Find the positive whole x and y values to which this equation is true?

OP

Original post by Gregorius

The reason that

\sqrt{D}

is important is that you can re-cast the problem as finding solutions to

\displaystyle (x + y \sqrt{D})(x - y \sqrt{D}) = 1

so that the problem is equivalent to finding a non-trivial unit of norm 1 in the ring

\mathbb{Z}[\sqrt{D}]

.

It sounds as though you need to brush up on the theory of continued fractions. There's a pretty good tutorial here - and it includes a treatment of the key fact about how the continued fraction of a quadratic surd (like

\sqrt{D}

) is repeating.

Thanks, that has cleared some things up. This is something I haven't learnt yet, it was a problem for the york applicant day. Could you explain what a non trivial unit of norm 1 in the ring z is?

Reply 5

Original post by BinaryJava

Thanks, that has cleared some things up. This is something I haven't learnt yet, it was a problem for the york applicant day. Could you explain what a non trivial unit of norm 1 in the ring z is?

OK. So

\displaystyle \mathbb{Z}[\sqrt{D}] = \{x + y \sqrt{D}: x, y \in \mathbb{Z} \}

with the obvious operations of addition and multiplication to make it a ring. The point is that we use numbers of the form

x + y \sqrt{D}

to study solutions of the Pell equation.

The norm of

x + y \sqrt{D}

is defined to be precisely

x^2 - D y^2

(if you think about the norm of a complex number being derived from the product of the number with its complex conjugate, you get an analogy of where this is coming from).

So if the norm of

x + y \sqrt{D}

is

x^2 - D y^2

then to solve the Pell equation is precisely to find a conjugate pair of elements satisfying

\displaystyle (x + y \sqrt{D}) (x - y \sqrt{D}) = 1

In other words, we want to find divisors of unity (so-called units) that are not equal to unity (i.e. they are non-trivial) and which have norm one.

Reply 6

Original post by BinaryJava

...it was a problem for the york applicant day...

Are you thinking of coming to York? Good department - but I'm biased as I work at York Uni!

Reply 7

OP

Original post by Gregorius

Are you thinking of coming to York? Good department - but I'm biased as I work at York Uni!

I really liked it there, the research from the maths department really impressed me. It is definitely going to be a university I choose as an option, just not sure whether insurance or firm.

Reply 8

OP

Original post by Gregorius

OK. So

\displaystyle \mathbb{Z}[\sqrt{D}] = \{x + y \sqrt{D}: x, y \in \mathbb{Z} \}

with the obvious operations of addition and multiplication to make it a ring. The point is that we use numbers of the form

x + y \sqrt{D}

to study solutions of the Pell equation.

The norm of

x + y \sqrt{D}

is defined to be precisely

x^2 - D y^2

(if you think about the norm of a complex number being derived from the product of the number with its complex conjugate, you get an analogy of where this is coming from).

So if the norm of

x + y \sqrt{D}

is

x^2 - D y^2

then to solve the Pell equation is precisely to find a conjugate pair of elements satisfying

\displaystyle (x + y \sqrt{D}) (x - y \sqrt{D}) = 1

In other words, we want to find divisors of unity (so-called units) that are not equal to unity (i.e. they are non-trivial) and which have norm one.

Ok, that makes sense.

Reply 9

OP

Is this on the right track, I have got stuck trying to get rid of the 12/5 on my

\sqrt 41

Do I then sub this into the top equation?

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Reply 10

Original post by BinaryJava

Is this on the right track, I have got stuck trying to get rid of the 12/5 on my

\sqrt 41

Do I then sub this into the top equation?

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I'm not entirely sure what you are doing here, but it looks as if you are calculating the general continued fraction rather than the simple form we need here. If you are following the link I posted earlier, you should be working through the algorithm in section 6.2.2.

So you should be getting something like

\displaystyle x_0 = \sqrt{41} = 6 + \frac{1}{x_1}

which will simplify out to

\displaystyle x_1 = \frac{\sqrt{41} + 6}{5} = 2 + \frac{1}{x_2}

and then

\displaystyle x_2 = \frac{\sqrt{41} + 4}{5} = 2 + \frac{1}{x_3}

and so on (but thankfully not for very far!)

Do you follow?

Reply 11

OP

Are these the correct workings and is my notation at the end correct?

How would I use this expansion to go about solving my equation?

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Reply 12