Harristotelian Logic

I figured out a simplified proof of an important assertion where it's obvious to anyone with a math degree or anyone who has bothered to study much mathematics at all that what I'm talking about is an automorphism(?) I think that's the right word.

So it's an easy proof with well-known ideas, but then I did something wacky.

I brainstormed a bit and said something wrong!!!

And then posters who argue with me, in particular Arturo Magidin and David Ullrich decided to try and exploit the mistake in replies in some of my threads.

Now that is deliberate behavior. Now it's an accident as I brainstorm as I see fit and part of the process is making mistakes as you try ideas, but it's a useful accident because they did reply in those threads showing their contempt for what is mathematically correct.

So what is the argument and what does it have to do with automorphisms (hoping I'm using the right word)?

In an integral domain I have an approach starting with what I want to be a factorization

175x^2 - 15x + 2 = 2(f(x) + 1)*(g(x) + 1)

where of course there is a simple non-monic quadratic on the left with the factorization on the right with functions f(x) and g(x) to be determined.

The next steps go against the gain immediately as I multiply through f(x) + 1 by 2 and multiply both sides by 7, introducing versus dividing away:

7*(175x^2 - 15x + 2) = 7*(2f(x) + 2)*(g(x) + 1)

then I fiddle with the left side some and multiply through g(x) + 1 by 7, still doing things against the grain:

(49x^2 - 14x)5^2 + (7x-1)(7)(5) + 7^2 = (2f(x) + 2)*(7g(x) + 7)

and now I bring in new functions, in a kind of odd way as I work towards a conclusion:

Let

2f(x) = 5a_1(x) + 5

and

7g(x) =5a_2(x)

then substitute to get

(49x^2 - 14x)5^2 + (7x-1)(7)(5) + 49 = (5a_1(x) + 7)*(5a_2(x) + 7)

where now I finally solve for the a's using

a_1(x)*a_2(x) = 49x^2 - 14x

and

a_1(x) + a_2(x) = 7x-1

so the a's are roots of

a^2 - (7x-1)a + (49x^2 - 14x) = 0.

And what do automorphisms have to do with anything?

Well f(x) and g(x) are indistinguishable in the original factorization, so I should be able to shift them around without changing anything.

But the a's are roots of that last quadratic like expression thingy, which has only TWO SOLUTIONS.

So the weird answer is that f(x) = g(x), and you can work things out from there, but most importantly you have that only one root of

a^2 - (7x-1)a + (49x^2 - 14x) = 0

can have 7 as a factor.

But if you solve for the functions using the quadratic formula, you find that a lot is wrapped up in that solution so that it's hard to pull out the actual solution for f(x) and g(x), which is where I made a mistake yesterday and posters worked hard to exploit it as you can see in those threads.

But there is no way that Ullrich and Magidin could have gotten their math Ph.D's and not known about automorphisms or whatever the right word is, so they DELIBERATELY lied in their recent posts trying to exploit my most recent brainstorming mistakes showing their absolute contempt for mathematics.

And to make matters worse, Magidin is supposedly a specialist in Galois Theory where automorphisms I think are kind of important, right?

So his contempt is especially acute. He tried to hide a simple conclusion that follows from the most basic idea in the area he is supposedly a specialist.