# Pell's equation

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• April 24th 2010, 08:07 PM
davismj
Pell's equation
I have an equation of the form

$2(2x-1)^2 -(2n-1)^2 = 1$

I know that this is a pell's equation, but I don't know what to do with it, since its not in the form.

$x^2 - dn^2 = 1$

Thanks for your help!
• April 24th 2010, 09:03 PM
mr fantastic
Quote:

Originally Posted by davismj
I have an equation of the form

$2(2x-1)^2 -(2n-1)^2 = 1$

I know that this is a pell's equation, but I don't know what to do with it, since its not in the form.

$x^2 - dn^2 = 1$

Thanks for your help!

You have $u^2 - D v^2 = 1$ where $u = \sqrt{2} (2x - 1)$, $v = (2n-1)$ and $D = 1$.
• April 24th 2010, 09:21 PM
davismj
Quote:

Originally Posted by mr fantastic
You have $u^2 - D v^2 = 1$ where $u = \sqrt{2} (2x - 1)$, $v = (2n-1)$ and $D = 1$.

Thanks for the response.

Hmm. Still confused cause if I apply the algorithm, I get:

$(\sqrt{2}-1)^2 = 3 - 2\sqrt{2}$

which implies $u = -2\sqrt{2}$. But that is not a solution since u is positive. What am I doing wrong?
• April 24th 2010, 11:02 PM
chiph588@
Quote:

Originally Posted by davismj
I have an equation of the form

$2(2x-1)^2 -(2n-1)^2 = 1$

I know that this is a pell's equation, but I don't know what to do with it, since its not in the form.

$x^2 - dn^2 = 1$

Thanks for your help!

Let $a=2n-1$ and $b=2x-1$. We then have $a^2-2b^2=-1$

Thus $(a^2-2b^2)^2=(-1)^2=1 \implies (a^2-2b^2)^2-2(2ab)^2=1$.

But $2b^2=a^2+1$, so we get $(2a^2+1)^2-2(2ab)^2=1$.

Summarizing, we see our original equation was transformed into Pell's equation.
• April 24th 2010, 11:05 PM
chiph588@
Quote:

Originally Posted by mr fantastic
You have $u^2 - D v^2 = 1$ where $u = \sqrt{2} (2x - 1)$, $v = (2n-1)$ and $D = 1$.

The $\sqrt2$ kind of scares me. Aren't we dealing with integers here?
• April 25th 2010, 05:06 PM
davismj
Quote:

Originally Posted by chiph588@
Let $a=2n-1$ and $b=2x-1$. We then have $a^2-2b^2=-1$

Thus $(a^2-2b^2)^2=(-1)^2=1 \implies (a^2-2b^2)^2-2(2ab)^2=1$.

I don't think I agree with this. This seems to be true iff 2(2ab)^2 = 0, yes?
• April 25th 2010, 05:10 PM
chiph588@
Quote:

Originally Posted by davismj
I don't think I agree with this. This seems to be true iff 2(2ab)^2 = 0, yes?

Sorry, I had a typo there. That line should read

$(a^2-2b^2)^2=(-1)^2=1 \implies (a^2+2b^2)^2-2(2ab)^2=1$
• April 25th 2010, 05:22 PM
davismj
Quote:

Originally Posted by chiph588@
Sorry, I had a typo there. That line should read

$(a^2-2b^2)^2=(-1)^2=1 \implies (a^2+2b^2)^2-2(2ab)^2=1$

Gotcha. You'll have to forgive me because our class isn't a number theory class nor does it have a textbook, so I have a wikipedia understanding of Pell's equation. So now I need to come up with a solution of a and b and then raise it to a subsequent power?