If it's a recurrence, just return the recurrence relation, and any
initial conditions to that relation (i.e., the same form as the input
to rsolve).

It might be useful to have a helper function that could take such a
relation and generate the nth term of each function, by the way.

Aaron Meurer

On Sun, Jun 30, 2013 at 12:42 PM, Ondřej Čertík <[email protected]> wrote:
> On Sun, Jun 30, 2013 at 3:00 AM, Thilina Rathnayake
> <[email protected]> wrote:
>> Looks like we will run into more and more trouble representing the
>> solutions.
>>
>> The latest is that when solving quadratic Diophantine equation,
>> A*x**2 + B*x*y + C*y**2 + D*x + E*y + F = 0, for the case B**2 - 4AC > 0,
>> when we know a basic solution, all the other solutions can be represented as
>> a recurrence relation.
>>
>> Suppose, we find that X0 = 9 and Y0 = 4 is a solution to the given equation,
>> we can find P, Q, K, R, S, L such that, Xn+1 = PXn + QYn + K and
>> Yn+1 = RXn + SYn + L  where Xn+1 and Yn+1 will also be solutions to the
>> equation given that Xn and Yn are solutions. How do we represent this in the
>> solution? Perhaps as a matrix?
>
> Or maybe just return the P, Q, K, R, S, L in the lowest API solver. How were
> you imagining returning it as a matrix?
>
> Ondrej
>
>>
>> Regards,
>> Thilina
>>
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