Math question - factoring a bivariate polynomial

[Max A.]

On 10/18/06, Andrew Plewe <aplewe at sbcglobal.net> wrote:

> Thanks for your reply. My only issue here is that D gets to be too large to
> factor.

The trouble here is that finding roots of your bivariate polynomial is
_equivalent_ to factoring (well, may be partial) of D.
If you somehow got a root (x,y) of your polynomial then you
immediately have a representation of D as the product of two factors:
D = (2x + b) * (4y - 2x + 2a + b)

> I think it may be easier to solve the system of bivariate polynomial

No, it is not easier (unless you implicitly come up with a new
factorization algorithm which is quite unlikely).

So, a straightforward factorization of D is more preferable in the
problem you are solving. Even for large D, it is possible to determine
its up to 60-digit prime factors rather quickly (thanks to the ECM
factorization algorithm). Take a look at
http://www.loria.fr/~zimmerma/records/ecmnet.html

Max