Harristotelian Logic

If any of you have wanted me to explain how to prove Fermat's Last Theorem in a very simple way then this post is for you.

I show that x^p + y^p = z^p, requires that

(a1 z^2 + b1 xy)…(a2 z^2 + b2 xy) = 0(mod x^2 + y^2 + vz^2), where v is an integer, and a1...ap = v^p+1, b1...bp = -2, are just two of the p+1 defining equations for the 2p unknowns.

Then I notice something spectacularly simple.

I notice that x^2 + y^2 - z^2 MUST share prime counting number factors with x and y, if you assume x, y and z are nonzero integers.

And then I make use of the indeterminancy of the a's to consider a1 = sqrt(v+1) and a2 = sqrt(v+1), where v = -1 plus a multiple of some integer factor of x or y that is shared with x^2 + y^2 - z^2 (you know, like if x has a factor 11 that is also a factor z-y, then 11 would be the type of factor I'm talking about).

I set v to -1 plus a multiple of such a factor so that I can force

x^2 + y^2 + vz^2 to have that factor to any power I wish (like with the above 11^100293849, if I wanted too).

That then OBVIOUSLY forces

(a1 z^2 + b1 xy)…(a2 z^2 + b2 xy) to have that factor to the SAME power(again, with the above, 11^1000293849).

BUT WAIT!!!! x or y HAS that factor, and z CANNOT since it's pairwise coprime to x and y, so that leaves the a's!!!!!

BUT, I just said I've set only a1 and a2 to have that factor since they both equal sqrt(v+1), so all those factors must come from (a1 z^2 + b1 xy) and (a2 z^2 + b2 xy).

Well, did you notice something else? I ARBITRARILY considered a1 and a2 equal to sqrt(v+1). What about a1 = v+1? Now NONE of the other a's has that factor so everything is forced on the one (a1 z^2 + b1 xy).

Some of you should be falling on the floor in amazement that such a simple and extraordinary proof could have been undiscovered for so long, but most of you are now completely befuddled.

I am not going to bother looking at replies to my other posts, but I will look to see who replies to this one.

I will also post the link to my website, which is

http://communities.msn.com/ProofofFermatsLastTheorem&naventryid=104

in case you want to look at interesting details.

Luckily for me, despite the fact that most of you are so limited that you're still just reacting by reflex to believe I'm just a nut, this argument is so damn simple that soon enough someone will realize it is correct, and then some of you will get that sinking feeling in your gut as you contemplate how you reacted when it was presented to you :-).

Thank God for that! I didn't realize how closeminded modern mathematicians apparently are. You folks have to be forced by extreme measures to accept even simple stuff that falls just a little bit outside what you're used to.

That is, none of you are much fun at all! Sure hope the world never depends on your type discovering anything, or humanity will surely perish.