From: "Iain Davidson" <Sttscitrans@tesco.net>
Subject: Re: Diophantine Equation solved by Fermat (not FLT)
Date: Tue, 25 Jul 2000 02:18:06 +0100
Newsgroups: sci.math
Summary: [missing]

James Wanless <james@jwanless.freeserve.co.uk> wrote in message
news:7EYe5.16696$cD.1206232@news3.cableinet.net...
> I've just read, on Encyclopedia Britannica.com, that Fermat showed that:
> x^3 - y^2 = 2 has only the two (equivalent) solutions.
> Anyone know anything more about this...?

If you are talking about integer solutions, Fermat only stated that 3 and 2
were the
solutions.

Euler attempted a proof by factoring
x^3 = y^2 +2
as
x^3 = [y+ sqr(-2)][y- sqrt(-2)]

and claimed that [y+ sqr(-2)][y- sqrt(-2)]
could only be a cube if  [y+ sqr(-2)] and
[y - sqr(-2)] were.

So [y+ sqr(-2)] = (a + b*sqrt(-2))^3
This gives conditions on the integers a and b
and the result "follows".

However, the same method does not work
with x^3 = 2y^2 -5   as it shows the
equation has no solutions, yet y = 4 is obviously
a solution.

Unique factorisation is the key, and a complete proof
would have to be based on the fact that numbers
like (a + b*sqrt(-2)) have unique factorisation.
As division with remainder can be defined for these integers,
this can be done using Euclid's Algorithm.

Bachet showed before Fermat that there were other rational
solutions.