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2005-04-11, 10:31
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#1 |
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Nov 2004
32 Posts |
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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2005-04-11, 11:37
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#2 | |
"Bob Silverman"
Nov 2003
North of Boston
53×61 Posts |
Quote:
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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2005-04-11, 18:46
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#3 |
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Nov 2004
118 Posts |
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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2005-04-11, 19:25
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#4 |
Jun 2003
31378 Posts |
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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