quaternion factorizationgreenspun.com : LUSENET : quaternions : One Thread
Suppose quaternion q=a+bi+cj+dk for integers a,b,c,d is such that |q|=AB for integers A and B. Do there always exist quaternions p and r having integer parts such that |p|=A, |r|=B and either pr or rp equals q?
More generally, if |q|^2=AB do there exist p and r as above with |p|^2=A, |q|^2=B and either pr or rp equal to q or even just equal to some quaternion whose parts are a plus/minus permutation of the parts of q, e.g. q=-b+ci-dj+ak.
-- m.a. freedman (firstname.lastname@example.org), December 24, 2002
Any integer can be represented by the square of four other integers: A = T^2 + X^2 + Y^2 + Z^2, where all numbers are integers, and A is any integer. One can take a quaternion that has all integers p=(T,X,Y,Z), and form this scalar by multiplying out the conjugate:
p p* = (T, X, Y, Z)(T, -X, -Y, -Z) = (T^2+X^2+Y^2+Z^2, 0, 0, 0)
So p p* r r* can represent an abritrary interger product A B with p and r being any permutation of T, X, Y, and Z. I doubt any thing can be said about just p r, but p p* r r* can do the task.
-- Douglas Sweetser (email@example.com), January 18, 2003.
I'd love to see the anwer to this question. The only answer that I've seen posted basically says that you can construct a |q|^2 = AB for any integer AB, but I've yet to see anything that really addresses the question that was asked. So can you always nontrivially factor a quatnernion of integers with composite squared magnitude, or not?
-- Bryan Reed (firstname.lastname@example.org), September 08, 2003.