Harristotelian Logic

So yeah, you can take the expression

a^3 + 3(-1+xf^2)a^2 - f^2(x^3 f^4 - 3x^2 f^2 + 3x) = 0

and instead of focusing on the a's multiply it out and focus on x like you normally do anyway to get a polynomial P(x).

It will be non-monic where the leading coefficient is f^2.

But, it will also be clear that if 'a' has f as a factor, the f^2 will divide off, giving you a monic polynomial, and thus, an algebraic integer value for x.

So you can map algebraic integer a to algebraic integer x that way.

You can also do that with an a coprime to f.

But you cannot map algebraic integer a to algebraic integer x if a does not have f as a factor nor is it coprime to it.

So it's easy to check me when you have the key.

I like that way of explaining which I just thought of yesterday to handle an objection from a poster.

It's oddly simple: Focus not on the a's but on solving x in terms of the a's, and then it all just falls into place so easily, you wonder how people could argue about it for years.

I am a bit curious to see if anyone will keep arguing at this point, and I'll check tomorrow.

But remember what I said about brilliance is as brilliance does, and also remember that certain people are getting serious gain from the world thinking that mathematical ideas I can show are false are correct, but their gain is at the expense of others, especially their students who are being taught and tested on the bogus stuff.

If that continues with the information readily available, then the natural thing for world society to do when it does figure it out, is to erase the gains and give penalties.