Re: x^2 - Ay^2 =1

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

On 22 May, 10:08, "Philippe 92" <nos...@xxxxxxxxxxxx> wrote:
If you say that... I don't have this book.
and yes, _binary form_
not "any real number".

Yes, but the point is that you can solve say

x^3 - 19y^3 =-1
without leaving the integers, by transforming
the equation.

You can arrive at the same result by expanding
r = cubrt(19) as a CF.

Of course you do not get an infinite number
of solutions from 8+3r as the powers of 8+3r
are ternary, not binary units. So a ternary
form F(x,y,z) with F(x,y,z)F(x',y'z') = F(x'',y''z'')
would have an infinite number of solutions.



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 :
That is Dario Alpern doesn't use the "general" algorithm but the
integer variant, whatever the name you give to it.
Variant which is suitable only for quadratic numbers.
(that is any number in Q(sqrt(d)), for any d>0 not a square)

It should be on his methods page.

(the original problem was to solve the normal Pell's equation
x^2 - A*y^2 = 1 in Z, using A, x, y = polynomials in n.
For any n, p(n) being an integer, p is in Z[X], not Q[X].
Searching for solutions in Q[X] is equivallent to searching for
_rational_ x, y in a Pell's equation...)

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.

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




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