Re: x^2 - Ay^2 =1

On 17 May, 18:41, "Philippe 92" <nos...@xxxxxxxxxxxx> wrote:
sttscitr...@xxxxxxxxx wrote :

On 17 May, 11:46, Vincenzo Librandi >

Yes, but you are not solving for A in
x^2-Ay^2 =1

The difficult question is if A = 23332322391

The fundamental solution of this one has 4648 digits.
(found in 1s by Dario Alpern's solver)

or 109

The one given by Fermat to Frenicle in 1657, hence this one can be
solved by hand calculation...

Yes, but it is the fact that there exists an algorithm
that solves x^2-Ay^2 =1 that is important.

I suppose you could use the "Vincenzo method"
on a polynomial identity version of Pell's Equation

The f,g,h etc are polynimials with integer coefficients

f^2-gh^2 =1
(f+1)(f-1) = gh^2
As GCD(f+1,f-1) =2, f+1 and f-1 can't share any
polynomial factors that aren't units.

f-1 = pq^2
f+1 = pq^2 +2 = rs^2 ?

So
[pq^2+1]^2 - ((pq)^2 +2p)q^2 =1

So this is a solution for A= (pq)^2 +2p

Is there a solution in non-trivial polynomials with
integer coefficients for rs^2-pq^2 = 2 ?
Then

[pq^2+1]^2 - (pr)(qs)^2 =1


.

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