$m + n \sqrt{2}$ 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.  

Introduce the map $\nu :A=\mathbb Z[\sqrt 2]\to \mathbb Z$ such that $\nu(a+b\sqrt 2) = a^2-2b^2$, let us call this the norm map. Then I claim that $\nu$ is multiplicative (exercise, and the most important part of all this proof, really!). Now suppose that $x$ is invertible in $A$, so there is $y$ such that $xy=1$. Applying $\nu$, we see that $\nu(x)$ is a unit in $\mathbb Z$. Since the units in this ring are $1$ and $-1$, we have shown that

If $x$ is a unit in $A$, then $\nu(x)$ is a unit in $\mathbb Z$, i.e. $\pm 1$.

Now you want to prove the converse of the above. Hence assume that the norm is a unit. If $x=a+b\sqrt 2$, let $\bar x=a-b\sqrt 2$, the conjugate of $x$. Then $x\bar x=\nu(x)$, and since $\nu(x)$ is $\pm 1$, we see that the inverse of $x$ is simply $\nu(x) \bar x$.
