Harristotelian Logic

In the ring of algebraic integers the following factorization is not possible:

P(x) = (g_1(x) + 2)*(g_2(x) + 1)

whern P(x) is a non-monic polynomial with integer coefficients and f_1(0) = f_2(0) = 0, if the f's are non-rational with rational x.

Took a while to get to all the conditions but that is water under the bridge.

Note then that given

d_1*d_2*P(x) = (f_1(x) + d_1)*(f_2(x) + d_2)

if f_1(0) = f_2(0) = 0 and the f's have non-rational values with rational x, then it's not possible to divide of d_1 and d_2 in the ring of algebraic integers in general as then you would end up with the impossible factorization in that ring.

Why?

Well, from that factorization you can get to

d_1*d_2*P(x) = (h_1(x) + d_1*d_2)*(h_2(x) + d_1*d_2)

and if you assume functions w_1(x) and w_2(x) where w_1(x)*w_2(x) = d_1*d_2, then you can divide off to get

P(x) = (h_1(x)/w_1(x) + w_2(x))*(h_2(x)/w_2(x) + w_1(x))

and then use

h_1(x)/w_1(x) + w_2(x) = j_1(x) + 2

and

h_2(x)/w_2(x) + w_1(x) =j_2(x) + 2

to get

P(x) = (j_1(x) + 2)*(j_2(x) + 1)

and the impossible factorization proving that you can't use functions to divide off the constants, contradicting the objection that Dik Winter originally produced.

Notice that prior posts on sci.math only failed in terms of conditions as if P(x) is monic—not surprising considering how algebraic integers are defined—then you CAN find the factorization as demonstrated by William Hughes.

To me the difference between me and most of you is that I care about what is true, so I just kept going to get to the final answer while too many of you only care to preserve what you believe without regard to mathematical truth, so you only look for an excuse to hold on.

That makes me a discoverer.

Kind of a neat exercise though any of you who bothered to read my last paper on non-polynomial factorization could have seen what the conditions had to be from the example in it, as you have a non-monic P(x).

But to see that you'd have to care about what is mathematical truth.

You'd have to be a real mathematician.