> Since the OP's problem has no inequalities (such as requiring that all > integers in question are non-negative), it is solved by using Hermite normal > form. > > If A is an m by n integer matrix, the Hermite normal form of A is an upper > triangular integer matrix H (also m by n), along with an m by m integer > matrix of determinant 1 (i.e. it's invertible) so that H = U A. So the > original equation A x = b becomes (after multiplying by U): H x = U b, for > which it's easy to get the general solution (just back solve. If you ever > get a non-integer by dividing there are no solutions). Since U is > invertible, multiplying the equation by U doesn't introduce spurious > solutions.
**Applause** Well... Thank you for teaching me that :-) Nathann -- You received this message because you are subscribed to the Google Groups "sage-support" group. To post to this group, send email to sage-support@googlegroups.com. To unsubscribe from this group, send email to sage-support+unsubscr...@googlegroups.com. Visit this group at http://groups.google.com/group/sage-support?hl=en.
