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bobbym
2012-10-21 17:25:54

No, there is a method to find the convergents.

Leroy
2012-10-21 17:18:42

So I have find the convergent by trial and error

bobbym
2012-10-21 15:13:38

There is a theorem that says that. This is not something I just cooked up, I wished I had.

Leroy
2012-10-21 14:17:31

I have understood everything quite well,but I don't understand what is the logic for taking the convergents,is there any rule for which convergent will be the solution?

bobbym
2012-10-21 12:28:50
Leroy
2012-10-21 12:12:57

Hi again,I have understood pell's equation,now I am curious about nagetive pell's equation,so is there any method to solve x^2-ny^2=-1

bobbym
2012-10-20 01:36:50

Hi Leroy;

No bother at all. You  remember that the fundamental solution is  x = 10 and y = 3. We use that to get all the rest.

With

Running the recurrences in the forward direction we get the first bunch:

Leroy
2012-10-20 01:29:07

What ways could you please explain.(sorry for disturbing you so much)

bobbym
2012-10-20 01:06:34

Hi;

Yes, that is one way if you have the convergents. x=numerator and y=denominator of the period.

There are other ways that do not require the convergents.

Leroy
2012-10-20 01:02:16

You mean every 2nd convergent after the fundamental one?

bobbym
2012-10-20 00:38:50

You use two recurrences to find more answers.

Leroy
2012-10-20 00:34:04

Thank you,now I can find a pell's equation's fundamental solution,but what is the method of finding the additional ones?please explain.

bobbym
2012-10-19 02:35:38

It would seem so. Vardi and others believe he at least formulated the problem correctly even if he was unable to solve for the 205 000 digit answer.

zetafunc.
2012-10-19 02:30:32

That is interesting, especially how they were only able to get the solution (all the digits) in 1965.

So, Archimedes probably knew that equations of that form were solvable.

bobbym
2012-10-19 02:23:55

The equation in post #37 is the Pell equation for that problem.