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<DIV>Leaving alpha out of it, this factorization is</DIV>
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<DIV>x^3+y^3+z^3-3xyz = (x+y+z)*(x^2+y^2+z^2-xy-xz-yz).</DIV>
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<DIV>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. </DIV>
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<DIV>Franklin T. Adams-Watters<BR> </DIV> <BR>-----Original Message-----<BR>From: Edwin Clark <A href="mailto:eclark@math.usf.edu">eclark@math.usf.edu</A><BR><BR>
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<DIV class=AOLPlainTextBody id=AOLMsgPart_0_e9d01dbc-e091-49c5-9dff-68798130e756><PRE><TT>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. ...</TT></PRE></DIV></DIV></DIV>
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