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Old 2005-04-11, 10:31   #1
jtavares
 
Nov 2004

32 Posts
Default Looking for solutions to w^2-n*x^2=z^2

How can i find solutions (w, x, z) to the following equation:

w^2-n*x^2=z^2


Can i use the continued fraction expansion? and how?
Is the solution related to the factorization of n?
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Old 2005-04-11, 11:37   #2
R.D. Silverman
 
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Nov 2003

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Quote:
Originally Posted by jtavares
How can i find solutions (w, x, z) to the following equation:

w^2-n*x^2=z^2


Can i use the continued fraction expansion? and how?
Is the solution related to the factorization of n?
We will assume n is squarefree (standard assumption; otherwise just do
a change of variables)

For z = 1, solution methods are well known. This is Pell's equation. For z > 1,
there are extensions of the cfrac method. See Henri Cohen's book. I don't
have it handy, so can't look up the exact reference.
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Old 2005-04-11, 18:46   #3
jtavares
 
Nov 2004

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No, Henri Cohen's books do not deal with it - "A course in computacional algebraic number theory" and "Advanced topics in computacional number theory". I could only find the solution to x^2+dy^2=p with d>0 and p prime (Cornachia). I am not sure if this could be used to solve w^2-n*x^2=z^2 by transforming it to z^2+n*x^2=w^2. Anyway there sould be a suitable continued fraction aproximation too but i can not find it.
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Old 2005-04-11, 19:25   #4
Citrix
 
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Jun 2003

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an easier method

w^2-n*x^2=z^2

then

w^2-z^2=n*x^2

or

(w-z)* (w+z) =n*x^2

factorize n*x^2 to get solutions of w and z.

Citrix
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