Saturday, September 11, 1999
JSH: Why I win.
I thought I'd add a bit more of the math side in this post. Some of you are about to have one of those "aha" experiences. Many of you will undoubtably still not have a clue and I hope it's all of you who've been endlessly criticizing me.
My "f polynomial" has an exponent of p(p-1). I treat it as having an exponent of (p-1) and look at roots to the p exponent.
What I've demonstrated in my current proof is the method for reducing this polynomial to one that has (p-1)/2 roots.
These roots are the two number combinations of the f^p roots.
For instance, with p=5, I get a polynomial with roots ab,cd
where ab are the reals and cd are the nonreals (a convention I've been using all along).
All well and fine for p=5, but for p>5 there are more than (p-3)/2 nonreal two number combinations (some of you are getting that "aha").
Since the calculations I use don't determine which of those I'm using, it shows that they are all equal. That is, the sums are equal. Ok, it's easier to show with an example.
e.g. for p=7, I start with a polynomial having 6 roots and reduce to one with 3, so I must have
cd + ef = ce + df = de + cf
(Now the lightbulbs should be really going off.)
With this alone I can produce a contradiction with p=7 because it forces
sqr(u^2 + y^2) to be an integer, but I needed something better for the general case, so I went to a modular relation.
The form of the modulus for the first coefficient of the reduced polynomial is easy to figure out. I'll think about explaining it, so that the necessary "rigor" is on my webpage.
It's been tons of fun folks. I think this has to be the greatest adventure of my entire life. And, I never would have guessed that it'd all end this way though.
I'm deeply appreciative of all the help.
I promise to be as forgiving as I can of the endless criticism over the past three plus years (can you imagine?) that I've received from a loud minority of you, but I won't forget.
My "f polynomial" has an exponent of p(p-1). I treat it as having an exponent of (p-1) and look at roots to the p exponent.
What I've demonstrated in my current proof is the method for reducing this polynomial to one that has (p-1)/2 roots.
These roots are the two number combinations of the f^p roots.
For instance, with p=5, I get a polynomial with roots ab,cd
where ab are the reals and cd are the nonreals (a convention I've been using all along).
All well and fine for p=5, but for p>5 there are more than (p-3)/2 nonreal two number combinations (some of you are getting that "aha").
Since the calculations I use don't determine which of those I'm using, it shows that they are all equal. That is, the sums are equal. Ok, it's easier to show with an example.
e.g. for p=7, I start with a polynomial having 6 roots and reduce to one with 3, so I must have
cd + ef = ce + df = de + cf
(Now the lightbulbs should be really going off.)
With this alone I can produce a contradiction with p=7 because it forces
sqr(u^2 + y^2) to be an integer, but I needed something better for the general case, so I went to a modular relation.
The form of the modulus for the first coefficient of the reduced polynomial is easy to figure out. I'll think about explaining it, so that the necessary "rigor" is on my webpage.
It's been tons of fun folks. I think this has to be the greatest adventure of my entire life. And, I never would have guessed that it'd all end this way though.
I'm deeply appreciative of all the help.
I promise to be as forgiving as I can of the endless criticism over the past three plus years (can you imagine?) that I've received from a loud minority of you, but I won't forget.
# posted by James Harris @ 13:03 
