A try for a simple proof referring to binomial equations, revealing a contradiction between an assumed minimum FLT disproof and any possible result.
Written by Juergen Buchmueller <pullmoll@t-online.de> in Nov 2005
Assume some integer (natural number) cube with a side length of (a + x), which resulted from adding the volumes of two other cubes, like say q3 + r3. In other words assume that Fermat's Last Theorem was wrong. The resulting cube with a total volume of
You have probably heard about Pascal's Triangle, and also the Binomial Theorem, which tells us that
Please don't regard the exact relation of a to x in this example. There is an infinite number of solutions for the binomial equation (in real numbers). In other words: any relation a : x is possible, and this special case (3:1) was chosen by me just to make drawing it a little easier.
Now what happens if we choose x such that it equals the initially specified 2nd cube's side length r? This choice is certainly possible, because r is always smaller than the resulting sum cube's side length (a + x). One solution for the binomial equation would have x3 equal to our initially given integer (natural number) r3.
And it would mean that there would have to be some remaining volume:
According to our initial assumption that FLT was wrong, now the volume q3 would have to be an integer cube and could always [*] also be represented by yet another binomial equation:
Does this leave room for any assumed minimum solution for a disproof of FLT? The q3 volume, if it really was an integer cube, would always contain some smaller cube y3 - for simplicity let's assume it was in the top, right corner just like x3 - plus the remaining volume b3 + 3by2 + 3b2y, which is not the volume of a cube. If b3 + 3by2 + 3b2y was again a cube, then the proposed solution could not have been the smallest possible solution for the disproof in the first place.
As a matter of fact the solution for any assumed minimum FLT disproof could in no case be the sum of just two cubes, but would always have to be the sum of three cubes plus some remaining non-cubic volume — or, looking at it from a little different angle, the sum of two cubes plus some remaining partial binomial formula.
Isn't this just plain contrary to the initial assumption that there is a minimum solution for a sum of just two cubes being equal to the volume of another cube? And if no minimum solution can actually exist, then how any solution could exist?
What else does the result tell but: there can be no minimum solution for a cube being the result of adding just two cubes?
I can not and I won't write down (or try to draw) these same thoughts for higher dimensions, while I am quite confident they are similiar for all higher dimensions, i.e. for the binomial formulas for every exponent greater than 3, too.
What do you think? Is this logical argument sufficient for a proof of FLT?
Let me know.
Ciao,
Juergen
Interesting links:
Peter Schorer's paper asking Is There a 'Simple' Proof of Fermat's Last Theorem? [1].
Eric W. Weisstein. "Pascal's Triangle." From MathWorld--A Wolfram Web Resource.
Eric W. Weisstein. "Binomial Theorem." From MathWorld--A Wolfram Web Resource.
[1] I was surprised and amused to read that in the first paper Schorer mentions Fermat was proud of his "method of infinite descent", which Fermat used in all of his proofs. And since he used it in all of his proofs, why then should his margin comment of a truely simple proof for his theorem have been going down a different road? Pierre de Fermat and Blaise Pascal, the inventor of the Pascal triangle, knew each other... I don't know if they probably talked about something related to binomial equations.
Anyway - even funnier is the fact that Schorer's paper is published by some company Occam press - at least to me this seems funny :-)
[*] Oh wait: the 13 cube as the remaining volume. Well, you obviously can not dissect the 13 cube into a binomial equation. But the fact that you can not 'attach' such a 13 cube to any other cube and get a new cube would be something I have to give proof for?
Here we go: You might have wondered, why my so called proof, as I tried to line it out above, then would not work for the a2 + b2=c2 case? The explanation is simple. There is a case for the partial binomial x2=a2 + 2 ⋅ a ⋅ b with the side a being equal to the side of the unit square 12, that is with a = 1.
The first existing example with a = 1 and b = 4 is this:
Therefore for the n=2 case of FLT there is indeed a minimum solution that is not necessarily itself again representable by some binomial formula like (b + y)2, because this minimal solution contains the unit square, while for every case with n>2 there can be no such solution — for geometrically obvious (to me) reasons.