> 1a. Re: some questions about algebraic factoring in the field of adjoine

Hello Bernhard, Hello David.

> --- In primenumbers@yahoogroups.com,

> "bhelmes_1" <bhelmes@...> asked:

> > d=4n+2 or d=4n+3

> > and the equation of Pell a^2-d*b^2=1

> > What is the fastest way to solve the equation of Pell

> Posted by: "djbroadhurst" d.broadhurst@... djbroadhurst

> Date: Fri Jun 21, 2013 5:10 pm ((PDT))

> In the case of positive d = 3 mod 4,

> used quadunit(4*d), in Pari-GP,

> and then take powers of this unit,

> which has norm = +1.

More generally, for any given integer d,

if a1^2 - d b1^2 = 1

and

a2^2 - d b2^2 = 1

then

(a1^2 + d b1^2)^2 - (2 d a1 b1)^2 = 1

and

(a1 a2 + d b1 b2)^2 - d (a1 b2 + b1 a2)^2 = 1

Thus if

[a1,b1,d] and [a2,b2,d]

are solutions to Pells equation,

then so are:

[a1^2 + d b1^2,2 d a1 b1,d]

and

[a1 a2 + d b1 b2, a1 b2 + b1 a2,d]

Then by finding, for a given integer d,

the smallest solution in a and b,

such that

a^2 - d b^2 = 1,

it is a simple matter to

repeated use the recursively generated solutions

[a1^2 + d b1^2,2 d a1 b1,d]

and

[a1 a2 + d b1 b2, a1 b2 + b1 a2,d]

to find all the solutions, within some bound on a and b, of the Pell

equation for a given integer d.

Kermit- --- In primenumbers@yahoogroups.com,

Kermit Rose <kermit@...> wrote:

> if

Indeed. Yet it is both clearer and more efficient to

> a1^2 - d b1^2 = 1

> a2^2 - d b2^2 = 1

> then

> (a1^2 + d b1^2)^2 - (2 d a1 b1)^2 = 1

> (a1 a2 + d b1 b2)^2 - d (a1 b2 + b1 a2)^2 = 1

take powers of a unit. I illustrated this for positive

d=4 mod 3, where one takes uses quadunit(4*d),

with positive norm. The extension to other

discriminants is straightforward. Moreover

"quadunit" yields the building block that Kermit

assumed to have been found, by undisclosed methods :-)

David - typo: I meant d=3 mod 4

--- In primenumbers@yahoogroups.com, "djbroadhurst" <d.broadhurst@...> wrote:

>

>

>

> --- In primenumbers@yahoogroups.com,

> Kermit Rose <kermit@> wrote:

>

> > if

> > a1^2 - d b1^2 = 1

> > a2^2 - d b2^2 = 1

> > then

> > (a1^2 + d b1^2)^2 - (2 d a1 b1)^2 = 1

> > (a1 a2 + d b1 b2)^2 - d (a1 b2 + b1 a2)^2 = 1

>

> Indeed. Yet it is both clearer and more efficient to

> take powers of a unit. I illustrated this for positive

> d=4 mod 3, where one takes uses quadunit(4*d),

> with positive norm. The extension to other

> discriminants is straightforward. Moreover

> "quadunit" yields the building block that Kermit

> assumed to have been found, by undisclosed methods :-)

>

> David

> - On Sun, Jun 23, 2013 at 7:34 AM, djbroadhurst <d.broadhurst@...>wrote:

> **

no : the 3rd equation is wrong.

> Kermit Rose <kermit@...> wrote:

> > if

> > a1^2 - d b1^2 = 1

> > a2^2 - d b2^2 = 1

> > then

> > (a1^2 + d b1^2)^2 - (2 d a1 b1)^2 = 1

> > (a1 a2 + d b1 b2)^2 - d (a1 b2 + b1 a2)^2 = 1

>

> Indeed.

>

M.

[Non-text portions of this message have been removed] - On 6/24/2013 12:07 PM,

"Maximilian Hasler"

wrote:> ________________________________________________________________________

(a1^2 + d b1^2)^2 - (2 d a1 b1)^2 = 1

> 1a. Re: Pell Equation

> Posted by: "Maximilian Hasler" maximilian.hasler@... maximilian_hasler

> Date: Sun Jun 23, 2013 12:37 pm ((PDT))

>

>

>> **

>> Kermit Rose <kermit@...> wrote:

>>> if

>>> a1^2 - d b1^2 = 1

>>> a2^2 - d b2^2 = 1

>>> then

>>> (a1^2 + d b1^2)^2 - (2 d a1 b1)^2 = 1

>>> (a1 a2 + d b1 b2)^2 - d (a1 b2 + b1 a2)^2 = 1

>>

> no : the 3rd equation is wrong.

>

> M.

>

should have been

(a1^2 + d b1^2)^2 - d (2 a1 b1)^2 = 1

Kermit