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

[Alec Mihailovs]

----- Original Message -----
From: franktaw at netscape.net
Sent: Monday, April 24, 2006 11:16 AM

> 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.
>
> Franklin T. Adams-Watters

The proof is simple - it equals 1/2*((x-y)^2+(x-z)^2+(y-z)^2).

Alec Mihailovs


> -----Original Message-----
> From: Edwin Clark eclark at math.usf.edu
>
> 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. ...