Re: x^2 - Ay^2 =1

sttscitrans@xxxxxxxxx wrote :
On 24 May, 00:51, "Philippe 92" <nos...@xxxxxxxxxxxx> wrote:
sttscitr...@xxxxxxxxx wrote :

On 22 May, 10:08, "Philippe 92" <nos...@xxxxxxxxxxxx> wrote:

It is also one of the methods that Dario Alpern
used to implement his solver for
AX^2 +BXY +CY^2 = N

yes but ...
abstract from Dario Alpern's program : [snipped]
That is Dario Alpern doesn't use the "general" algorithm but the
integer variant, whatever the name you give to it.

It should be on his methods page.

Such low level details are not... the method page says : "find the CF
expansion of", not says how to find this.



sqrt(n^2 +n +1) = an +b + c/n + d/n^2 + ..

The '"integer part" of the expression can be thought of
as an +b the fractional part" c/n + d/n^2 + ..

You can use synthetic division to obtain
1/(c/n + d/n^2 + ) = a'n +b' + c'/n + ...

The coefficients are rational in general.

Yes, that is precisely the problem, we *must* use Q[X] not Z[X]...
this solves some x^2 - A*y^2 = K for some K, from which we can say
nothing about the solutions of equation with K = 1.


The convergents in this CF approximate
in some sense sqrt(n^2+n+1)

[P(n)/Q(n)]^2 a.e. n^2+n+1

It is another matter whether the approximation is good enough to give
x^2 -(n^2+n+1)y^2 = 1 or -1

Yes... and if the approximation is too good, it doesn't work either, it
must be exactly good enough...

May be there is no solutions in polynomials, although for any n there
is allways a solution in Z for x^2 - A*y^2 = 1, with A = n^2+n+1

Regards.

--
Philippe C., mail : chephip+news@xxxxxxx
site : http://chephip.free.fr/ (recreational mathematics)


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