Monday, March 02, 1998
Before I write FLT proof one more post
Because it'll take me a bit of time to write a satisfactory proof (assuming I can do so), I thought I'd make one more post outlining the method and the evidence that makes me so certain now that I have proven Fermat's Last Theorem.
As I've stated before, x^p + y^p = z^p can be rewritten as
(x+y-z)^p = p(z-x)(z-y)(x+y)Q
I have to introduce the Q term because of the complexity of the general expansion. Also, I continue to use a positive z because I get signs confused when I use
x^p + y^p + z^p = 0
Now, it's obvious enough in the above that the right and left must be divisible by 2 raised to the same power. Let's assume that x is even. Then (z-y) must be divisible by 2 raised to a power that is divisible by p.
Now then, what if I write
(x+ay^b - z)^p = y^p[(ay^{b-1})^p - 1] + p(z-x)(z-ay^b)(x+ay^b)Q{a,b}
It should be obvious to anyone that I can find an 'a' and 'b' such that (z-ay^b) is divisible by 2 raised to the same power as (z-y) i.e. if (z-y)=2^s F then I can have (z-ay^b)=2^s G
Well that's all well and good unless I tell you that I can make y^p[(ay^{b-1})^p - 1] also divisible by 2^s. (The sum of the two terms is then divisible by 2^{s+1})
To do so, all that is necessary is that I show that x must be divisible by a 2 raised to a higher power than y-1 or z-1 are. i.e. if y-1 is divisible by 4 and not 8 then x must be divisible by at least 8.
And that is all that is necessary to prove Fermat's Last Theorem.
Of course, n=2 must not be invalidated by the above.
Consider, x^2 + y^2 = z^2 which is the same as (x+y-z)^2 = 2(z-x)(z-y)
Using the same method as above, I'd have
(x+ay^2 - z)^2 = y^2[(ay)^2 - 1] + 2(z-x)(z-ay^2)
In no case is (y-1) allowed to be divisible by a higher power of 2 than x unless (z-y) is divisible by 2 and not by 4 (a possibility not open to p odd prime).
For instance, 13^2 + 84^2 = 85^2. Notice that 13-1 is divisible by 4 at most as is 84.
Why is this? To prevent what proves FLT. If the even term is divisible by a greater power of two than the odd minus 1 then there comes a gap through which a contradiction must follow.
Astute readers can quickly figure out why. I would just as soon explain it in detail but it would make too long of a post. That's where all the legwork I've done over the past couple of years comes in. Basically, all I do is show that z+x=(z-y)+(x+y) trivial as it may seem helps prove that the even term must be divisible by a greater power of 2 than either odd minus 1. And that by writing x^p = (z-y)(z^{p-1}+…+y^{p-1}).
That much is simple and hopefully many of you understand it from that. In a complete proof it's necessary to fill in the details.
Last thing. Above I have an 'a' and 'b' which I say can be chosen such that (z-ay^b) is still divisible by the same powers of 2 as (z-y) as is [(ay^{b-1})p -1]. To do so, I only need b = 2^s where s>t where t is (z-y)=2^t F.
That actually makes this post a proof but not in the form most mathematicians would consider complete. Oh well. I guess that means I have work to do.
As I've stated before, x^p + y^p = z^p can be rewritten as
(x+y-z)^p = p(z-x)(z-y)(x+y)Q
I have to introduce the Q term because of the complexity of the general expansion. Also, I continue to use a positive z because I get signs confused when I use
x^p + y^p + z^p = 0
Now, it's obvious enough in the above that the right and left must be divisible by 2 raised to the same power. Let's assume that x is even. Then (z-y) must be divisible by 2 raised to a power that is divisible by p.
Now then, what if I write
(x+ay^b - z)^p = y^p[(ay^{b-1})^p - 1] + p(z-x)(z-ay^b)(x+ay^b)Q{a,b}
It should be obvious to anyone that I can find an 'a' and 'b' such that (z-ay^b) is divisible by 2 raised to the same power as (z-y) i.e. if (z-y)=2^s F then I can have (z-ay^b)=2^s G
Well that's all well and good unless I tell you that I can make y^p[(ay^{b-1})^p - 1] also divisible by 2^s. (The sum of the two terms is then divisible by 2^{s+1})
To do so, all that is necessary is that I show that x must be divisible by a 2 raised to a higher power than y-1 or z-1 are. i.e. if y-1 is divisible by 4 and not 8 then x must be divisible by at least 8.
And that is all that is necessary to prove Fermat's Last Theorem.
Of course, n=2 must not be invalidated by the above.
Consider, x^2 + y^2 = z^2 which is the same as (x+y-z)^2 = 2(z-x)(z-y)
Using the same method as above, I'd have
(x+ay^2 - z)^2 = y^2[(ay)^2 - 1] + 2(z-x)(z-ay^2)
In no case is (y-1) allowed to be divisible by a higher power of 2 than x unless (z-y) is divisible by 2 and not by 4 (a possibility not open to p odd prime).
For instance, 13^2 + 84^2 = 85^2. Notice that 13-1 is divisible by 4 at most as is 84.
Why is this? To prevent what proves FLT. If the even term is divisible by a greater power of two than the odd minus 1 then there comes a gap through which a contradiction must follow.
Astute readers can quickly figure out why. I would just as soon explain it in detail but it would make too long of a post. That's where all the legwork I've done over the past couple of years comes in. Basically, all I do is show that z+x=(z-y)+(x+y) trivial as it may seem helps prove that the even term must be divisible by a greater power of 2 than either odd minus 1. And that by writing x^p = (z-y)(z^{p-1}+…+y^{p-1}).
That much is simple and hopefully many of you understand it from that. In a complete proof it's necessary to fill in the details.
Last thing. Above I have an 'a' and 'b' which I say can be chosen such that (z-ay^b) is still divisible by the same powers of 2 as (z-y) as is [(ay^{b-1})p -1]. To do so, I only need b = 2^s where s>t where t is (z-y)=2^t F.
That actually makes this post a proof but not in the form most mathematicians would consider complete. Oh well. I guess that means I have work to do.
# posted by James Harris @ 21:29 
