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