Re: x^2 - Ay^2 =1

Vincenzo Librandi wrote :
Philippe wrote:

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


Repeat:
For Pell's equation x^2-Ay^2=1 if is knowed A and only A.

Hi, Vincenzo,

No, not given A, but given a specific form of polynomial P(n),
you get from n both A=P(n) and a solution X(n), Y(n).
And your polynomials have been choosen such as it works, not to give
some value of A given in advance.

We are waiting for your polynomials giving A = 61, 109 or 421...

In this change, how in any my other's case, A is gived how form
polynomial:

A=9n^2-n with n>1
Y=216n-12
X=648n^2-72n+1
===================
A=9n^2+n
Y=216n+12
X=648n^2+72n+1 (with n>1)

This is again false, they don't give the _fundamental_ solutions
(hence they are useless). I'm too lazzy to correct,
the CF expansion period seems to be 2 and 4, hence calculations
should be not too hard.


Trovero tutte le forme polimoniali di A;
anche quelle che legano i fatidici A=109,
or A=421, o anche ad altri A che sono numeri
primi e che hanno soluzioni di X anche con
centinaia di migliaia di cifre.

Per cio che attiene le primitive vi ho
fatto vedere come si ottengono da una semplice scomposizione.

Excuse me for language.

I tried to translate this through Google...
then a little bit arranged, as direct translation was a mess.


I will find all the polynomial forms of A.


Good luck, endless task...


Also those that are related with the fatidic A=109, or A=421,


I understand that you intend to find the solutions for
A = n^2 + n + 1 for instance ?
Greeat...

All the polynomial for A you have given, up to now, are very
specific ones, for which you know the solutions in advance.
And you never prooved they give the fundamental solutions for that A.
(and even several examples where they don't, see two more above)


or also to other A that are prime numbers
or that give solutions for X with even hundred of [thousands ?]
of digits.


And sure there are endless specific cases.
That's why we have a _method_ that works perfectly well to
solve any given Pell's equation (Lagrange's algorithm)...
a method, not an endless list of cases.

And A being prime or not is of little concern here, as you just can't
deduce the solutions of x^2 - (p*q)*y^2 = 1 from those of
x^2 - p*y^2 = 1 and x^2 - q*y^2 = 1.


In what concerns the primitive, I have to say
they are obtained from a simple decomposition.


No they are not.
Also I'm not quite sure you don't mix up 'prime' and 'primitive',
or may be it's Google...

Solving a Pell's equation is finding it's fundamental solution,
that is the one from which all others result, for this value of A.
That is also the smallest >0 solution for this A.
This can't be obtained from any 'simple decomposition' of whatever.

I give you a solution, which is may be or not the fundamental one,
of this equation :
x^2 - 21*y^2 = 1
X = 665335, Y = 145188
Could you say if this solution is fundamental or not ?
Could you find from this the fundamental solution ?
show me your 'simple decomposition'...

Another example
x^2 - 85*y^2 = 1
X = 285769, Y = 30996
same question.

Regards.


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


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