1. ## 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.

$x^2 - dn^2 = 1$

2. 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.

$x^2 - dn^2 = 1$

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

3. 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?

4. 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.

$x^2 - dn^2 = 1$

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.

5. 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?

6. 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?

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