Re: Is this a valid approach to Fermat's Last

On Tue, 17 Oct 2006 18:47:22 EDT, Josef <josefg@xxxxxxxxx> wrote:


Josef wrote:
Hello all,

This is my approach to proving Fermat's Last
Theorem. It is a bit different, but hopefully it is
interesting. Any feedback would be more than
welcomed.

FLT states that: a^n + b^n = c^n has no non-zero
integer solutions for a, b and c when n > 2. This
could be thought of as a vector problem.

Looking at the n = 1, 4^1 + 5^1 = 9^1. Two vectors
added together equals another vector.

Looking at n = 2, 3^2 + 4^2 = 5^2. Two squared
vector added together equal a vector squared. A right
triangle can be formed from these three vectors.

For n > 2, there are still three vectors. These
vectors can still form a triangle. The agurment that
there is no triangle if a+b < c is true. But then
those integers do not even satisfy the case of n =
1(See above). This triangle is an obtuse triangle.


Where have you used the requirement that a, b and c
be integers?

The requirement that a, b and c be integers is in the initial FTL statement. I am trying to prove by contradiction. That should answer your answer your question.

No, that doesn't answer the question at all. We know that the
hypothesis is that a, b and c are integers. And we know that
it's a proof by contradiction. The question is this:

At what point _in the proof_ do you use the fact that a, b, and
c are integers?

(The point to the question is this: If, as it appears, the fact
that a, b, and c are integers is never used in the proof, then
you have a proof of the stronger result that a^n + b^n = c^n has
no non-trivial solutions for real numebers a, b, c. But the
stronger result is false, and that gives an easy way to show
that the proof is wrong. Now, if you can say where in the proof
you _do_ use the fact that a, b, c are integers that would be
different.)

There is always a solution c for any positive a, b if
you do not
require c to be an integer.

I agree.

- Randy

Thanks for the input,
Josef


************************

David C. Ullrich
.

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