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**zetafunc.****Guest**

So, generally you require a calculator to solve a Pell equation...?

**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 107,155

Depends on the Pell equation. For some of them like

you can tell just by looking at them. There is the method of Brahmagupta and Bhaskara two ancient Indian mathematicians.

Generally though, in computational number theory you will require the use of at least a good scientific calculator.

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**zetafunc.****Guest**

You can tell just by looking at them? You mean, finding the fundamental solution using their method, or by trial and error?

I found a useful recurrence relation which works for the OP's example. Unfortunately it does not help unless you have the fundamental solution.

**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 107,155

Yes, once you have the fundamental solution ( or any solution other than the trivial one ) you use a recurrence to get more solutions.

You can tell just by looking at them?

By inspection x = 24 and y = 5.

For that one you must compute the answer.

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**zetafunc.****Guest**

I must compute the answer for that one? I don't understand...

That Pell equation (the OP's original one) is even easier to solve by inspection, and once you have an (x,y) pair, it's easy to get the next solution (x = 199, y = 60). Why is it different to x^2 = 1 + 23y^2?

**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 107,155

When I say by inspection I mean without computing the convergents. The OP's one requires some computation. The other is as simple as x +2 = 4.

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**zetafunc.****Guest**

I am able to get a solution without convergents...

**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 107,155

What did you get?

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**zetafunc.****Guest**

x = 199, y = 60.

**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 107,155

You used the recurrence to get that. But if you want to get the fundamental solution convergents of the continued fraction is best.

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**zetafunc.****Guest**

But how can I compute them by hand? These problems often come to me when I have no computer or calculator near me, such as in an olympiad, or if someone comes up to me with a problem like it.

**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 107,155

They never come up in an olympiad. The only time someone asks that is when they know nothing and think the problem is easy. Guys like that think everything is as easy as plugging into the quadratic formula.

For instance, find the fundamental solution to:

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**zetafunc.****Guest**

How would Brahmagupta and Bhaskara approach that problem?

**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 107,155

They could not approach it. It is a problem that is historically famous. This is how the problem begins.

Archimedes wrote:

Page 1

If thou art diligent and wise, O stranger, compute the number of cattle of the Sun, who

once upon a time grazed on the ffields of the Thrinacian isle of Sicily, divided into four herds of

different colours, one milk white, another glossy black, the third yellow and the last dappled. In

each herd were bulls, mighty in number according to these proportions: Understand, stranger,

that the white bulls were equal to a half and a third of the black together with the whole of the

yellow, while the black were equal to the fourth part of the dappled and a fifth, together with,

once more, the whole of the yellow. Observe further that the remaining bulls, the dappled, were

equal to a sixth part of the white and a seventh, together with all the yellow. These were the

proportions of the cows: The white were precisely equal to the third part and a fourth of the

whole herd of the black; while the black were equal to the fourth part once more of the dappled

and with it a fifth part, when all, including the bulls, went to pasture together. Now the dappled

in four parts were equal in number to a fifth part and a sixth of the yellow herd. Finally the

yellow were in number equal to a sixth part and seventh of the white herd.

If thou canst accurately tell, O stranger, the number of Cattle of the Sun, giving separately

the number of well-fed bulls and again the number of females according to each colour, thou

wouldst not be called unskilled or ignorant of numbers, but not yet shalt thou be numbered

among the wise.

But come, understand also all these conditions regarding the cows of the Sun. When the

white bulls mingled their number with the black, they stood rm, equal in depth and breadth,

and the plains of Thrinacia, stretching far in all ways, were filled with their multitude. Again,

when the yellow and the dappled bulls were gathered into one herd they stood in such a manner

that their number, beginning from one, grew slowly greater till it completed a triangular figure,

there being no bulls of other colours in their midst nor one of them lacking.

If thou art able, O stranger, to find out all these things and gather them together in your

mind, giving all the relations, thou shalt depart crowned with glory and knowing that thou hast

been adjudged perfect in this species of wisdom.

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**zetafunc.****Guest**

How do you form the Pell equation from that?

I would not have liked to sit an exam set by Archimedes, that is for sure.

**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 107,155

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

Read this for more:

http://www.sciencenews.org/sn_arc98/4_1 … thland.htm

It took more that 1800 years for someone to partially solve the problem. It was only solved fully in the 20th century with the aid of computers.

Incidentally, the Pell equation is incorrectly named. Pell had nothing to do with it. It should be called Fermat's equation.

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**zetafunc.****Guest**

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****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 107,155

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.

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**Leroy****Guest**

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****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 107,155

You use two recurrences to find more answers.

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**Leroy****Guest**

You mean every 2nd convergent after the fundamental one?

**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 107,155

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.

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**Leroy****Guest**

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

**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 107,155

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:

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**Leroy****Guest**

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