Wednesday, August 29, 2001
FLT Proof, simple after all
I made an interesting mistake while discussing my proof of Fermat's Last Theorem on this newsgroup when I was overly restrictive with my own argument.
Based on my holding on to some overly restrictive assumptions, I cast doubt on one of my latest arguments, and even believed it was wrong myself, when it was not and the proof is correct, and quite simple.
Explaining the above won't take long, and the explanation will also show why there are no valid mathematical objections to the proof.
Ok, I take x^p + y^p = z^p, and show that
(a1 z^2 + b1 xy)…(ap z^2 + bp xy) = 0(mod x^2 + y^2 + vz^2),
and what's new there are the a's, b's and that v. Well the v is an integer, and the a's and b's are somewhat defined by an expression that I won't show here, since I don't have to, to explain my argument.
What I will add about them is that
a1…ap = v^p + 1 and b1…b2 = -2.
Also, I said somewhat defined because it's the crux of this argument and the relates to the overly restrictive assumption I made. I kept forgetting that the a's and b's weren't completely determined yet.
The explanation is simple. There are p+1 coefficients in the main expression and as you can see 2p unknowns. That means that if you, say, determine all the b's, like b1 = -2, and have all the other b's equal 1, you have now determined the a's.
To highlight this fact, I've kept producing the example
fx^2 + gx + h = (a1 x + b1)(a2 x + b2), with f, g and h integers, and even gave a more detailed example using
x^2 + 4x + 4 = (a1 x + (1+sqrt(3)i)/2)(a2 x + (1-sqrt(3)i/2)
in addition to the (x+2)(x+2) that you're all familiar with.
In the case above I CHOOSE b1 = (1+sqrt(3)i)/2 and that forces everything else.
So, back to
(a1 z^2 + b1 xy)…(ap z^2 + bp xy) = 0(mod x^2 + y^2 + vz^2).
For any factor of x^2 + y^2 - z^2, I can force v^p + 1 to have that same factor by picking v = -1 + mF, where m is some counting number, and F is that factor.
But here's the key point that was the basis of my proof until I forgot what I just told you above and bowed to the negative pressure, my a's aren't determined above, so I can have each one of them have a factor of F as a factor.
Yup, it's that simple because that forces the b's to have the same factor.
Not sure?
Consider 2x + m, where we'll deal with integers for simplicity. Is it possible for that expression to be divisible by 2 if m is not?
That's the lever that I'm using here, but some have essentially claimed that what's true for integers doesn't apply for other numbers.
And I said it was, and what made me really upset is that they then essentially claimed the issue wasn't resolvable, and claimed it was up to me to resolve it if it was.
Ok, I said that I could answer any mathematical objections in this post, and the simple resolution actually goes back to what these people have been claiming as support.
Think about
x^2 + 4x + 4 = (a1 x + (1+sqrt(3)i)/2)(a2 x + (1-sqrt(3)i/2)
again, and consider that letting x = b1/a1 or b2/a2 gives you that familiar 2.
If a1, a2, b1, or b2 is in any way a fraction, however you define that past rational numbers, then then they can be put into a non-fractional form without changing the value of the expression from x^2 + 4x + 4.
Looking at the primary polynomial that fact should be trivial to you, but if you accept it, you have to accept that my proof of Fermat's Last Theorem is correct.
You see, the same kind of fulcrum that I have to make the proof is now being applied to you and your beliefs because of its simplicity.
Will you lie to yourself? Will you lie to the world by posting against me?
Oh yeah, in case you're saying, but what if x^2 + y^2 - z^2 only has 2 as a prime factor, since it has to be even, well then z^2 + y^2 - x^2 would have to be the same way, and so would z^2 + x^2 - y^2.
They can't all have only 2 as a prime factor, so they aren't all integers.
By the way, that fact is supported by the actual values for x,y and z that have been given, which I've talked about so much. Remember x=sqrt(6)-i, y = sqrt(6)+i, and z=sqrt(6), with x^3 + y^3 = z^3? You'll notice that factor of 2 popping up, but no other integer factors, as you will with any such example.
Based on my holding on to some overly restrictive assumptions, I cast doubt on one of my latest arguments, and even believed it was wrong myself, when it was not and the proof is correct, and quite simple.
Explaining the above won't take long, and the explanation will also show why there are no valid mathematical objections to the proof.
Ok, I take x^p + y^p = z^p, and show that
(a1 z^2 + b1 xy)…(ap z^2 + bp xy) = 0(mod x^2 + y^2 + vz^2),
and what's new there are the a's, b's and that v. Well the v is an integer, and the a's and b's are somewhat defined by an expression that I won't show here, since I don't have to, to explain my argument.
What I will add about them is that
a1…ap = v^p + 1 and b1…b2 = -2.
Also, I said somewhat defined because it's the crux of this argument and the relates to the overly restrictive assumption I made. I kept forgetting that the a's and b's weren't completely determined yet.
The explanation is simple. There are p+1 coefficients in the main expression and as you can see 2p unknowns. That means that if you, say, determine all the b's, like b1 = -2, and have all the other b's equal 1, you have now determined the a's.
To highlight this fact, I've kept producing the example
fx^2 + gx + h = (a1 x + b1)(a2 x + b2), with f, g and h integers, and even gave a more detailed example using
x^2 + 4x + 4 = (a1 x + (1+sqrt(3)i)/2)(a2 x + (1-sqrt(3)i/2)
in addition to the (x+2)(x+2) that you're all familiar with.
In the case above I CHOOSE b1 = (1+sqrt(3)i)/2 and that forces everything else.
So, back to
(a1 z^2 + b1 xy)…(ap z^2 + bp xy) = 0(mod x^2 + y^2 + vz^2).
For any factor of x^2 + y^2 - z^2, I can force v^p + 1 to have that same factor by picking v = -1 + mF, where m is some counting number, and F is that factor.
But here's the key point that was the basis of my proof until I forgot what I just told you above and bowed to the negative pressure, my a's aren't determined above, so I can have each one of them have a factor of F as a factor.
Yup, it's that simple because that forces the b's to have the same factor.
Not sure?
Consider 2x + m, where we'll deal with integers for simplicity. Is it possible for that expression to be divisible by 2 if m is not?
That's the lever that I'm using here, but some have essentially claimed that what's true for integers doesn't apply for other numbers.
And I said it was, and what made me really upset is that they then essentially claimed the issue wasn't resolvable, and claimed it was up to me to resolve it if it was.
Ok, I said that I could answer any mathematical objections in this post, and the simple resolution actually goes back to what these people have been claiming as support.
Think about
x^2 + 4x + 4 = (a1 x + (1+sqrt(3)i)/2)(a2 x + (1-sqrt(3)i/2)
again, and consider that letting x = b1/a1 or b2/a2 gives you that familiar 2.
If a1, a2, b1, or b2 is in any way a fraction, however you define that past rational numbers, then then they can be put into a non-fractional form without changing the value of the expression from x^2 + 4x + 4.
Looking at the primary polynomial that fact should be trivial to you, but if you accept it, you have to accept that my proof of Fermat's Last Theorem is correct.
You see, the same kind of fulcrum that I have to make the proof is now being applied to you and your beliefs because of its simplicity.
Will you lie to yourself? Will you lie to the world by posting against me?
Oh yeah, in case you're saying, but what if x^2 + y^2 - z^2 only has 2 as a prime factor, since it has to be even, well then z^2 + y^2 - x^2 would have to be the same way, and so would z^2 + x^2 - y^2.
They can't all have only 2 as a prime factor, so they aren't all integers.
By the way, that fact is supported by the actual values for x,y and z that have been given, which I've talked about so much. Remember x=sqrt(6)-i, y = sqrt(6)+i, and z=sqrt(6), with x^3 + y^3 = z^3? You'll notice that factor of 2 popping up, but no other integer factors, as you will with any such example.
# posted by James Harris @ 08:35 
