Let me see if I've got this right then:

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

$\displaystyle \sqrt{2} = [1; 2, 2, 2, ~ ... ~ ]$

Thus $\displaystyle l = 1$ and l is odd.

So I have

$\displaystyle C_{2l - 1} = C_{2 \cdot 1 - 1} = C_1 = 1 + \frac{1}{2} = \frac{3}{2} = \frac{x}{y}$

Thus $\displaystyle y = \frac{2}{3}x$.

So

$\displaystyle x^2 - 2\left ( \frac{2}{3} x \right )^2 = 1 ==> x = 3$. Thus $\displaystyle y = 2$. Which is a correct solution.

-Dan