Harristotelian Logic

Considering continuing debates over the latest simplified approach to non-polynomial factorization I've concluded that some of you got confused on the specificity of the solution I gave for the a's, which actually is a conclusion helped by some responses from mathematicians I emailed.

So here is a generalized approach which removes that area of potential confusion.

This time starting with

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

in an integral domain, begin determining the functions f(x) and g(x) with

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

and

7g(x) = 5a_2(x)

and now for the generalized approach I introduce another polynomial Q(x):

(49x^2+7Q(x))5^2 - (3x-1+5Q(x))(5)(7) + 7^2 = (5a_1(x)+7)*(5a_2(x)+7)

so that I have as solutions for the a's:

a_1(x)*a_2(x) = 49x^2 + 7Q(x)

and

a_1(x) + a_2(x) = 3x - 1 + 5Q(x)

so they must be roots of

a^2 - (3x - 1 + 5Q(x))a + (49x^2 + 7Q(x)) = 0.

That is the generalized solution for the a's rather than a highly specific one, so assume that an f(x) and g(x) exist in the ring of algebraic integers with the start

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

and if such exist, then the a's will be algebraic integers as well, and ALL possible solutions for the a's are defined by

a^2 - (3x - 1 + 5Q(x))a + (49x^2 + 7Q(x)) = 0

meaning variation is just with Q(x), and with the a's algebraic integers it must be true that Q(x) is an algebraic integer; therefore, it must be true that ONE of the roots for a Q(x) that works, will have 7 as a factor, while the other will not.

So what if f(x) and g(x) can't be algebraic integers when x is an integer?

Well, why not? And even if they aren't, at best you can hope they'd be ratios of algebraic integers with a denominator having non-unit factors in common with 2, not 7.

So that wouldn't change anything important.

Now some of you may refuse to acknowledge that there can exist functions f(x) and g(x) that behave as needed but then you're not people who are thinking rationally on the subject.

My original posts used Q(x) = -2x and I think doubt was cast on the specificity of THAT solution as maybe it seemed too arbitrary, while putting in the general solution to cover all possible solutions for the a's no matter what f(x) and g(x) are, should take away any concerns in that area.

So how might some of the people who obsessively argue with me still try to disagree?

Well, considering

a^2 - (3x - 1 + 5Q(x))a + (49x^2 + 7Q(x)) = 0

they'd have to claim that Q(x) does not have integer coefficients if neither f(x) nor g(x) would divide off factors from 7 as they wish, so that they could claim that if you get to a monic with integer coefficients it would be reducible over Q, and they'd have to claim that NEVER when Q(x) has integer coefficients, except when Q(x) = 0, of
course, could you have 7f(x) and 7g(x) both be algebraic integers.

---

Maybe the only way for some of you to understand what is happening and to comprehend how big a mistake it makes in your mind to declare the sqrt() function has only one solution and start down that road to some really big errors is to just figure out a Q(x) that will work with an integer x so that you can SEE the freaking 7 as a factor—once you resolve the square roots.

So with

(49x^2+7Q(x))5^2 - (3x-1+5Q(x))(5)(7) + 7^2 = (5a_1(x)+7)*(5a_2(x)+7)

where the a's are roots of

a^2 - (3x - 1 + 5Q(x))a + (49x^2 + 7Q(x)) = 0

and Q(x) is some polynomial YOU can pick so that with some integer x, you get integer roots, then I want you to do something simple, pause with your solution for the quadratic formula, and consider, what if?

What if you couldn't just resolve that square root?

Now go back to

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

where now you'll have integer values for f(x) and g(x) as well, and just ask yourself, how would you write that in terms of square roots?

So how did this error in thinking with the square roots propagate and flourish?

I say because of simple failure as mathematicians had no idea how to figure out something like one root having 7 as a factor if they didn't have integers with even something as simple as a monic quadratic with integer coefficients.

So they made stuff up.

Maybe it was a lot of ego, or maybe it was just a refusal to accept a limitation that there was something they didn't know with math they thought was so simple, but for whatever reason, they started making stuff up, and part of doing that was declaring the sqrt() had only one solution so they could escape the reality of two solutions and run off into error.

And now holding on to the errors in thinking so that I STILL get in arguments with people declaring that -2 is not a solution for sqrt(4) is just willful stupidity and foolish pride.

So you have all kinds of machinery with Galois Theory that will work the EXACT SAME WAY with integers, if you just refuse to do things like resolve sqrt(25).

If you just use the quadratic formula, and refuse to evaluate the square root when you get integers, everything works the exact same way.