TITLE: Something screwy going on in $\mathbb Z[\sqrt{51}]$
QUESTION [10 upvotes]: In $\mathbb Z[\sqrt{6}]$, I can readily find that $(-1)(2 - \sqrt{6})(2 + \sqrt{6}) = 2$ and $(3 - \sqrt{6})(3 + \sqrt{6}) = 3$. It looks strange but it checks out.
But when I try the same thing for $3$ and $17$ in $\mathbb Z[\sqrt{51}]$ I seem to run into a wall. I can't solve $x^2 - 51y^2 = \pm3$ in integers, nor $x^2 - 51y^2 = \pm17$. I've had six decades in which to get rusty at solving equations in two variables, so maybe I have managed to overlook solutions to both of these. Or could it really be possible that $3$ and $17$ are actually irreducible in this domain?

REPLY [7 votes]: Both your equations can be transformed to $3u^2-17v^2=\pm1$ (because one of $x,y$ must be a multiple of $3$ resp. $17$). Then 
$$-u^2-v^2\equiv \pm1\pmod4 $$
is impossible with the plus sign and
$$v^2\equiv \pm1\pmod3 $$
is impossible with the minus sign. Hence there is no solution.