If α is a unit in $\mathbb{Z}[\sqrt2]$ then $a^2 - 2b^2 = \pm1$

Question

$$\mathbb{Z}[\sqrt2] = \{a + b√2: a,b \in \mathbb{Z}\}$$

The question is asking to prove that if $\alpha$ is a unit in this set, then: $$a^2 - 2b^2 = \pm1$$ I've hit a dead end already: Attempt at proof

I wanted to use the fact that there must exist another element in $\mathbb{Z}[\sqrt2]$ such that $\alpha\beta = 1$.

Now I'm lost, since I've just accidentally shown that the difference of $a^2$ and $2b^2$ should be equal to $1$, not that they are both equal to $1$. Am I going about this the wrong way?

Are you sure the question's written properly? How can $2b^2 = 1$ if $b \in \mathbb Z$? — Théophile, Jan 30 '15 at 20:59
I'm starting to think that it is a typo in the question. I cannot figure this out for the life of me lol. — James Norton, Jan 30 '15 at 21:01
The norm of $a+b\sqrt{2}$, in this ring, is $a^2-2b^2$. A quick characterization of units can be given by verifying that the norm of a product is the product of the norms (Brahmagupta Identity). And your question should say $a^2-2b^2=\pm 1$, or something equivalent. — André Nicolas, Jan 30 '15 at 21:03
Let me add an element is a unit if and only if its norm is a unit in $\mathbf Z$, i.e. is equal to $\pm1$. — Bernard, Jan 30 '15 at 21:07
possible duplicate of The units of $\mathbb Z[\sqrt{2}]$. See also http://math.stackexchange.com/questions/438913/determine-all-units-of-the-domain-mathbbz-sqrt2?rq=1. — Dietrich Burde, Jan 30 '15 at 21:29
@DietrichBurde: The linked questions are harder, for there one is asked to describe the units explicitly. For this question, one does not need to do that. — André Nicolas, Jan 30 '15 at 22:06

Dietrich Burde · Answer 1 · 2015-01-30T21:34:08.297

0

Since $(1+\sqrt{2})(-1+\sqrt{2})=1$, we see that $\pm (1+\sqrt{2})^n$ is a unit for all $n\in \mathbb{Z}$. Using the norm it is easy to see that every unit in $\mathbb{Z} [\sqrt{2}]^{\times}$ is of this form. In fact $\alpha=a+b\sqrt{2}$ is a unit iff $N(\alpha)=a^2-2b^2=\pm 1$.

edited Jan 30 '15 at 21:34

answered Jan 30 '15 at 21:28

Dietrich Burde

130,978

If α is a unit in $\mathbb{Z}[\sqrt2]$ then $a^2 - 2b^2 = \pm1$

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