Hello,

First of all, I was not able to run your code. It fails on the line

 sage: L = [list(u^i) for i in [0..3]]

Hopefully, with list(K(u^i)) instead of list(u^i) it works fine.

I did not check the reference but the units of a quadratic number
fields are the solution of Pell equation with either 1 or -1 (which
are the invertible elements in Z). And actually, the subgroup of
solutions which corresponds to 1 always form a subgroup (which is
either the whole group or a subgroup of index 2). As you can see with
the same example, taking the square of a solution with -1 you get 1
because the norm is multiplicative

sage: x,y = K(u)
sage: x^2-1621*y^2
-1
sage: K(u).norm()
-1

sage: x,y = K(u)**2
sage: x^2-1621*y^2
1
sage: K(u**2).norm()
1

I am not sure it solves the issue from the reference.

Vincent

2014-10-31 10:14 UTC−06:00, Michael Beeson <profbee...@gmail.com>:
> Here I attempt to solve Pell's equation with d = 1621 following the method
> on page 93 of Stein's book.
> But the solution produced is instead a solution of the negative Pell
> equation x^2-y^2 = -1  (instead of 1).
> Actually, the example on page 93 (after correcting the typo "v" to "u") has
>
> the same problem:  it claims
> that [-2,1]  solves Pell's equation with d=5,  whereas, it really solves
> the negative Pell equation.
>
> sage: K.<a> = QuadraticField(1621)
> sage: G = K.unit_group()
> sage: u = G.1
> sage: L = [list(u^i) for i in [0..3]]
> sage: L
> [[1, 0], [4823622127875/2, 119806883557/2], [23267330432525342852015627/2,
> 577903134597288688851375/2], [56116404965454319198851772383057215250,
> 1393793173905903098261469193463230841]]
> sage: x = L[2][0];
> sage: y = L[2][1];
> sage: x
> 23267330432525342852015627/2
> sage: x = L[3][0];
> sage: y = L[3][1];
> sage: x
> 56116404965454319198851772383057215250
> sage: y
> 1393793173905903098261469193463230841
> sage: x^2-1621*y^2
> -1
>
>
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