TITLE: Why does the equation $x^2-82y^2=\pm2$ have solutions in every $\mathbb{Z}_p$ but not in $\mathbb{Z}$?
QUESTION [10 upvotes]: I have been working on an exercise in H. P. F. Swinnerton-Dyer's book, A Brief Guide to Algebraic Number Theory. The question is like this: 

Show that $x^2-82y^2=\pm2$ has solutions in every $\mathbb{Z}_p$ but not in $\mathbb{Z}$.What conclusion can you draw about $\mathbb{Q}(\sqrt{82})$?

I thought it might be solved by using the Hensel's lemma. But I can't give an answer.
Thanks in advance!

REPLY [2 votes]: Regarding the last part of the question, this tells you that the principal genus of discriminant $328$ contains a non-trivial class, and hence that the class number of $\mathbb Q(\sqrt{82})$ is divisible by $2$.