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I would like to determine all units of the domain $\mathbb{Z}[\sqrt{2}]$.

Recall that if $D$ is an integral domain, then a unit of $D$ is an associate of the multiplicative identity $1$, i.e. $u \in D$ is a unit iff $u$ divides $1$ and $1$ also divides $u$. Equivalently, $u \in D$ is a unit iff it is a divisor of $1$, i.e. iff $\exists x \in D$ such that $ux=1$. Thus, units are also called invertible elements.

The set $\mathbb{Z}[\sqrt{2}] := \{a+b \sqrt{2}: a,b \in \mathbb{Z} \}$, under the usual addition and multiplication operations, forms a domain. How do I find all its units? Suppose $a+b \sqrt{2}$ is a unit. Then $(a+b \sqrt{2})(x+y \sqrt{2})=1$ for some $x,y \in \mathbb{Z}$. Thus, $ax+2by=1$ and $ay+bx=0$. Solving for $x,y$, we obtain that $x=\frac{a}{a^2 - 2b^2}$ and $y=\frac{-b}{a^2 - 2b^2}$ must be integers. How do I obtain the set of values of $a,b$ for which $x$ and $y$ are integral?

Since $(\sqrt{2}-1)(\sqrt{2}+1)= 2-1^2=1$, the two factors $\pm 1 + 1 \sqrt{2}$ are units. Can we systematically find all the units of $\mathbb{Z}[\sqrt{2}]$?

This answer given here to a related question shows only that if $a+b \sqrt{2}$ is a unit and is > 1, and if $a,b>0$, then $a+b \sqrt{2}$ is of the form $(1+\sqrt{2})^n$. The case where $a,b$ are of opposite signs, and the case where the unit is between 0 and 1 (such units exist, for eg, $3-2 \sqrt{2}$) are open.

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    They are $\pm$ the powers of $\sqrt{2}-1$. That these are units is easy. That we get them all this way can be a slightly messy calculation, or a consequence of more general results. – André Nicolas Jul 08 '13 at 15:04
  • Do you know Dirichlet's unit theorem? –  Jul 08 '13 at 15:05
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    The quantity $N(a+ b \sqrt{2}) = a^2-2b^2$ is called the norm of $a+ b \sqrt{2}$. Since it is multiplicative ($N(xy) = N(x) N(y)$ for $x, y \in \mathbb{Z}[\sqrt{2}]$), you can show the norm of a unit must be invertible in $\mathbb{Z}$, so is either $1$ or $-1$. Then you have to solve $a^2 - 2b^2 = \pm 1$. – Joel Cohen Jul 08 '13 at 15:10
  • @AndréNicolas: Could you please elaborate as an answer? – AG. Jul 08 '13 at 15:57
  • @BenjaLim: Nope, what is Dirichlet's unit theorem? – AG. Jul 08 '13 at 15:58
  • @JoelCohen: could you please elaborate on your answer - on why $a^2-2b^2=\pm 1$ and how to solve this? – AG. Jul 08 '13 at 15:58
  • This question is marked as duplicate, but I do not believe this question has a complete answer in the given link. The answer in the given link shows only that if $a+b \sqrt{2}$ is a unit and is > 1, and if $a,b>0$, then $a+b \sqrt{2}$ is of the form $(1+\sqrt{2})^n$. The case where $a,b$ are of opposite signs, and the case where the unit is between 0 and 1 (such units exist, for eg $3-2 \sqrt{2}$) are open. – AG. Jul 13 '13 at 03:02
  • Why voted for reopening this question??? Obviously the answer in the original question's topic is enough to finish the problem: for example, if the unit $u$ is positive, but $<1$, then for some $n$ big enough $u(1+\sqrt 2)^n>1$ and it's still invertible, so there is $m\ge 1$ such that $u(1+\sqrt 2)^n=(1+\sqrt 2)^m$, and so on. (Dear OP, try to think to your problem a little bit more before giving slim arguments for reopening. Meanwhile, I leave you the plesure to finish the problem by considering $u<0$.) –  Jul 13 '13 at 08:10
  • The remaining case is if $a,b$ are of opposite signs. Here, suppose $a+b \sqrt{2}$ is a unit, $>1, a>0, b<0$. Then $a-b \sqrt{2}$ is also larger than 1, with $a,-b>0$, whence from the previous case $a-b\sqrt{2}$ can be expressed as $(1+\sqrt{2})^n$ for some $n \in N$. Multiplying both sides by $a+b \sqrt{2}$, we get $a+b \sqrt{2}=(a^2-2b^2)(-1+\sqrt{2})^n = (\pm 1)(-1+\sqrt{2})^n$, which seems impossible because $(-1+\sqrt{2})^n$ can't be $>1$. So, it looks like if $a,b$ have different signs, $a+b \sqrt{2}$ must lie in $(-1,1)$. So this remaining case is impossible. Is this correct? – AG. Jul 13 '13 at 14:09
  • To summarize, if $u=a+b \sqrt{2}$ is a unit, it is of the form $(\pm 1 \pm \sqrt{2})^n$ for some positive integer $n$. If $u$ is a unit, so is $-u$, so it suffices to consider $u>0$. If $u \in (1,\infty)$, we showed $a,b>0$ and $u=(1+\sqrt{2})^n$, while if $u \in (0,1)$, as you explained we can multiply to get a unit $>1$ and show that $u=(-1+\sqrt{2})^n$ for some $n$. – AG. Jul 13 '13 at 14:16

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