x^3 + y^3 + z^3 = 3xyz
A.N.W.Hone
A.N.W.Hone at kent.ac.uk
Tue Apr 25 12:15:50 CEST 2006
That's a neater proof than the diagonalization I just did!
Andy
On Mon, 24 Apr 2006, Alec Mihailovs wrote:
> ----- 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. ...
>
>
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