## Pell Equation

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• ... Hello Bernhard, Hello David. 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
Message 1 of 5 , Jun 22, 2013
> 1a. Re: some questions about algebraic factoring in the field of adjoine

> > 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

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

> In the case of positive d = 3 mod 4,
> and then take powers of this unit,
> which has norm = +1.

Hello Bernhard, Hello David.

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
• ... 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),
Message 2 of 5 , Jun 23, 2013
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
• typo: I meant d=3 mod 4
Message 3 of 5 , Jun 23, 2013
typo: I meant d=3 mod 4

>
>
>
> 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
>
• ... no : the 3rd equation is wrong. M. [Non-text portions of this message have been removed]
Message 4 of 5 , Jun 23, 2013

> **
> 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.
>

no : the 3rd equation is wrong.

M.

[Non-text portions of this message have been removed]
• On 6/24/2013 12:07 PM, Maximilian Hasler ... (a1^2 + d b1^2)^2 - (2 d a1 b1)^2 = 1 should have been (a1^2 + d b1^2)^2 - d (2 a1 b1)^2 = 1 Kermit
Message 5 of 5 , Jun 24, 2013
On 6/24/2013 12:07 PM,

"Maximilian Hasler"

wrote:
> ________________________________________________________________________
> 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.
>

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

should have been

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

Kermit
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