Harristotelian Logic

Well, it looks like there was this easy enough resolution, which oddly enough can be said to rely on the impossibility of an algebraic integer being the root of a non-monic polynomial with integer coefficients irreducible over Q.

Starting with the simple polynomial

P(x) = 175x^2 - 15x + 2

I multiplied by 7 and did my creative thing with the terms to get

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

to factor it as

7*P(x) = (5a_1(x) + 7)(5a_2(x)+ 7)

where the a's are roots of

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

But I finally got around to just dividing that damn 7 off, to get

P(x) = (7x^2 - 2x)5^2 + (7x-1)(5) + 7

giving the factorization

P(x) = (5b_1(x) + w_1)(5b_2(x)+ w_2)

sticking in b's and w's, and with the 5's still around I could work out that

b_1(x)*b_2(x) = 7x^2 - 2x,

w_1*b_2(x) + w_2*b_1(x) = 7x-1

and w_1*w_2 = 7

so I could so some substitutions and I chose to focus on b_1(x) and w_1 to get

7*b_1(x)^2 - (7x-1)w_1*b_1(x) + (7x^2 - 2x)w_1^2 = 0

which is a non-monic polynomial—unless w_1 has 7 as a factor!!!

And with non-zero integer x, it is irreducible over Q, proving that b_1(x) can't be an algebraic integer, unless w_1 has 7 as a factor.

That is a nice absolute relying on the very result that has often been used against my assertion that 7 is properly a factor of only one root as, hey!!! NOTHING but 7 as a factor can even give you algebraic integer functions if you divide off that pesky 7 from both sides of

7*P(x) = (5a_1(x) + 7)(5a_2(x)+ 7).

Try any w_1 other than one with 7 as a factor with x a non-zero integer, and b_1(x) cannot be an algebraic integer function as it is then the root of a non-monic polynomial with integer coefficients irreducible over Q.

Damn strange, but it is definitive.

Astute readers may notice that with w_1=7, you just get back a solution for the a's.

Also you may notice this approach doesn't deal with the issue of what is the factor of the a's, it just shows they can't have algebraic integer factors, since 7 itself can't be a factor of just one of them, bringing into question claims that you can find algebraic integer factors from 7 for both.

My guess is that old problem in mathematics where when you have false assumptions you can appear to prove anything, when really you don't have proof.

The ring of algebraic integers gives you all kinds of problems and weirdness.