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

Question

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.

Answer 1

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 $m^2-2n^2 = \pm 1$.

Answer 2

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 $\nu(x) \bar x$.

Answer 3

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 $1,\sqrt 2$.