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• ## some questions about algebraic factoring in the field of adjoined square

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• A beautiful day to all, i have tried to express the following ideas as good as possible. Nevertheless it is not perfect :-( Let f:=p*q the number which should
Message 1 of 2 , Jun 12
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A beautiful day to all,

i have tried to express the following ideas as good as possible.
Nevertheless it is not perfect :-(

Let
f:=p*q the number which should be factorized, p and q primes
d:=4n+2 or d=4n+3 with jacobi (d, f)=-1
D:=sqrt (d)
F: a+bD with a, b element Z
and the equation of Pell a²-db²=1

As i know there are always infinite solutions for this equation

1. Question
What is the fastest way to solve the equation of Pell.
Is continued fraction preferable
or is it better to use Hensel lifting
for the polynomial f(b)=db²+1 and to look for a solution
with a²=f(b)

2. Question
Are the solutions of the equation of Pell always useful
for the factoring of f
if you reduce the solutions to a²-db²=1 mod f

you get a1²-db1² = a2²-db2² mod f
if a1=a2 or b1=b2 you get the equation of the form
x^2=y^2 mod f

3. Question
Does it make a difference if you look for a factoring solution in N
or in F(D)

Examples:
5. f=35
d=2
solutions of Pell a²-db²=1 mod 35 :
1, 6, 29, 34, 21+5D, 14+5D, 21+30D, 14+30D
gcd (6-1, 35) = 5
gcd (34-29, 35)= 5
gcd (21, 35) = 7
gcd (14, 35) = 7 and so on

6. f=55
d=3
solutions of Pell a²-db²=1 mod 55 :
1, 21, 34, 54, 44+20D, 44+35D, 11+20D, 11+35D
gcd (44, 55)=11
gcd (20, 55)=5
gcd (11, 55)=11

Perhaps this application helps more for small values f<100
http://devalco.de
I. Primes
4. cycle structur of primes
c) mit adjungierter Wurzel / with adjoined square
Choose p=35 Adjoined square root=2
You find the solutions of Pell by clicking on the |1+A|=1

Nice greetings from the primes
Bernhard
• ... 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. Here is an example with d =
Message 1 of 2 , Jun 21
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> 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

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.

Here is an example with d = 139:

for(n=1,100,r*=q;print([n,a=real(r),b=imag(r)]))}

[1, 77563250, 6578829]
[2, 12032115501124999, 1020550716868500]
[3, 1866499965285267079810250, 158314460780301358671171]
...

NB: At n=100, "a" has 819 decimal digits.

David
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