Saturday, September 29, 2001


Why I call Magidin an incompetent liar

I was reading through replies to my posts on the newsgroup, when I noticed a reply from Arturo Magidin. He started the post by explaining how he was replying since he'd claimed to have "killfiled" me, so he wouldn't see my posts.

Some have noticed I've been particularly hard on this person. I continually maintain that he's either incompetent as a mathematician, a liar or both.

You may believe you know why.

Not too long ago I made a post stepping through my proof of Fermat's Last Theorem for p=3 that's not that much different from my Part 2. I felt that it was clearly interesting enough mathematics that a competent mathematician or an honest one would have to admit the truth, so that's what I said.

I ended that post by saying that only an incompetent mathematician or a liar could post against it. I said that because the argument is so simple, direct, and especially because it has to do with Fermat's Last Theorem.

Yet Magidin posted against it. He insinuated that the argument was false, and continued that position through repeated posts, only to finally claim that he would not discuss things further as he was killfiling me.

Some of you may think I call Magidin a liar because of the question of whether or not rational numbers find their way into the ring I use in my proof.

I'm an amateur. I can make small mistakes. I don't mind it when honest, competent people point that out to me. But I'm not such an amateur or so stupid that I will be pushed off of a correct proof by a meaningless objection.

I maintain that Magidin is indeed either incompetent as a mathematician, or more likely a liar, or both.

His objection is not a true objection. It's an observation that doesn't matter.

Some of you may have noticed that I changed the definition of my ring slightly to accomodate the points that he raised. The changes involved were small.

Yet, Magidin apparently maintains there is no value in my arguments because how else could he be off somewhere claiming to ignore my posts if he's not saying my arguments ultimately don't have any value?

He is either incompetent as a mathematician, a liar or both.

If you feel that's too harsh. If you wish to test what I'm saying then feel free to do so. As I've said before I don't mind it when people point out to me when I'm wrong.

But now you know just how angry I get when people lie about that.

Tell the truth. It's that simple people. Just tell the truth.

Tuesday, September 25, 2001


Understanding short FLT Proof requires paradigm shift

One thing that's clearly emerged from replies to some of my posts on the newsgroup is a problem created by people looking at the math just one way.

To explain I like to give the simple quadratic

fx^2 + gx + h = (a1 x + b1)(a2 x + b2), where f, g and h are integers.

Apparently, a lot of you immediately want to solve that, and it might take you a little while to realize you can't with the information given.

Now if I give you

x^2 + 4x + 4 = (a1 x + b1)(a2 x + b2),

you may jump to a conclusion based on ONE solution out of infinity.

What also seems clear is that most of you by reflex think I'm talking about

fx^2 + gx + h = 0, and x^2 + 4x + 4 = 0, when there isn't any evidence to support that conclusion.

So, what's the point of my going over the above?

The point is that some of you have made replies claiming flaws in my short proof of Fermat's Last Theorem that depend on your not understanding clearly factorization as an ABSTRACT and NOT just as a way to solve polynomials set to 0.

Why do I say this?

Some on the newsgroup have accepted an argument against the proof that depends on something like saying f cannot equal 0, with fx^2 + gx + h because then they claim everything is undefined.

Sound puzzling? Solve for x, with fx^2 + gx + h = (a1 x + b1)(a2 x + b2), using the quadratic formula and it should be clear to you then.

If you think that f cannot equal 0 because you get a divide by 0 error, watch this:

gx + h = b1(a2 x + b2).

Is that undefined because of a divide by 0 error?

My proof, like the examples above, depends on only the ring operations of multiplication and addition, so it CANNOT have divide by 0 errors.

So why are there people on this newsgroup so confident that the proof is wrong based on an objection that depends on the assertion of a divide by 0 error?

I don't know, but I wish you'd tell me. I find the illogic strange and discomfiting, as well as surprising.

In any event, I won't post the link to the proof here because I have a post today that not only links to it, but explains briefly the idea behind it.

Eventually the proof will be accepted as I push people making mathematical replies to using objectivity instead of emotion, but considering how things look now, it could take a while.

Monday, September 17, 2001


Final issue with FLT proof: The ring

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.

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