Z(t,t^-1) solutions to a(t)p(t)+b(t)q(t)=1
magi
with integer coefficients. I am looking for Laurent polynomials with
integer coefficients for which a(t)p(t)+b(t)q(t) = 1.
Over rationals, not integers, this could be unambiguously
determined by the GCD algorithm. However, in integers, the GCD
algorithm will not work (example: p(t) = t+1, q(t) = 3).
Can someone tell how to find (a) if integer coefficient Laurent
polynomial solutions exist, and (b) what they are? Thanks.
Gerry Myerson
wrote:
If you run what you call the GCD algorithm (what I'd call the Euclidean
algorithm), and it doesn't work, isn't that proof that solutions don't
exist?
--
Gerry Myerson (ge...@maths.mq.edi.ai) (i -> u for email)
magi
wrote:
> In article <i5lkh0$k4...@news.acm.uiuc.edu>, magi <zbx.m...@gmail.com>
I think not. Consider p(t) = t^2 + 5t + 1 and q(t) = 5t + 1. In
integers,
the Euclidean algorithm (thanks for the correction by the way)
couldn't proceed,
but t^(-2) p(t) - t^(-2) q(t) = 1. I think that this problem is a lot
harder than
it looks.
Gerry Myerson
<489343d6-7aff-468d...@q2g2000yqq.googlegroups.com>,
magi <zbx....@gmail.com> wrote:
In this case, the Euclidean algorithm, run over the rationals,
gives you (1) p(t) - ((1/5)t + (24/25)) q(t) = 1/25,
so you multiply by 25 to get
(25) p(t) - (5 t + 24) q(t) = 1.
But I do accept that there is more to this problem
than meets the eye. If, say, p(t) = 2 t + 1, and q(t) = 2 t + 17,
then there are integer coefficient polynomials a(t) and b(t)
such that a(t) p(t) - b(t) q(t) = 1; however, they have degree 4.
This example is from my paper, On resultants,
Proc Amer Math Soc 89 (1983) 419-420,
and you might find something useful there.