Pell's equation

[Michael Beeson]

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

[Vincent Delecroix]

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 <profb...@gmail.com>:

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[John Cremona]

I was not sure whether your question was mathematical or
computational. Usually "Pell equation" means the +1 equation and ones
says "negative Pell equation" for the -1 version. (I once wrote a
paper "On the negative Pell equation"!). As Vincent says if you have
a solution x+y*sqrt(d) to the negative equation, square it to get a
solution to the positive equation.

But perhaps you just meant that the instructions were confusing?

John Cremona

[Michael Beeson]

This question was simply a result of my misreading the example on page 93, where it says

"We first solve Pell's equation x2 􀀀 5y2 = 1 with d = 5 by nding the units of

the ring of integers of Q(sqrt(5) using Sage."  Of course I should have realized

that just finding those units will give me the solutions of x^2 - 5 y^2 = plus or minus 1, not just 1.

I'm sorry for taking your time to point that out.