# [seqfan] Re: Help with sequence

andrew at nevercenter.com andrew at nevercenter.com
Tue Aug 25 20:35:23 CEST 2009

```Quoting Ignacio Larrosa Cañestro <ilarrosa at mundo-r.com>:
> n = y^2 - a^2x^2 = (y + ax)(y - ax)
>
> and reduces to a system of linear equations
>
> y + ax = d1
>
> y - ax = d2
>
> where d1*d2 = n, and easy to solve.

Yes, this is assuming you know the factors of n. I'm curious, though,
if something along these lines could be used to "profile" a^2*x^2,
without having to directly factor n.

If, via the Hasse-Minkowski theorem, you can show that a^2*x^2 + n =
y^2 has solutions for some set A = {a_1, a_2, ... a_n}, where a_i is
prime, then these would be very good candidates for inclusion in your
factor base when running the quadratic sieve algorithm. For values of
a where there are no solutions, there's a simple way to extend the
approach so that you can generate a bunch of linear congruences of the
form x == r_j mod (a_i), which can be joined to make a sort of sieve
to narrow down possible values for x.

```