Part (b) of this question:
a) Express $\sqrt{17}$ as a continued fraction.
b) Hence, or otherwise, find the smallest positive integers $x$,$y$ such that $x^2 - 17y^2 = 1$.
I'm fine with part (a) and I know that the answer to (b) is probably $x =33$ and $y = 8$ and that it has something to do with the convergents of the continued fraction I found in (a). But I don't really know the relation between $x$, $y$ and these convergents and as a result can't really provide an adequate answer to the question. Any help would be greatly appreciated. Thank you!
The continued fraction for $\sqrt{17}$ is indeed $[4; \bar{8}]$
From 1, the convergents are: $$ 4 = \dfrac{4}{1} ,\qquad 4 + \dfrac{1}{8} = \dfrac{33}{8} ,\qquad 4 + \dfrac{1}{8+\dfrac{1}{8}} = \dfrac{268}{65} $$ Since $x^2 - 17y^2 = -1$ for $(x,y)=(4,1)$, you need the even-numbered convergents and so $(x,y)=(33,8)$ is the smallest solution of $x^2 - 17y^2 = 1$. The next one is $(x,y)=(2177,528)$.