24 Feb 2008 @ 09:25, by John Grieve
Two Conjectures concerning Fermat’s Last Theorem
For most of my life, I have been an amateur mathematician and interested in the problem called Fermat’s Last Theorem. Like most other people I was delighted when Prof. Wiles came up with his proof in 1994 and I only regretted that I was not able to understand the advanced mathematics which it contained. I have always believed that Fermat had a proof of his own and that it has to be based on the mathematics available in his time. So I continued to struggle with this problem.
My recent researches have led me to two interconnected hypotheses which, if proven, lead to a solution of this problem and which arguably could be the same approach which Fermat used.
My first conjecture concerns the equation:
X^3 + Y^3 = Z^3
If we assume that there are indeed two cubes which when added together equal a third cube then we realize that since a cube has six sides, each of which is a square, then we have the interesting fact that the first cube has a square equal to X^2, the second cube has a square equal to Y^2 and the third cube has a square equal to Z^2. This obvious fact leads me to a question which I believe is original, as I have never come across it in conversation with other mathematicians or seen it in the literature. Is it possible, I ask myself, that given the hypothetical existence of these sides x, y, and z and these squares X^2, Y^2 and Z^2, that there is a definite relationship between these three squares.
The surface area of the first cube is 6x^2, that of the second cube is 6y^2 and that of the third cube 6z^2.It is clear to me that since cubes are based on squares and squares are based on Pythagorean triangles, that it is the underlying Pythagorean triangles which make it impossible for two cubes to add to another cube or two fourth powers to add to another fourth power and so on. In fact I would say that here X^2 + Y^2 = Z^2 though this implies a contradiction.
This is paradoxical and clearly impossible. Prof. Wiles, whatever the merits or demerits of his proposed solution, does nothing to clarify the amazing contradiction at the heart of the Fermat problem.
If this is indeed so, the fact that X, Y and Z must be Pythagorean triples leads me to my next conjecture. I believe, though I cannot prove, that the same relationship between N=2 and N=3 in the Fermat equation
X^N + Y^N = Z^N
Exists between N=3 and N=4 and in general between any two consecutive powers. In other words, the values which satisfy the equation for any N must also satisfy the equation for N+1. So the solutions are like a set of Russian dolls or nested boxes in that each is contained in the previous one. Since we know that there are no solutions for N=3 then establishing these two conjectures would immediately prove Fermat’s Last Theorem.
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