Fermat's Last Theorem

Well, he didn't write it down, and look where that's left us! Perhaps he was just imagining things; even disciplined thinkers succumb to the euphoria of a false AHA moment. Or maybe the answer is sitting in front of us and we just can't see it ...

In modern terms, prove

an + bn = cn
has no solution for integers a,b,c and n, where n > 2.


Our current state of affairs has culminated in an apparent solution by a post-doc named Andrew Wiles - after seven years of mulling in his house - a solution only grasped by handfuls of mathematicians on the planet. For the amateur, this still leaves hope that a simpler proof may exist, one that Fermat could have imagined in his day. What idea did Fermat's 17th century mind grasp? Was it perhaps stimulated by the text he scribbled in, a translation of Diophantus' Arithmetica that had begun life in Alexandria, Egypt, a millenia before? (Here's an interesting english review of that material).

N-Dimensional Space

My first attempt to solve this follows.

I do problem solving in a fairly visual way, looking at charts or framing a problem as a composition of lines, power curves, squares and building blocks. The FLT problem's n exponent suggests the progression of dimensions of space, namely, 0 dimensions: a point; 1 dimension: a line; 2 dimensions: a surface; 3 dimensions: volume; 4 dimensions: well, some kind of space that a 4D hypercube is at home in. Etc. Like most, I have trouble getting past 3d + time.

Can this problem be represented in such a way that each successive dimension places more constraints on the possible {a,b,c} solutions? Can it be shown that the a3 + b3 = c3 problem has an a2 + b2 = c2 part that must also be satisfied? And by extension, does the N'th dimension problem literally have a cross-section which is exactly the N-1 dimension problem, and which supplies the domain of solutions from which the N'th dimension problem must choose? I'll try this rough direction, starting from the first dimension. (I skip dimension 0, since the addition and power operations don't quite apply to it.)

For n=1, we have D1: a + b = c, which is solved by, well, any three integers that actually do add up!

a2 + b2 = c2

For n=2, we have the famous pythagorian theorem, D2: a2 + b2 = c2, for which pythagorean triples (e.g. (3,4,5), (5, 12, 13) etc.) are solutions.

The diagram below shows a2 + b2 = c2, reinterpreted slightly as square a and square b inside square c, with the overlap of square a and b called square e. This overlap area e must also be equal to the area of c minus the area made by union of squares a and b, namely,

i) e2 = 2d(a-e)

A basic check on this is with {a=3,b=4,c=5}; then d = (c - a) = 2, so e = 2, and e2 = 4 = 2*2(3-2).

Are there constraints on a, d, and e?

If we join a2 + b2 = c2 with b=(e+d) and c2 = (a+d)2, we get:

a2 + (e+d)2 = (a+d)2

a2 + e2 +2ed + d2 = a2 + 2ad + d2

and from i) above, substituting 2d(a-e) for e2:

2ad - 2ed + 2ed = 2ad

1=1

So there appear to be no constraints on a, d and e. Next, when we try this approach in 3 dimensions, it appears we do run into a significant constraint.

a3 + b3 = c3

For n=3 we have cube a and cube b inside cube c, with the overlap of cube a and cube b equalling cube e, and this cube also equalling the volume of c minus the union of cubes a and b. So the green volume must equal the brown volume in the following diagram if a3 + b3 = c3 holds:

The area outside of cubes b and a consists of various smaller cubic volumes:

e3 = 3d(a+e)(a-e) + 3d2(a-e)

        = 3d(a-e)(a+e+d)

OR   = 3d(a2 - e2) + 3d2(a-e)

Here is a 3d perspective flash video animation of the volume relationships (also available in .avi format).

Revolving Cube (Firfox Version)

This freeCAD 3D animation was captured using Microsoft Expression. freeCAD is pretty easy to use, but it has a few glitches in rendering semi- transparent surfaces.

3D Algebra!

I read that the n=3 problem had been proven unsatisfiable over 200 years ago (1753, Leonhard Euler), but I felt the strange compulsion to take a crack at it myself. After a number of false starts I've arrived at this. Can you spot a mistake?

Can we satisfy both 2D and 3D problems?

If we force the cross-section of this c3 cube to be the 2D problem above (a2 + b2 = c2) then (a-e) can be replaced by e2/(2d) from step (i) above:

e3 = 3d(a-e)(a + e + d)

From (i), (a-e) = e2/(2d)

e3 = 3de2/(2d)(a + e + d)

e = 3(a + e + d)/2

2e = 3a + e + d

e = 3a + d

Now d has to be positive, and e can't be larger than a, so we have a contradiction. We can't satisfy both the n=2 and n=3 problems with the same set of a,b,c.


I believe the above cubic diagram/video shows that the n=3 solution set must be a subset of the n=2 problem, and therefore there is no solution to the n=3 problem. If the argument that all n>3 problems eventually have a cross-section n=3 problem, then this should also prove there is no solution for all n>3 ?!?

Alternate n > 3

I was hoping to find a pattern by this time for n=4+ (by examining the n-dimensional space that en takes up), but this may not be a viable approach. Next vacation ...