,3,0, so it is positive
definite.

Andy

On Mon, 24 Apr 2006 franktaw at netscape.net wrote:

> Leaving alpha out of it, this factorization is
>
> x^3+y^3+z^3-3xyz = (x+y+z)*(x^2+y^2+z^2-xy-xz-yz).
>
> A proof escapes me at the moment, but I'm sure the second term is >= 0 for all real x,y,z, with zeros only for x=y=z.  This implies that the two classes of solutions described below are the only integer solutions.
>
>
>
> -----Original Message-----
> From: Edwin Clark eclark at math.usf.edu
>
>
> On Mon, 24 Apr 2006, Emeric Deutsch wrote:
>
> > Dear Seqfans,
> > Do you know any literature on the solutions of
> > x^3 + y^3 + z^3 = 3xyz in integers?
> > Any other solutions than x=y=z=1,2,3,... ?
>
> Since
>
> x^3+y^3+z^3-3*x*y*z = (x-z-z*alpha+y*alpha)*(x-y-y*alpha+z*alpha)*(z+x+y)
>
> where alpha is a root of x^2 + x + 1,
>
> any roots of x+y+z=0 will be a solution. ...
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```