Monday, December 09, 2002
JSH: Problem with Magidin's positio
I do read all the posts made in reply to my survey threads, though I try not to reply in those threads. I mention that as every time I ask about Magidin—who is one of the principle people who not only claims that the short FLT Proof is wrong, but he actually talks math—people indicate they believe him.
Well, I thought you'd like to know of an interesting problem with his position, and I'm curious about whether or not he will acknowledge it or give a resolution.
You see, his position from what I see leads to an infinite descent which causes a contradiction.
To understand that you need to know what his position is, and I'll show it for p=3, which you see get used a lot. Obviously, p=3 is used a lot because it's the simplest case for p odd prime, though my FLT Proof covers
x^p + y^p = z^p, for p odd prime, and not just p=3,
but let's get to more details.
A key expression for p=3 that follows from my approach is
(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),
where a1, a2, a3, b1, b2, b3 are all algebraic integers, as the ring is algebraic integers, and v is an arbitrary integer, while, of course, x, y and z are assumed integer counterexamples to FLT.
(Which means I'm asking "What if…?" and looking to see if that doesn't create a problem, which it does, which proves FLT, which is the point of contention as people like Magidin claim that I haven't found that problem.)
It so happens that I've focused on v+1, which is a factor of v^3+1 as
v^3 + 1 = (v+1)(v^2 - v + 1),
and I've given v a value (it's arbitrary remember) that gives it a prime factor of f, which just means that f could be 5, or 7, or 11 or some other prime number other than 3, as, in general, p itself is excluded, to handle a complication that would arise if x were divisible by p.
And my proof depends on only two of the a's, say a1 and a2, having a factor of f, where things are a little more complicated because the a's are themselves not integers, though they are algebraic integers.
Magidin has disputed that only two of the a's have a factor of f, and has instead claimed that ALL the a's have non unit algebraic integer factors of f, and he has based his claim, among other things on Galois Theory.
However, people seem to forget that my choice of v+1 was from a math perspective arbitrary in that you can consider some other factor of v^3 + 1, or you can even pick some other algebraic integer factor of v+1, and apply Magidin's argument.
That creates the infinite descent, which would force all of the a's to have all the factors of v^3 + 1.
To break it, Magidin needs some algebraic integer factor of v^3 + 1 that is shared by ONLY TWO of the a's.
However, how do you differentiate my choice of f, from any other algebraic integer factor of v^3+1?
I'm curious to see if anyone gets my point immediately, or if anyone simply has a gut feeling that it's wrong, which means I'm asking for knee-jerk replies.
Well, I thought you'd like to know of an interesting problem with his position, and I'm curious about whether or not he will acknowledge it or give a resolution.
You see, his position from what I see leads to an infinite descent which causes a contradiction.
To understand that you need to know what his position is, and I'll show it for p=3, which you see get used a lot. Obviously, p=3 is used a lot because it's the simplest case for p odd prime, though my FLT Proof covers
x^p + y^p = z^p, for p odd prime, and not just p=3,
but let's get to more details.
A key expression for p=3 that follows from my approach is
(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),
where a1, a2, a3, b1, b2, b3 are all algebraic integers, as the ring is algebraic integers, and v is an arbitrary integer, while, of course, x, y and z are assumed integer counterexamples to FLT.
(Which means I'm asking "What if…?" and looking to see if that doesn't create a problem, which it does, which proves FLT, which is the point of contention as people like Magidin claim that I haven't found that problem.)
It so happens that I've focused on v+1, which is a factor of v^3+1 as
v^3 + 1 = (v+1)(v^2 - v + 1),
and I've given v a value (it's arbitrary remember) that gives it a prime factor of f, which just means that f could be 5, or 7, or 11 or some other prime number other than 3, as, in general, p itself is excluded, to handle a complication that would arise if x were divisible by p.
And my proof depends on only two of the a's, say a1 and a2, having a factor of f, where things are a little more complicated because the a's are themselves not integers, though they are algebraic integers.
Magidin has disputed that only two of the a's have a factor of f, and has instead claimed that ALL the a's have non unit algebraic integer factors of f, and he has based his claim, among other things on Galois Theory.
However, people seem to forget that my choice of v+1 was from a math perspective arbitrary in that you can consider some other factor of v^3 + 1, or you can even pick some other algebraic integer factor of v+1, and apply Magidin's argument.
That creates the infinite descent, which would force all of the a's to have all the factors of v^3 + 1.
To break it, Magidin needs some algebraic integer factor of v^3 + 1 that is shared by ONLY TWO of the a's.
However, how do you differentiate my choice of f, from any other algebraic integer factor of v^3+1?
I'm curious to see if anyone gets my point immediately, or if anyone simply has a gut feeling that it's wrong, which means I'm asking for knee-jerk replies.
# posted by James Harris @ 11:03 
