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Topic: Pell of the form a*x^2 - b*y^2 (Read 689 times) |
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Christine
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Posts: 159
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Pell of the form a*x^2 - b*y^2
« on: Mar 15th, 2016, 2:35pm » |
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Could you pleases show how to solve Pell equations of the type:
a*x^2 - b*y^2 = +/- 1
for example,
27*x^2 - 343*y^2 = -1
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pex
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Posts: 880
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Re: Pell of the form a*x^2 - b*y^2
« Reply #1 on: Mar 16th, 2016, 3:03am » |
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As far as I know, you want x/y to be a convergent of the continued fraction of sqrt(b/a), because the only hope of getting ax2 - by2 close to zero is to try and make (x/y)2 close to b/a. I don't know if there's anything more fancy that can be done than "keep trying until it works"; for your example, I find the following, letting x/y be successive convergents to sqrt(343/27):
Code: ......x .....y .rhs (annoying dots added to keep columns aligned)
......3 .....1 -100
......4 .....1 ..89
......7 .....2 .-49
.....25 .....7 ..68
.....57 ....16 .-85
.....82 ....23 .101
....139 ....39 .-36
....638 ...179 .125
....777 ...218 .-49
...2969 ...833 ..20
..27498 ..7715 .-67
..57965 .16263 .108
..85463 .23978 .-49
.314354 .88197 ..45
1342879 376766 ..-1 |
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so x=1342879, y=376766 is a solution.
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| « Last Edit: Mar 16th, 2016, 3:09am by pex » |
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Grimbal
wu::riddles Moderator
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Re: Pell of the form a*x^2 - b*y^2
« Reply #2 on: Mar 18th, 2016, 7:33am » |
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For approximating fractions, there is the Stern-Brocot tree, which is probably the same in different terms.
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