Quasi-Fermat Equasion

[Enola Straight]

Is there a solution for X^e+Y^e=Z^e, where X,Y,and Z are whole numbers and e is the base of the natural log?

I know, due to Fermat, that no whole number solutions exist for whole number exponents greater than 2, but e isn't a whole number (2.71828182856...).

I also know that certain rational fraction and decimal exponents work, but what about
irrational/trancendental exponents?

[Grog]

i thought it was untrue for Rational exponents in general cant remember why tough. It is false in general for irrational numbers tough (and quite easy to prove). I would guess that it is hard to prove/unknown it for a specific number like e.

[Grog]

i thought it was untrue for Rational exponents in general cant remember why tough. It is false in general for irrational numbers tough (and quite easy to prove). I would guess that it is hard to prove/unknown it for a specific number like e.

quote sorry my proof idea does not prove that there are irrational solutions ignore me

[Enola Straight]

Parallel discussion on Straight Dope.
http://boards.straightdope.com/sdmb/sho ... st14915085