From: "Iain Davidson" Subject: Re: A Fermat Proof and Number Theory Date: Sat, 11 Sep 1999 03:15:55 +0100 Newsgroups: sci.math Keywords: x^2 + 2 = y^3 MAppell917 wrote in message news:19990910125950.01182.00000454@ng-bg1.aol.com... > Fermat had a proof of the following number theory problem: > > Prove that 26 is the ONLY number sandwiched between a square and a cube. That > is to say, 25 is 5^2 and 27 is 3^3. Fermat proved that 26 was the ONLY number > in all of the integers that were sandwiched like this. I'm wondering if anyone > knows how to prove this or if there is a proof anywhere on the internet. I'd > like to see it. In Dickson's "Theory of numbers" it says that "Fermat stated that he could give a rigorous proof that 25 is the only integral square which is less than a cube by 2" . So Fermat did not actually publish a proof. x^2 +2 = y^3 x,y odd s^2 + 2t^2 = (s + t*sqrt(-2))(s - t*sqrt(-2)) Euler argued that s^2 + 2t^2 could only be a cube if both (s + t*sqrt(-2)) and (s - sqrt(-2)) were cubes. (s + t*sqrt(-2) = (a + b*sqrt(-2))^3 = a^3 + 3a^2*b*sqrt(-2) + -6a*b^2 -2b^3*sqrt(-2) = (a^3 - 6a*b^2) +(3a^2*b -2b^3)sqrt(-2) s = x, t = 1 b(3a^2 -2b^2) = 1 b = 1, (3a^2 -2b^2) = 1 b = -1, (3a^2 -2b^2) = -1 etc. Euler's assumption is not generally valid however. It works in this case because Z(sqrt(-2)) has unique factorisation, 1, -1 are the only units and the factors are relatively prime. x^2 + 2 = (x + sqrt(-2))(x - sqrt(-2)) = y^3 Assume there is prime p in Z[sqrt(-2)] that divides both (x + sqrt(-2)) and (x - sqrt(-2)). p then divides 2sqrt(-2) = (sqrt(-2))^3 sqrt(-2) is prime, so only sqrt(-2) and its associate -sqrt(-2) can be a common divisor. So sqrt(-2) divides (x - sqrt(-2)) So sqrt(-2) divides x, but x is an integer and so must be even. x and y cannot both be even so (x + sqrt(-2)) and (x - sqrt(-2)) are relatively prime. As factorisation in Z[sqrt(-2)] is unique then (x + sqrt(-2)) and (x - sqrt(-2)) can only have the form uv^3, where u is a unit. The only units are 1, -1. So Euler's assumption is correct in this case