# Thread: Pell's equation

1. ## Pell's equation

I have an equation of the form

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

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.

$\displaystyle x^2 - dn^2 = 1$

Thanks for your help!

2. Originally Posted by davismj
I have an equation of the form

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

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.

$\displaystyle x^2 - dn^2 = 1$

Thanks for your help!
You have $\displaystyle u^2 - D v^2 = 1$ where $\displaystyle u = \sqrt{2} (2x - 1)$, $\displaystyle v = (2n-1)$ and $\displaystyle D = 1$.

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

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

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

which implies $\displaystyle u = -2\sqrt{2}$. But that is not a solution since u is positive. What am I doing wrong?

4. Originally Posted by davismj
I have an equation of the form

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

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.

$\displaystyle x^2 - dn^2 = 1$

Thanks for your help!
Let $\displaystyle a=2n-1$ and $\displaystyle b=2x-1$. We then have $\displaystyle a^2-2b^2=-1$

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

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

Summarizing, we see our original equation was transformed into Pell's equation.

5. Originally Posted by mr fantastic
You have $\displaystyle u^2 - D v^2 = 1$ where $\displaystyle u = \sqrt{2} (2x - 1)$, $\displaystyle v = (2n-1)$ and $\displaystyle D = 1$.
The $\displaystyle \sqrt2$ kind of scares me. Aren't we dealing with integers here?

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

Thus $\displaystyle (a^2-2b^2)^2=(-1)^2=1 \implies (a^2-2b^2)^2-2(2ab)^2=1$.
I don't think I agree with this. This seems to be true iff 2(2ab)^2 = 0, yes?

7. 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

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

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

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