Harristotelian Logic

For me there have been two perspectives as I work to figure out how to explain the definition problem in mathematics with LOTS of opposition, and I wonder about mathematicians so dedicated to attacking an argument that is clearly correct.

I remind of that as I present what should finish their ability to distract, as I've seen a strange and dedicated effort to ignore the actual math, and simply toss up just about anything rather than face the truth.

All variables are in the ring of algebraic integers unless otherwise stated.

Let

P(m) = f^2((m^3 f^4 - 3m^2 f^2 + 3m) x^3 - 3(-1+mf^2 )x u^2 + u^3 f)

and let

R(m) = (m^3 f^4 - 3m^2 f^2 + 3m) x^3 - 3(-1+mf^2 )x u^2 + u^3 f

so P(m) = f^2 R(m).

Now consider

P(m) = (a_1 x + uf)(a_2 x + uf)(a_3 x + uf)

where the a's are given by the following cubic:

a^3 + 3(-1+mf^2)a^2 - f^2(m^3 f^4 - 3m^2 f^2 + 3m).

Then it must be true that the following factorization exists

R(m) = (b_1 x + u)(b_2 x + u)(b_3 x + uf)

where the b's are given by the following cubic:

b^3 + 3(-1+mf^2)b^2 - (m^3 f^4 - 3m^2 f^2 + 3m),

where a_3 = b_3, and at m=0, b_3 = 3.

PROOF:

From

R(m) = (m^3 f^4 - 3m^2 f^2 + 3m) x^3 - 3(-1+mf^2 )x u^2 + u^3 f

at m=0, R(0) = u^2(3x + uf), and the cubic for the b's

b^3 + 3(-1+mf^2)b^2 - (m^3 f^4 - 3m^2 f^2 + 3m),

gives b^3 - 3b^2 = 0, which gives that two of the b's are 0, while the third is 3.

That gives

R(0) = (0 x + u)(0 x + u)(3 x + uf) = u^2( 3x + uf)

as required.

No other primitive (non-primitive is like 2x+2 which has a factor of 2) cubic has the required roots, so the cubic

b^3 + 3(-1+mf^2)b^2 - (m^3 f^4 - 3m^2 f^2 + 3m)

is the correct one.

Then it follows that two of the a's have a factor that is f.

However, it is possible to show that for certain integer values of f, and integer values of m, the a's do not in the ring of algebraic integers have f as a factor, which is a contradiction.

Mathematicians can continue to run from the truth, but they must keep attacking algebra itself to do so.