Friday, May 31, 2002


Oh yeah, short FLT Proof is STILL correct

I think those of you who played with my Java program that counts primes—and couldn't quite figure out how it works—which is basically ANY of you who played with it should be a little less arrogrant or cocksure about your ability to determine whether or not the short proof of Fermat's Last Theorem is correct, as it's far simpler than the short proof.

I'm sure there are many of you out there who realize you don't have a clue whether it's right or not, but you've probably felt that if it were right then there's somebody out there smart enough to know it's right, who'd be nice enough to tell you (or somebody).

Sorry, wakeup to the real world. You're on your own dependent on me as your guide.

Luckily for you, I'm self-correcting to a large extent, so if the proof were wrong, I'd tell you.

It's not wrong.

The short proof of Fermat's Last Theorem is at

Wednesday, May 15, 2002


When mathematicians make mistakes.

For a while now I've been trying to sound an alarm about problems in the mathematical establishment. The trouble is that I've made mistakes in explaining my position. But now I've got all my evidence together and can present it simply.

The relevance is the dependency of the physics community on the work of mathematicians. My fear is that there are fewer safeguards within that community against even gross error in accepted work than anyone is admitting or realizing.

Now to the evidence that supports my position.

A while back I posted about the polynomial-like expression

(v^3+1)W^3 - 3vW - 2,

where it's like a polynomial because v is not constant, though v is an integer.

I've been factoring that expression as

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

where a1, a2, a3, b1, b2 and b3 are algebraic integers.

(Algebraic integers are just numbers that can be the roots of a monic polynomial with integer or algebraic integer coefficients. For instance, X^2 - 2 has the root sqrt(2), so sqrt(2) is an algebraic integer.)

From looking at v=-1, which is 3W - 2, I quickly came to the conclusion that two of the a's had a factor of sqrt(v+1), and that's where this saga starts because it took me a long time to prove it, while mathematicians argued with me, and STILL argue—which is very important—that that's wrong.

Which almost brings us completely back to my recent posts here because I started one thread a while back with an assertion about algebraic integers (I was in error), and later I posted an attempted proof of my statement about the a's having a factor of sqrt(v+1)(again I was in error).

Some of you may wonder why you should care as the problem probably seems esoteric to you, but I think it should intrigue you if mathematicians are wrong about such a thing, especially if some of them have claimed—which they have—that they have PROOF that none of the a's has a factor of sqrt(v+1).

The reason is that if mathematicians can as a group claim "proof" when there is none, I suggest that maybe you should be concerned.

Now it turns out that it's rather easy to prove the mathematicians wrong here, though it took me a little while to see how.

The answer comes by making things a little less abstract by adding more the polynomial like expression, which has to do with where it comes from.

That expression is

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

which has a factor of x^2 + y^2 + vz^2, when x^3 + y^3 + z^3.

(Hope I didn't lose those of you who didn't realize until now that this relates back to Fermat's Last Theorem, where you probably have a kneejerk reaction to assume that anyone talking about it is a nut. Oh well, if I lost you, I lost you.)

Now the relation there is quite old and not in contention, as again, what has been in contention is the question about factors of the a's, and now we have

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

and interestingly enough, the value of the a's and b's has NOT changed from before (that's an important point).

Ok, remember that v is not constant, so I now use v = -1 + mf^{2j}, where m is some integer, and f is a prime factor of x and z-y, and j is a counting number (the exact form f^{2j} relates back to FLT, though the point is to remove all factors of f from x in certain situations).

That gives me

m^3 f^{6j}z^6 - 3 m^2 f^{4j}z^6 + (3 f^{2j}z^6 - 3f^{2j} x^2 y^2 z^2)m + 3 x^2 y^2 z^2 - 2x^3 y^3 ,

which looks a LOT more complicated, which has a lot to do with why I hadn't really looked at it closely until recently.

If you do look at it closely, you'll notice that I've grouped with a focus on m, which makes sense as it's the variable, and what I have here is a polynomial, since x, y, z, f, and j are all constants.

Since it's a polynomial, it factors.

Now notice the constant term, which is

3x^2 y^2 z^2 - 2x^3 y^3 = x^2 y^2(3z^2 - 2xy),

and for some of you that "x^2 y^2" may be leaping out at you as you quickly realize it supports my position.

Well here's the coup de grace.

Remember how before I had

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

Well after my substitution for v, I have

(m^3 f^{6j} - 3 m^2 f^{4j} + 3 f^{2j}m )z^6 + (3x^2 y^2- 3f^{2j} x^2 y^2 m)z^2 - 2x^3 y^3 = (a1 z^2 + b1 xy)(a2 z^2 + b2 xy)(a3 z^2 + b3 xy),

and look at the coefficient of z^6.

It is

m^3 f^{6j} - 3 m^2 f^{4j} + 3 f^{2j}m,

which has a factor of m.

Well that means that a1, a2 or a3 shares some factor with m as well, which explains the x^2 y^2 in the constant term as apparently TWO of the a's share some factor of m, while the third does not.

(And yes, a1, a2, a3, b1, b2 and b3 are functions of m.)

I demonstrate that clearly by pointing out that separating off f^{2j} from the polynomial leaves the constant term without any factors of f.

Here's that constant term again:

x^2 y^2(3z^2 - 2xy)

and remember that f is a prime factor of x shared with z-y, and j is actually chosen such that x^2 divided by f^{2j} no longer has any factors of f.

It's all rather simple but it devastates any "proofs" against my claim that two of the a's have a factor of sqrt(v+1) because for any of them to be true there'd have to be factors of f left in the constant term.

(Don't see how that relates? Well remeber that v = -1 + mf^{2j}, so v+1 = mf^{2j}, so sqrt(v+1) = sqrt(m) f^j.)

What should REALLY bother you isn't so much that the mathematicians are wrong.

You should start being bothered at hearing that they claimed proofs to something that is clearly false.

What should make you worry is the fact that as we speak, they are STILL maintaining that I'm wrong and they're right.

And I'm not talking about just any posters on the sci.math newsgroup (yes, most of the discussions have been there), these are mathematicians with established positions (one has a Ph.d from Berkeley).

Now, why should you care?

Well, just for giggles imagine that it's slightly important to humanity for mathematicians to produce "proofs" that actually are proofs. And imagine, just for the sake of argument, that if mathematicians can steadfastly maintain such a clearly false position, what might it mean if they did so with something a little more

And imagine that you have research that uses some mathematical tool backed by a "proof", which is nonesuch.

Luckily experimental results will probably ultimately show that your work is bogus, but do you really want to depend on that?

I think I should have demonstrated why I've raised questions that should be answered. The question is what will the physics community do.

If you want more detail than I've presented here look at, though I warn you that it's a short proof of Fermat's Last Theorem, and yes, the mathematicians need their false "proofs" to claim that it's not, which should scare you more than anything else I've said here.

This page is powered by Blogger. Isn't yours?