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

> --- 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.

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