$m + n \sqrt{2}$ is invertible $\iff$ $m^2 - 2n^2 =\pm 1$ 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.  

As you said, the only candidate for the inverse is clearly $\frac1{m+n\sqrt{2}}$.
We have
$$\frac1{m+n\sqrt{2}} = \frac{m-n\sqrt{2}}{m^2-2n^2} = \frac{m}{m^2-2n^2} + \frac{-n}{m^2-2n^2}\sqrt{2}$$
This is an element of $\mathbb{Z}[\sqrt{2}]$ if and only if both $\frac{m}{m^2-2n^2}$ and $\frac{-n}{m^2-2n^2}$ are in $\mathbb{Z}$.
Let $d = \gcd(m,n)$. Assume that $m^2-2n^2 \mid m$ and $m^2-2n^2 \mid n$. Then also $m^2-2n^2 \mid d$. On the other hand, $d \mid m$ and $d \mid n$ so $d^2 \mid m^2-2n^2$. Therefore $d^2 \mid d$ so $d = 1$.
Therefore if $\frac{m}{m^2-2n^2}, \frac{-n}{m^2-2n^2} \in \mathbb{Z}$ then $m^2-2n^2$ divides $d = 1$ so we conclude $m^2-2n^2 = \pm 1$.
