Tuesday, April 22, 2003
Why I can't be wrong, FLT argument key claim
Believe it or not, I have NO interest in continuing with the claim that I have a proof of Fermat's Last Theorem if I do not, and I'm considering what some may try to call irrefutable evidence that I don't (um, even me), but in the meantime, maybe SOMEONE could tell me what's wrong with the simple argument which still has me convinced!!!
And yes I'm desperate. I need to put this behind me if I'm wrong. But I can't let it go if I'm right, as that would really suck.
Ok, I have this expression I can factor as
(a1 z^2 + b1 f^j uy)(a2 z^2 + b2 f^j uy)( a3 z^2 + b3 f^j uy),
where
a1 a2 a3 = (-1+mf^{2j})^3 + 1, and b1 b2 b3 = -2.
But by keeping a lot of variables, deliberately, that same expression is also
f^{6j} z^6 m^3 -3f^{4j} z^6 m^2 + 3(f^{2j} -u^2 f^{3j} y^2 z^2) m + 3 (uf^j)^2 y^2 z^2 - 2 u^3 f^{3j} y^3
and I consider it as a polynomial with respect to m, as m is the only thing I vary in my FLT argument.
(Expression derived in detail at
http://groups.msn.com/AmateurMath/algebraoffactorizations.msnw )
Here's where I need help with the following statement:
Now each factor can be separated into factors of the constant term with respect to m, versus those terms with a factor of m.
If that is true then I can't see how I can be wrong. It's the simple idea that lies at the heart of my claim of proof. I need to see if anyone can refute it.
If it's true then there's nothing wrong with my proof. It is true; therefore, there is nothing wrong with my proof.
At least one of the a's must have a factor of m, from their product, so assuming that a1 does, consider
(a1 z2 + b1 f^j uy)
and since a1 has a factor of m, let's split b1 up into possible factors of m, which I'll call w, and factors of the constant term which I'll call c. Then I have
(a1 z2 + (w+c) f^j uy) = a1 z^2 + w f^j uy + c f^j uy
showing the factor of the constant term times a factor of f, which is f^j, but when f^{2j} is divided off from the original expression it is removed from the constant term, proving that a factor of f^j must be removed here, if a1 has a factor of m, proving that a1 has a factor of f^j if it has a factor of m.
So can I split b1 up that way or not? Aren't a's that have factors of m forced out of the constant term?
It's so simple. That's it. So if you can prove that all wrong then you'll do me a favor.
Then again, it seems to befuddle people and I think all those variables make people dizzy.
Of course, I'm not wrong, which is why things are really frustrating for me now.
And yes I'm desperate. I need to put this behind me if I'm wrong. But I can't let it go if I'm right, as that would really suck.
Ok, I have this expression I can factor as
(a1 z^2 + b1 f^j uy)(a2 z^2 + b2 f^j uy)( a3 z^2 + b3 f^j uy),
where
a1 a2 a3 = (-1+mf^{2j})^3 + 1, and b1 b2 b3 = -2.
But by keeping a lot of variables, deliberately, that same expression is also
f^{6j} z^6 m^3 -3f^{4j} z^6 m^2 + 3(f^{2j} -u^2 f^{3j} y^2 z^2) m + 3 (uf^j)^2 y^2 z^2 - 2 u^3 f^{3j} y^3
and I consider it as a polynomial with respect to m, as m is the only thing I vary in my FLT argument.
(Expression derived in detail at
http://groups.msn.com/AmateurMath/algebraoffactorizations.msnw )
Here's where I need help with the following statement:
Now each factor can be separated into factors of the constant term with respect to m, versus those terms with a factor of m.
If that is true then I can't see how I can be wrong. It's the simple idea that lies at the heart of my claim of proof. I need to see if anyone can refute it.
If it's true then there's nothing wrong with my proof. It is true; therefore, there is nothing wrong with my proof.
At least one of the a's must have a factor of m, from their product, so assuming that a1 does, consider
(a1 z2 + b1 f^j uy)
and since a1 has a factor of m, let's split b1 up into possible factors of m, which I'll call w, and factors of the constant term which I'll call c. Then I have
(a1 z2 + (w+c) f^j uy) = a1 z^2 + w f^j uy + c f^j uy
showing the factor of the constant term times a factor of f, which is f^j, but when f^{2j} is divided off from the original expression it is removed from the constant term, proving that a factor of f^j must be removed here, if a1 has a factor of m, proving that a1 has a factor of f^j if it has a factor of m.
So can I split b1 up that way or not? Aren't a's that have factors of m forced out of the constant term?
It's so simple. That's it. So if you can prove that all wrong then you'll do me a favor.
Then again, it seems to befuddle people and I think all those variables make people dizzy.
Of course, I'm not wrong, which is why things are really frustrating for me now.
# posted by James Harris @ 12:08 
