Is it possible to factor $27$ in $\mathbb Z[\sqrt7]$ other than in $27=3\cdot3\cdot 3$? I need to know if there exists an element of norm$27$ in $\mathbb Z[\sqrt 7]$, that is, whether there are any integer solutions to $a^2 - 7b^2 = 27$. When I used modular arithmetic, I find out that $a^2 \equiv b^2$ (mod 3). Now I'm not exactly sure what this is supposed to tell me. 
Edit: I just found out that $27 = -(2 + \sqrt(7))^3 (2 - \sqrt(7))^3$ so how can we just say that 27 doesn't factor? 

Note that the only squares mod $7$ are $1,2,4$.  This means that $a^2 \not\equiv 6 \equiv 27 \pmod{7}$ for any $a$, so the desired equation is impossible.
