Re: x^2 - Ay^2 =1

On 19 May, 10:08, "Philippe 92" <nos...@xxxxxxxxxxxx> wrote:
(supersedes <mn.9a9b7d75de1a712a.22...@xxxxxxx>)

sttscitr...@xxxxxxxxx wrote :

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,

do you think so ?
even in integers it is false : f = 4, GCD(3,5) = 2 ???
GCD is 1 or 2.

What I should have said was
If d is a polynomial that divides f+1 and f-1
then it must divide the difference (f+1)-(f-1) =2
so GCD is -2,-1,1, 2.

f+1 and f-1 can't share any
polynomial factors that aren't units.

OK

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

This suffers the same flaw that Vincenzo's : knowing in advance a
solution you get an equation whose solutions are given, but this is of
no help in solving Pell's equation, that is knowing A find the
solutions.

Yes, I agree. It doesn't help you find solutions in integers.

The problem with polynomial Z[X] is worse as you you can't define an
euclidian division :

There does not have to be a Euclidean algorithm
on Z[sqrt(d)], but the continued fraction method
still works.

you need Q[X], that is polynomial with _rational_ coefficients.
Hence many properties required to solve a Pell's like equation are no more available.

Yes, you would think the cyclic method would work
on polynomoal Pell equations

.

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