Tuesday, July 31, 2001


Issue is factorization

What has become painfully clear in discussions on the newsgroup is that many of you that post in reply to my postings don't understand factorization.

Worse, of that same group, many of you are apparently confused by conventions like algebraic numbers or the FIELD of complex numbers versus complex number RINGS.

I've tried to guide you to a more general understanding by talking about fractions, and by demonstrating the use of the ring bridge, but I still get the feeling that most of you are still just sticking with what you know.

Ok, I'll just come out and say it without hinting:

I think most of you lack intellectual flexibility, logic, and mathematical rigor. That you have a sense of some mathematical understanding based on what you were taught, but that parrotlike you repeat it unable to do even simple extensions into territory that is somewhat new to you.

So what IS a factorization?

A factorization is just a statement that what's on both sides of the equal sign is the same thing.

12 = 3(4) because 3(4) = 3(4) and 3(4) = 12 = 3(2)(2).

(x+2)(x+2) = x^2 + 4x + 4 is another example.

What's so complicated about that?

Yet, some of you have been arguing strange things like, what if x is a complex number? Then couldn't "factors" be meaningless?

Factors can't be meaningless if a factorization exists, as long as x is in a ring

x+2 is a factor of x^2 + 4x + 4 PERIOD.

And it doesn't matter if I put the entire thing in the field of complex numbers, or in something funkier and say that x is an octonion!

x+2 will STILL be a factor of x^2 + 4x + 4 because (x+2)(x+2) = x^2 + 4x + 4.

Folks, this is basic math.

Now I've said to mathematicians in general that probably all you lose is prestige when it looks like the math community is denying something so simple, but hey, don't any of you care about the prestige of the math community?

Isn't there a single mathematician out there who isn't a little embarrassed to have people claiming they're mathematicians arguing against such truths on sci.math, who is willing to speak up to keep them from making you all look stupid?

Like I pointed out yesterday, a lot of these people posting apparently have nothing to lose. For some of them, this newsgroup is their chance to get a little attention, and they say stupid, rude or shocking things in attempts to get that attention.

But when it's clear that my proof is true, for reporters, that's news, and they are you.

Monday, July 30, 2001


FLT Proof: Power of the ring bridge

For a while now I've been talking about how you can actually test my proof of Fermat's Last Theorem.

Of course, the problem with that is that it can look a little hairy after you pass p=3.

So how do you test the proof with actual numbers?

Well, the proof requires that 2 have the factor of x^2 + y^2 - z^2 not shared with x and y as a factor, which keeps x, y and z from being integers (remember x^p + y^p = z^p).

Some of you may have noticed that for p=3, x=1, and y=1, you get

2^{1/3}/(2^{1/3} - 1), which is easily shown to not be a fraction.

Others may have looked at the gaussian integer solution that's been posted and noticed it worked as well, and it's a LOT easier to work with since x^2 + y^2 - z^2 gives an integer (though then x^2 + z^2 - y^2 is a bit of a pain).

But anyway, past p=3, the easiest example gets a good bit harder to evaluate.

After all, is 2^{(p-2)/p}/(2^{(p-2)/p} - 1) a fraction or not?

Strangely enough, or I'd think you'd know the proof is correct by now, your professors can't help you with that one, but the truth CAN be determined.

You can just use a quadratic ring bridge.

Using fx^2 + gx + h = (a1 x + b1)(a2 x + b2), and letting h = 2^{p-2}-1, b1 = 2^{(p-2)/p} - 1, and

a1 b2 = m, a2 b1 = 2, it's just a matter of finding m and f, such that

2 = (-(m+2) + sqrt((m+2)^2 - 4(2^{p-2}-1)f))/2


4 + (m+2) = sqrt((m+2)^2 - 4(2^{p-2}-1)f))


((m+2) + 4)^2 = (m+2)^2 - 4(2^{p-2}-1)f


f = -(((m+2) + 4)^2 - (m+2)^2)/(4(2^{p-2}-1))


f = -((m+2) + 4 - m - 2)((m+2) + 4 + m + 2)/(4(2^{p-2}-1))


f = 2(m + 4)/(2^{p-2}-1).

Of course, there's an integer m such that m+4 is divisible by 2^{p-2}-1.

And that's it.

I've proven that b2 IS indeed a factor of 2.

Note: I don't know what a1 is, and don't care.

Sunday, July 29, 2001


Test of credibility

Would any of you agree that Fermat's Last Theorem is a famous enough math problem that ALL professional mathematicians should have at least heard of it?

Would any of you agree that a simple proof of it that primarily used secondary school algebra should be accessible to ANY professional mathematician?

Would any of you suppose that evidence that there was a simple proof of Fermat's Last Theorem might zip through the mathematical community like a thunderbolt?

Now what if that proof was correct, and even a kid with a calculator could test some of its predictions and see that?

Do ANY of you really believe that professional mathematicians would not know this?

Mathematics of all the disciplines demands our trust because its own practitioners tell us what's valuable in the field.

Sure, to a certain extent, say physicists and biologists, do the same, but then again, it's hard to not realize something is important in physics when it produces the nuclear bomb, or in biology when it brings in the possibility of human cloning.

Now you know a good reason for why Nobel did not create a prize for mathematics.

He realized that in a field where major results could lack practical application so that only its practitioners decide what's valuable, there was never a guarantee of TRUE value.

I'm just here with the proof, and you know what's truly sad?

DESPITE this post, we will all wake up to a world tomorrow where the math folks still deny the obvious truth, when they could just get together, see the obvious and then admit to the world…the truth.

But dammit, it's going to be like pulling teeth. But guess what?

God gave me all the tools necessary for that, so at some point I will stop hesitating.

Why do I hesitate?

Because I have a job and a mortgage and I'm not excited about doing things that may affect other people's families because if it were me I'd like to be given the benefit of the doubt myself. I would like ample opportunity to do my job before someone lowered the boom on me.

So I will give ample opportunity and this drama may continue for a while until I am told that I can give no more time.

And then I will not, and it will be over so fast that you will think you've come instantly to another world.

But if I have to force that, then it will be a world that some would rather not be in.

A world where their families may pay a price for them not doing their jobs.

And that's not what I want, so I will wait, and hope as long as I can, until I have to finish my job.

Sunday, July 22, 2001


Proof that the math world is sick

I've been posting for a while that I have a simple proof of Fermat's Last Theorm that's only about a page, which is quite simple.

Lots of other people have been posting that it's not really math, and they say NO ONE in the world believes I have anything at all.

I want you to remember all of that when you get to the end of this post.

Just because I feel a little guilty, remember I posted what my intent has been: I want to show the world that the math community is being rotted out by the strange attitude that mathematicians can be "pure" by producing useless information, and refusing to have the discipline address as a priority the mathematical issues at the frontiers of physics.

My proof of FLT starts with x^2 + y^2 + vz^2 = 0(mod x^2 + y^2 + vz^2)

which is exactly the same as

x^2 + y^2 + vz^2 = x^2 + y^2 + vz^2

(which didn't keep one poster from arguing that I was dividing by zero in the first).

The above is a tautology. (My use of that phrase in this context brought objections and derision.)

Of course, I use x^p + y^p = z^p, in there, and eventually come to

(v^p+1)z^{2p}- px^2 y^2 (vz^2)Q'{z^2,x^2 y^2} - 2x^p y^p = 0(mod (x^2+y^2+vz^2))

which is a rather neat expression that says a lot.

(In case you'd like to look at the main expression for a specific case, with p=5

(v^5+ 1)z^10 - 5x^2 y^2 (vz^2)(v^2 z^4 - x^2 y^2) - 2x^5 y^5)).

What it says is that as long as x^p + y^p = z^p, that expression has

x^2 + y^2 + vz^2 as a factor.

Because I'm proving FLT, I look at the case where v is an integer, and quickly noticed that setting v=-1 gave me an interesting result—

px^2 y^2 z^2 Q'{z^2,x^2 y^2} - 2x^p y^p = 0(mod (x^2 + y^2 - z^2)).

So I consider an integer factor of x^2 + y^2 - z^2 that is NOT shared with x or y, which is odd, and I called it F.

So now, I decide to consider v's, where v = -1 + mF, where m is a counting number, which shouldn't be a surprise.

Obviously then, my primary expression—

(v^p+1)z^{2p}- px^2 y^2 (vz^2)Q'{z^2,x^2 y^2} - 2x^p y^p

is divisible by F.

Now I factor that expression into (a1 z^2 + b1 xy)…(ap z^2 + bp xy).

Here it should be obvious enough that

a1…ap have F as a factor.

So at this point, we now have

(a1 z^2 + b1 xy)…(ap z^2 + bp xy) = 0(mod x^2 + y^2 + vz^2)

(Some of you may be wondering why I don't substitute in -1 + mF for v. Well, it just seems like effort for nothing to me.)

Now I'll pause a bit because it finally became clear that many on the newsgroup weren't sure you could even do that, and then even if they figured such an expression MIGHT exist, they kept bugging me about the ring.

You see, their professors have been teaching them for a long time that you have to have a particular ring, and they are having problems with that programming.

In short, it is confusing them.

Once I finally understood the problem I brought them back to the quadratic case, which I figured would be a good place to clear everything up, so I gave

fx^2 + gx + h = (a1 x + b1)(a2 x + b2), and talked about it a LOT.

Didn't help much.

Today I saw a post from a guy who believes that if the main expression has a degree greater than 4, then you're stuck. I kindly noted that he was trying to refute Gauss. Why?

a1 a2 = f, a1 b2 + a2 b1 = g, and b1 b2 = h, so

you have four unknowns with f,g and h which are the knowns.

In general, you'll have 2n unknowns with n+1 knowns, which allows you to always solve for each 'a' using all of the b's.

That gives you a monic polynomial, which Gauss proved ALWAYS has a solution.

BUT all of the math community on sci.math has had problems with what I just presented, and there's still a big issue that you might not have noticed I glossed over—I didn't give the ring for a1,a2,b1, and b2.

Uh oh!!! According to the math community that means everything is INVALID!!!

Ohmigod, we just wasted our time, oh well.

Folks, this is the math that they're telling you is junk.

Translation: They don't understand it because most of them are parrots, so they call it junk rather than accept mathematical truth.

So, why not specify the ring?

Because it interferes with the proof, getting back to it:

Then for a given factor a1 of F

a1 z^2 + b1 xy = = 0(mod a1), and in general

an z^2 + bn xy = = 0(mod an), 0<n<p+1, so b1…bp = 0(mod F),



and there is a contradiction if F is a counting number greater than 2 (if F = 1, then I'd just use x^2 + z^2 - y^2).


Ooh that was fast! I bet most of you are wondering what all of that meant, which is a pity because it's basic math.

Ok, enough rhetoric, let's go back through it line by line.

"for a given factor a1 of F

a1 z^2 + b1 xy = = 0(mod a1)"

Ok, remember we have that a1…ap = F (for p=5, a1 a2 a3 a4 a5 = F), so obviously a1 is a factor of F.

Oh yeah, they've been telling you that I'm using some funky definition of factor.

Translation: They're confused, so I must be doing something wrong.

No, they're just confused.

I'm using factor the same way everybody does. You know 2(3)=6, so 2 is a FACTOR of 6.

Now remember, we have that

(a1 z^2 + b1 xy)…(ap z^2 + bp xy) = 0(mod x^2 + y^2 + vz^2)

which means that

(a1 z^2 + b1 xy)…(ap z^2 + bp xy) = 0(mod F)

(Why didn't I just write that above? It seems SO obvious that I just figured it'd be a waste.)


(a1 z^2 + b1 xy)…(ap z^2 + bp xy) = 0(mod a1…ap).

Wait! You may holler. How do I know that F factors that way on the right?

Because I didn't specify a1…ap, that's why.

Understand why I don't want to specify a ring at the start now?

(In case you missed it, the ring is specified when I set my a's. For instance, with x^2 + 4x + 4, a1=a2=1, forces b1=b2=2, but you could have a1 = 1+sqrt(3)i, and a2 = 1-sqrt(3)i, the math doesn't care which you use but some of you get confused.)

Now I'm going to give you an example with integers and see where we can go from there,

(3x + 2)(5x+a) = 0(mod 3(5)), and let me specify that must be true for ALL x's.

But, it's not, now is it?

'a' must be divisible by 5, but 2 isn't divisible by 3, so I can stick in an x where 5x+a isn't divisible by 3, and the expression is false.

The math is basically just a use of the distributive property, which should be basic enough.

In the proof I talk about sharing of factors but it might just confuse people here at this point.

Besides, it occurs to me that some of you are saying, well that's integers, other types of numbers can be really different.

Well that's why I've been trying to educate the community about discrete sets and the continuous field. That is, I hate to break it to you, but mathematicians got things a bit wrong. Just a little bit wrong, so don't get too excited. In any event, they ARE supposed to be taught about the difference ring operations make, so it shouldn't matter, they should still be able to get on the same page with me.

(Any of you use p-adics? Then you have NO excuse for not getting it.)

"a1 z^2 + b1 xy = = 0(mod a1), and in general

an z^2 + bn xy = = 0(mod an), 0<n<p+1, so b1…bp = 0(mod F)"

Well, a1 z^2 + b1 xy = 0(mod a1) means b1 xy = 0(mod a1), and I said that F is coprime to x and y, so it's a no-brainer that b1 = 0(mod a1).

Go down the line and b1…bp = 0(mod F).




and there is a contradiction if F is a counting number greater than 2 (if F = 1, then I'd just use x^2 + z^2 - y^2)."

That is, I arbitrarily picked x^2 + y^2 + vz^2 at the start of the proof, so if necessary I can switch things around a bit and start over, which would force F to be something other than 1.

So why are people still running around telling you there's nothing to my proof?

You'd think they'd want to be in on something like this, wouldn't you?

Well, I put a lot of it on bad training, and the rest is on the human need for consistency.

Think about it, if you see a bum on the street begging for change, you don't want to deal with him being an oil tycoon the next day.

Sure, we can handle it if he won the lottery or something but there MUST be an explanation like that or people get VERY unsettled.

I'm very unsettling. Ok. Get over it.

Your feelings are not a reason but an excuse, and the consequences will be the same because I believe bad training is the PRIMARY reason, and incompetence is not a good reason.

Friday, July 13, 2001


The clock is ticking

A little earlier I made a challenge post, which I suspect will go by like all the others as you folks seem to be so ignorant of your own deficiencies that it takes a sledgehammer to get your attention.

Since I'm a sporting person I will try and help you out by explaining to you some of why your past training was insufficient.

Most of you apparently have been trained to make proofs in one ring.

I've given you an example that crosses rings, or I should say it BRIDGES rings, and you all happily fall over each other saying dumb stuff because you refuse to leave your little conceptual boxes.

People, I'm sorry to tell you that it gets worse from here.

Mathematics doesn't really care about how you LIKE to think things should work, so some of the details of how things actually CAN work will be beyond the mental circuitry of many of you.

Others of you will get just enough to really mess you up.

(Unfortunately, that non-technical sentence is what will happen to some of you who may lose the ability to handle reality in the same way once you start part way down these paths.)

The test of you is to look at my challenge post and form an opinion.

I feel an urge to give you the paragraph proof now, but I said after a couple of hours, so I will stick to that.

I want all of you to form an opinion because you need to get out from those little useless mental shelters that are keeping you from the truth.

Besides, I actually need some of you and I don't want to take a lot of time in this first culling process.

When I give the proof later this evening I will give the MSN community name, some of the ground rules, and a description of the type of people I'm looking for.

I AM serious about this being a short route to a Ph.d for some of you, but just remember, I'm the guy who proved Fermat's Last Theorem in just a bit over 6 years after all of your mathematicians took around 360. My standards are kind of high, so it won't be a walk in the park, and nothing will just be given to you.

Everything will be earned by tremendous effort and through more emotional pain than you might have believed you could take.

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