TITLE: $m + n \sqrt{2}$ is invertible $\iff$ $m^2 - 2n^2 =\pm 1$
QUESTION [5 upvotes]: I'm having trouble proving that, for $m, n \in \mathbb{Z}$, the existence of a multiplicative inverse for $m + n \sqrt{2}$ implies that $m^2 - 2n^2 = \pm 1$. 
The first step, I believe, is to solve for the inverse, which is clearly $\frac{1}{m + n\sqrt{2}}$, provided that $m  +n \sqrt{2}$, as $m + n \sqrt{2}$ would otherwise not be invertible. From here,  I'm unsure on how to piece together this proof. I read through some hints on another answer on here, so it seems that a plausible step is to use the fact that $m^2 - 2n^2$ is a difference of squares and factors into $(m - \sqrt{2} n)(m + \sqrt{2}n)$. One of these factors is invertible, but we don't have any information on whether it's conjugate is, or ability to equate the product with, say, $1$ or $-1$, so I can't quite figure out how to get there. 
In the opposite direction, the first step seems to be  factoring into $(m + n \sqrt{2})(m - n \sqrt{2}) = \pm 1$. If this product is equal to $1$, then, $m - n \sqrt{2}$ is clearly the inverse of $m + n \sqrt{2}$, as we end up with a product of $1$. If not, we could scale both sides by $-1$, which should give us the same inverse. 
Any helpful thoughts and hint swould be greatly appreciated.

REPLY [3 votes]: Here is a different take.
The ring $\mathbb Z[\sqrt 2]$ is isomorphic to a subring of $M_2(\mathbb Z)$, the ring of $2 \times 2$ integer matrices:
$$
\phi: m + n \sqrt 2 \mapsto \pmatrix{ m & 2n \\ n & m }
$$
Then, $m + n \sqrt 2$ is invertible iff $\phi(m + n \sqrt 2)$ is invertible iff $\det\phi(m + n \sqrt 2)$ is invertible iff $\det\phi(m + n \sqrt 2)=\pm 1$.
The isomorphism $\phi$ comes from the maps $\mu : x \mapsto x (m + n \sqrt 2)$ in the $\mathbb Z$-basis $1,\sqrt 2$.