Clarification on algebraic integer solution existence claims

Can be very difficult to escape error in a mathematical argument, and have realized that some earlier posts talking a generalized quadratic factorization were in error with regard to claims about existence of solutions in ring of algebraic integers. Have corrected more than one post and may have missed others, so will also have this post to clarify. Actually had corrected to current thinking by November 2015, but didn't realize needed to update prior posts until today. Guess just forgot was in them. It happens.

The generalized factorization is in complex plane:

P(x) = (g₁(x) + 1)(g₂(x) + 2)

where P(x) is a primitive quadratic with integer coefficients, g₁(0) = g₂(0) = 0, but g₁(x) does not equal 0 for all x.

Introduced a mathematical approach which involves multiplying both sides by an integer k, which I choose to not be a unit in an important proof which is correct.

Introduce k, where k is a nonzero integer, and new functions f₁(x), and f₂(x), where:

g₂(x) = f₂(x) + k-2 and g₁(x) = f₁(x)/k,

multiply both sides by k, and substitute for the g's, which gives me the now symmetrical form:

k*P(x) =  (f₁(x) + k)(f₂(x) + k)

However, for a bit I forgot that detail of my choice with at least one post, which I've updated, but had further flawed thinking, and in the past claimed that the g's could not both be algebraic integers in general when non-rational. Which seemed so dramatic to me! And was wrong. Which again reminds me, emotion and math? Do not mix well.

Have now noted that they both are when k = 1 or -1, with approach for finding solutions for them that I discovered.

Is important enough should note have already posted full resolution explanation, which have had since 2015 but hadn't so elaborated, and will include pertinent information here for what happens if you try to force things with a different non-unit k using g's that are algebraic integers:

The algebra doesn't care, but will NOT let the f's be roots of the same quadratic with integer coefficients then with k not 1 or -1. They will disjoint from each other. What is key is the process where they are roots of the same quadratic with integer coefficients. As far as the math is concerned, you simply have one of the f's multiplied by k, so its removal to get one of the g's is as easy here as anywhere else.

So yeah, my thinking before was off, and actually discovered by looking at a case, and going, hey. Am SO lucky could find a proof with quadratics, as you can just put in some numbers and see an actual case. Then realized that yeah, with quadratics when you have something like 1+sqrt(3) and 1-sqrt(3), you need them to be the roots, for a quadratic with integer coefficients. Fiddle with one, and you disjoint the results, like add something to one, like 2, and have 3+sqrt(3) and 1-sqrt(3), and end up with roots at best of the same quartic with integer coefficients. Guess may as well fully explain how I noticed something actually kind of simple.

So realized it by doing the above. Looking at an actual case and then looking at the f's that resulted and seeing they couldn't be roots of the same quadratic with integer coefficients. Then it's like, duh.

My apologies for the error. If there are posts which still have it, please disregard. Or if you point them out will correct. And am still checking. Problem with something wild is that urge to plaster over a lot of posts, which have learned to try NOT to do.

Oh, and how did I notice finally had prior posts which were not updated to my current view? The usual, as was checking Blogger stats noted interest in a particular post. When looked at it to see why, realized mistake was there while reading through. Is a cool process. Not sure if was deliberate help, but in case it was? Thanks!!!

So yeah still rely on constant communication through the web, though now is primarily indirect. Am constantly getting feedback based on which posts draw interest, as well as from country stats as get an idea of who is paying attention. Is a daily thing. Now yeah, things are more, popular lately.

My earlier claim was that the algebra was blocking that form for algebraic integers, and that is in error.

This post is simply notice of that past mistake and that it has been corrected in current arguments.


James Harris