Search All of the Math Forum:
Views expressed in these public forums are not endorsed by
NCTM or The Math Forum.



Proof of FLT
Posted:
Sep 7, 1996 12:56 PM


I've typed that before and been very wrong but here I go again. As for the people who won't read this because of previous posts, who cares since there are always new folks around who might be more openminded.
Task.
Prove x^n + y^n = z^n false with x,y,z relatively prime integers and n prime Natural greater than 2.
lemma: Neither x,y or z can be divisible by n.
That proof is simple and I would expect part of the old lore on FLT. If not I have the proof.
Method.
Take x^2n + 2x^n y^n + y^2n = z^2n which is just x^n + y^n = z^n squared
Use the substitution x + delta x = my, y + delta y = mx, z + delta z = mz
to give
(x+delta x)^2n + 2(x+delta x)^n (y+delta y)^2n = (z+delta z)^2n
Expand and subtract off the first equation and then use the substitution
(delta z) = (delta y) which allows you to subtract off (delta y)^2n.
But that last substitution gives values for the deltas in terms of x,y, and z.
delta x = (x+yz)(xy)/(zx), delta y = z(xy)/(zx)
Note that the numerator of delta x must be divisible by n.
Now collect the terms with delta y and not delta x to the right.
The terms on the right will be like (2n)!(z^ay^a)(delta y)^a/a!
Since n is prime all of those terms will be divisible by n EXCEPT for the nth term
(2n)!(z^ny^n)(delta y)^n/n! = (2n)! x^n (delta y)^n/n!
Using our values for delta y gives
(2n)! x^n [z(xy)/(zx)]^n/n!
If we multiply our entire equation by (zx)^2n we get for the term above  (2n)! x^n (zx)^n z^n (xy)^n/n!  Note that the above has factor of x,y and z in the numerator.
Now looking at the left side, all terms are either multiplied by (delta x) or by n, so the left side is divisible by n. But that requires that the right by divisible by n as well.
*All terms on the right are automatically divisible by n, EXCEPT the one I've highlighted above. It is only divisibly by n if x,y,z or (xy) is divisible by n.
(xy) can't be divisible by n because that would give a proof of FLT in the first case, so x, y or z must be divisible by n.
And by the lemma given at the top that can't be the case either.
I just want to dedicate this one to the writers of Star Trek The Next Generation and of Deep Space Nine for your lack of faith in human creativity.
James S. Harris



