Harristotelian Logic

With

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

and

7g(x) = 5a_2(x)

along with

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

it is reasonable to suppose that a_2(x) has 7 as a factor, but critics who have argued with me for some time over approaches like this have continually maintained it does not in the ring of algebraic integers, and for that to be true g(x) must not be an algebraic integer.

With my generalized solution:

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

giving the a's as roots of

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

that claim can be rigorously tested, as you can simply shift to

7f(x) = 5a_2(x)

and

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

so that if g(x) is not an algebraic integer as the critics need then a_1(x) is not either, which forces Q(x) to not be, so assume instead you have Q'(x)/w(x) where w(x) is some factor of 7.

Then you'd have

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

which solves as

a(x) = (3x - 1 + 5Q'(x)/w(x) +/ sqrt((3x - 1 + 5Q'(x)/w(x))^2 - 4(49x^2 + 7Q'(x)/w(x)))/2

but you'd still need the radicals to be the same as all you're doing is switching from multiplying one solution by 7, to multiplying the other, while with the other you multiply by 5 and add 5, which can't change the radicals.

But that forces Q'(x)/w(x) to equal the original Q(x), forcing w(x)=1 and

Q'(x) = Q(x)

so you just get back what you had before.

So it's absolute mathematics with perfect argument, otherwise known as a proof, and no room for a mathematical objection.