Re: x^2 - Ay^2 =1

On 25 May, 11:13, Gerry <Gerry...@xxxxxxxxx> wrote:
On May 24, 4:25 pm, "sttscitr...@xxxxxxxxx" <sttscitr...@xxxxxxxxx>
wrote:

On 24 May, 05:00, Gerry <Gerry...@xxxxxxxxx> wrote:

On May 24, 3:19 am, "Philippe 92" <nos...@xxxxxxxxxxxx> wrote:

Gerry wrote :

<snip unrelated quoting>
I don't see why one equation would be easier
to solve than another.

If x^2 -Ay^2 = 4 , gcd(x,y) = 1 and 4 <sqrt(A)
the solution, if there is one, would be given by the
convergents of the CF expansion of sqrt(A)- Hide quoted text -


i guess it is a matter of efficiency for sure the (d,x,y)
solutions are a lot smaller versus the (D,X,Y) solutions
and i believe these also
can be found by the method of Vincenzo.

How do you know what squares to try ?

also i thought that there always exists a solution for
x^2 -Ay^2 = 4 or am i wrong?

There is always the trivial solution (2,0,A) =4
but this is just a multiple of the even more
trivial solution (1,0,A) = 1.

(x,y,7) =4 , gcd(x,y) =1
means that
(x,y,7) = 0 mod 4
(x,y,3) = 0 mod 4

As the squares mod 4 are 0,1
(x,1,3) = 0 mod 4
but this equation has no solutions.

Best rational approximations x/y to sqrt(A)
solve (x,y,A) =1 (-1), less good approximations
are needed to solve (x',y',A)= 20 so
generally speaking y > y'.

If (x,y,A) = 2 =>
(x,y,A)(x,y,A) = 4 =>
(xx +Ayy,2xy,A) = 4 =>
xx+Ayy = 2m
(2m,2xy,A) =4
(m.xy,A) =1

But you still have to find this solution for a given A

.

Quantcast