How to prove $x^2-34y^2=17$ has no integer solutions? My question is how might it be shown that the equation $x^2-34y^2=17$ has no solutions where $x$ and $y$ are both integers?
I used this website to check whether there were solutions: http://www.numbertheory.org/php/patz.html. 
There is a linked paper which details the methods, but it is far over my head. I suppose I am asking for how it might be shown more simply! Any help would be greatly appreciated.
 A: We must have $17\mid x$, hence the equation boils down to:
$$ 17 x^2 = 1 + 2y^2 = (1+y\sqrt{-2})(1-y\sqrt{-2}).\tag{1}$$
Since $\mathbb{Z}[\sqrt{-2}]$ is an euclidean domain and $17=3^2+2\cdot 2^2$, we just have to show that no square in $\mathbb{Z}[\sqrt{-2}]$ can have the form:
$$(1+y\sqrt{-2})(3+2\sqrt{-2})=(3-4y)+(2+3y)\sqrt{-2},$$
that is the same as saying that 
$$ a^2-2b^2 = 3-4y,\qquad 2ab = 2+3y\tag{2}$$
is impossible. $(2)$ implies:
$$ 3a^2 + 8ab - 6b^2 = 17, $$
but $a$ and $b$ cannot be both even, so $3a^2+8ab-6b^2\in\{2,3,5,6\}\pmod{8}$, while $17\equiv 1\pmod{8}$, hence $(2)$, then $(1)$, have no integer solutions.
A: Note that $35^2-34\cdot 6^2=1$.  Therefore the real curve $x^2-34 y^2=17$ (a hyperbola) is invariant under the linear transformation $$\begin{pmatrix}x \\ y\end{pmatrix}\mapsto\begin{pmatrix}35&-204\\-6&35\end{pmatrix}\begin{pmatrix}x \\ y\end{pmatrix}.$$  This is a hyperbolic rotation.  A (closed) fundamental domain of the transformation group generated by this rotation is given by $|y| \leq \sqrt{17/2} < 3$. The rotation has integral coefficients.  Therefore if $x^2-34 y^2=17$ has a solution over the integers then it has one with $|y|\leq 2$.  A direct computation shows that it has no such solutions.
