Existence of integer solution of $a^2 -17b^2 = $ any constant When checking whether if $9-\sqrt{17}$ in the ring $\{a+b\sqrt17: a,b \in \mathbb{Z}\}$ is a prime. 
Suppose 
$$\alpha\cdot \beta = 9-\sqrt{17},$$
using norm argument
$$N(\alpha)N(\beta) = N(9-\sqrt{17}) = -8.$$
If say $N(\alpha) = N(a+b\sqrt17) = a^2 - 17b^2$, I need to know if there exist integer solutions for 
$a^2 - 17b^2= 2$ or $a^2 - 17b^2 = 4$. So in general, how can I check if there exist integer solutions of $a^2 -17b^2 = $ any constant?
Thank you very much!
 A: I don't know of any general procedure.  However, for the example you give,
$$a^2-17b^2=\pm2$$
implies
$$a^2-b^2\equiv2\pmod4\ ,$$
which is impossible since the squares modulo $4$ are $0$ and $1$ only.  Furthermore, as the product of norms is $-8$, if one of them is $\pm4$ then the other must be $\mp2$, and we have just shown that this is impossible.
A: Let the "any constant" be $\mu$, $\mu \geq 0$. Then $a^{2} - 17 \, b^{2} = \mu$ becomes
$$x^{2} - 17 \, y^{2} = 1 \tag{1}$$
where $a = \sqrt{\mu} \, x$ and $b = \sqrt{\mu} \, y$. Equation (1) is a Pell equation, see Pell Equations, and has solutions
\begin{align}
x_{n} &= \frac{1}{2} \, \left[ (33+8 \, \sqrt{17})^{n} + (33 - 8 \, \sqrt{17})^{n} \right] \\
y_{n} &= \frac{1}{2 \, \sqrt{17}} \, \left[ (33+8 \, \sqrt{17})^{n} - (33 - 8 \, \sqrt{17})^{n} \right]
\end{align}
and in terms of $(a,b)$ the result becomes
\begin{align}
a_{n} &= \frac{\sqrt{\mu}}{2} \, \left[ (33+8 \, \sqrt{17})^{n} + (33 - 8 \, \sqrt{17})^{n} \right] \\
b_{n} &= \frac{\sqrt{\mu}}{2 \, \sqrt{17}} \, \left[ (33+8 \, \sqrt{17})^{n} - (33 - 8 \, \sqrt{17})^{n} \right]
\end{align}
for $n\geq 0$. For the case of negative "any constant" a similar path may be taken. 
