Harristotelian Logic

A simple exercise with basic quadratics and a quartic shows a problem with some standard views in modern number theory.

Start with

x^2 - 6x + 35 = 0

and solve using the quadratic formula to get

x = (3 +/- sqrt(-26)

and now comes a simple yet clever trick—multiply the last two coefficients of the original polynomial with one solution for x and I'll use x = 3 + sqrt(-26), so I have

x^2 - 6(3+sqrt(-26))x + 35(3 + sqrt(-26)) = 0

where the purpose here is to make sure that BOTH roots must have 3 + sqrt(-26) as a factor, but notice that one root must have 7 as a factor as well, as the last coefficient has 7 as a factor, so this is just a tricky way of forcing one root to have 7 as a factor without forcing them both.

Now I can group my radicals to one side and everything else on the other to get

sqrt(-26)(-6x + 35) = -x^2 + 18x - 105

and square both sides to get

(-26)(36x^2 -420x + 1225) = x^4 - 36x^3 + 534x^2 - 3780x + 11025

which is

x^4 - 36x^3 +1470x^2 - 14700x + 42875 = 0

where at least one root must have 7 as a factor, as long as I got the algebra right. Easier to do this sort of thing with math software.

Assuming that is right now it's just a matter of checking it for reducibility over Q. I only went that far for those of you who might wonder how you'd get to a monic with integer coefficients.

Remember that trivially you have that one of the roots must have 7 itself as a factor because I forced that with

x^2 - 6(3+sqrt(-26))x + 35(3 + sqrt(-26)) = 0

but if that polynomial is irreducible you can just use x = 7y, substitute and find you have a non-monic polynomial.