Harristotelian Logic

I'll talk using the p=3 case because it's simple, but there's nothing really different from the general case, so that shouldn't be a problem.

I will be going through to the conclusion, as I'll point out the issues with rings, and give y'all an answer.

What I have using my argument is that

(v^3+1)z^6 - 3v x^2 y^2 z^2 - 2x^3 y^3 = 0(mod x^2 + y^2 + vz^2)

when x^3 + y^3 = z^3.

At this point in the proof, the x,y and z are assumed to be nonzero integers, v is an arbitrary integer, so the ring is integers.

I then go use a factorization of the expression on the left to get

(v^3+1)z^6 - 3v x^2 y^2 z^2 - 2x^3 y^3 = (a1 z^2 + b1 xy)(a2 z^2 + b2 xy)( a3 z^2 + b3 xy),

and then go on to have

(a1 z^2 + b1 xy)(a2 z^2 + b2 xy)( a3 z^2 + b3 xy)= 0(mod x^2 + y^2 + vz^2).

But now several things have happened, and some of those things have been controversial on the newsgroup. That is, someone has objected strenuously to them.

First off, I'm now in complex numbers for my a's and b's.

Second, as given here, I don't necessarily know that the a's and b's aren't somehow dependent on x,y and z in some way that would mean I don't have a factorization.

To help out with that, I'll add that the a's and b's would be the same as the following:

(v^3+1)W^3 - 3v W - 2 = (a1 W + b1)(a2 W + b2)( a3 W + b3 ),

where I have polynomials.

Now then, I have that a1 a2 a3 = v^3 + 1, and b1 b2 b3 = -2, while there are two more defining expressions that I won't give here (I have given them in other posts).

I notice that with v=-1, my earlier expression

(v^3+1)z^6 - 3v x^2 y^2 z^2 - 2x^3 y^3 gives

x^2 y^2(3v z^2 - 2xy), where you can see that

a1 = a2 = 0, a3 =3, b1 b2 = 1, and b3 = -2.

I further notice that a1 = a2 = sqrt(v+1), a3 = v^2 - v+1, will give this result, so I now specify a ring where I have integers plus sqrt(v+1), and my b's.

And then I look at my modulus x^2 + y^2 + vz^2, and notice that with v=-1, I have x^2 + y^2 - z^2.

Since x^3 = (z-y)(z^2 + zy + y^2), I know that any prime factor of z-y would have to be a prime factor of x, so I pick one and call it f. (If x doesn't have one that it shares with z-y, then I'd just use y and z-x).

Further, I know that z-y, would have to have a factor of f^3, as long as f is not 3, so I look at x^2 + y^2 - z^2, and I know that I have a factor of at least f^2. Since x/f might still have a factor of f, I use a counting number j to handle all x's f factors, and let v = -1 + mf^{2j}.

I do that because then I have x^2 + y^2 - z^2 +mf^{2j}z^2, and I can now find some m to make that have a factor of f^{2jn}, where n is some arbitrarily high counting number.

Notice, that my choice of v's means that the a's and b's have the same modulus with respect to f that they had with v=-1.

I remind you that then I had that a1 = a2 = 0, a3 =3, b1 b2 = 1, and b3 = -2.

There was an ambiguity in the b's, but now, with v+1 nonzero, I know they have SOME modulus with respect to f, but I don't care what it is.

So, now I substitute in with my new v and get

(sqrt(m)f^j z^2 + b1 xy)(sqrt(m)f^j z^2 + b2 xy)( (v^2-v+1) z^2 + b3 xy) =
0(mod x^2 + y^2 - z^2 + mf^{2j}z^2),

and since x^2 + y^2 - z^2 + mf^{2j}z^2 has a factor of f^{2jn}, I have that

(sqrt(m)f^j z^2 + b1 xy)(sqrt(m)f^j z^2 + b2 xy)( (v^2-v+1) z^2 + b3 xy) =
0(mod f^{2jn}).

Now I've said that x has a factor of f^j, so it's simpler at this point to do a substitution like x = kf^j, so I have

(sqrt(m)f^j z^2+b1 ykf^j)(sqrt(m)f^j z^2 + b2 ykf^j)((v^2-v+1) z^2 +b3 ykf^j)=
0(mod f^{2jn}),

and now I can separate out f^{2j} and get

(sqrt(m) z^2+b1 yk)(sqrt(m) z^2 + b2 yk)((v^2-v+1) z^2 + b3 ykf^j) = 0(mod f^{2jn-2j}).

Ok, at this point, despite my ring containing elements that are noninteger, the modulus would have an integer as a factor, but that's not really an important issue because I know that my b3 has a modulus of 3 from before.

So, dropping that term with it I get

(sqrt(m) z^2 + b1 yk)(sqrt(m) z^2 + b2 yk) = 0(mod f^{2jn-2j}).

Now, some have debated whether or not some member of the ring could multiply times the modulus to remove factors of f. Well, we already saw b3 handled, so that just leaves b1 and b2. Well, if one had some factor of 1/f, and the other had some factor of f, then you could multiply back through by f, and find that the one with a factor of 1/f, would end up having to have a factor of f.

Besides, the ambiguity with v=-1, is only between whether or not b1 = b2 = 1 or -1.

I guess if any of you wish to start hollering that this isn't a proof on that point, I'll go into more detail, but I'll let you do that first because I figure it'll tell the world something about you.

Oh yeah, m is sort of arbitrary. So I can pick some m'. On my proof page, I basically assume that sqrt(m) and sqrt(m') are integers, but if they're not, it's actually simpler because then sqrt(m) - sqrt(m') = 0(mod f^j} is false.

I wanted to make this post to make it clear enough why the proof is true, not so much in the hopes that I'd end all objections, but to highlight to you just how weak the ground is for any objection.

Now I feel that a competent mathematician given the information I've presented will find the proof of Fermat's Last Theorem to be clear and direct.

Therefore, I will be arguing that those who make claims that there is no proof here are either liars or incompetent.

And if you're going to be one of those, I'd recommend that you be clearly incompetent, because, make no mistake, there might be more than just a cost to your pride if you're a liar, even if it's just because you look like one, when you're really incompetent.