Well, I'm not 'math folk' (grey beard programmer), but I love the Horizon documentary on Andrew Wiles and his solution, and I'd love to hear why my intuitive understanding is inapplicable from people who know better than myself. (Note that this will not in any way be a proof, but just the train of thought I believe to be the line of thinking Fermat may have used to construct his proper mathematical proof.)
My idea here is based upon physical/visual intuition, starting with why it works for n=2 (squares) and then why it cannot work for n=3 (cubes) and then that n>3 is necessarily more complex than n=3 thus cannot work either.
[Note that I will use lower case letters for the sides/roots and their uppercase letters to denote the areas or volumes.
Thus, the full equation is Z=Y+X, with X = x^n, resulting in z^n = y^n + x^n.
I also use (for n=2), dy = z - y, and Dy = 2(y(dy)) + dy^2,
and dx = z - x, and Dx = 2(x(dx)) + dx^2.
I'm sorry my dx and dy conflict with calculus notation but my notation means dx is "the difference between z and x" which is the same as "the length that must be added to x to equal z" and Dx is "total amount that must be added to X to get Z".
Therefore (for n=2), Dx = Y = 2(x(dx)) + dx^2, and Dy = X = 2(y(dy)) + dy^2.
]
For n=2, Z=Y+X works because (what can be visualized as a square) X can be "smushed" over two sides and their joining corner of (the other square) Y evenly, such that Z = Y + Dy = Y + 2(y(dy)) + dy^2. The term "2(y(dy))" is the amount that must be added along each of the two sides, and the term "dy^2" is the amount that must be added at the corner to complete the perfect square Z.
So, for example, 5^2 = 4^2 + 3^2 because both
3^2 = 9 = 2(4(1)) + 1^2 = 2(4) + 1 = 8 + 1, and
4^2 = 16 = 2(3(2)) + 2^2 = 2(6) + 4 = 12 + 4.
Now, for n=3, we must visualize the situation where the cube X is smushed over the cube Y's three faces and its joining corner. (Now X=x^3 and Y=y^3.)
The equations for dx and dy are the same, but Dx and Dy have expanded by a dimension:
Dx = Y = 3(x^2)(dx) + 3(x)(dx^2) + dx^3, and likewise
Dy = X = 3(y^2)(dy) + 3(y)(dy^2) + dy^3.
The term "3(x^2)(dx)" is the amount that must be added to three faces of the cube Y, the term "3(x)(dx^2)" is the amount that must be added along the three edges joining those three faces of the cube Y, and the term "dx^3" is the amount that must be added at the corner.
Now, I haven't the maths to prove why Dx and Dy for n=3 won't have integer solutions but my intuition says it has something to do with the fact that it's three dimensions and, therefore, a couple of odd numbers multiplying around in there (the first two terms) and the fact that there are only two cubes being smushed together to try and reconstitute another perfect cube. I also imagine Fermat could actually mathematically prove why it's impossible. Perhaps it can be shown that Dx and Dy cannot both have diophantine solutions. These are just guesses.
As for n>3, the terms (and physical/visual representations) will only become more complex and there will be still only two terms with which to reconstitute the hypercube.
Anyway, that's my intuition about the entire problem and I have to imagine that a proof that Fermat can easily intuit yet is (a bit?) too large to fit in the margin must surely tread down a simple path, perhaps even one that relies on a physical/visual interpretation of what the equations can be likened to.
I look forward to this being eviscerated or flatly rejected, if appropriate, or at least corrected for inconsistencies. If it serves anyone in their exploration of this insidiously complex yet apparently simple-seeming problem, my joy would only grow. If my name would someday appear in a mathematical paper that a real mathematician produces as a result of this, well that would be out of this world for this poverty-striken math wannabe.
[Edited to fix my n=3 equations.]