# The units of $\mathbb Z[\sqrt{2}]$

How can I show that the units $u$ of $R=\mathbb Z[\sqrt{2}]$ with $u>1$ are $(1+ \sqrt{2})^{n}$ ?

I have proved that the right ones are units because their module is one, and it is said to me to do it by induction on $b$ and multiplication by $-1+\sqrt{2}$. I have already shown that the units of this ring has norm $1$ and all the numbers with norm $1$ are units, this may help.

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Since this is equivalent with finding integer solutions of a specific Pell's equation $x²-2y²=1$, it could be done by use of continued fractions expansion of $\sqrt2$. In addition, this question is also equivalent with determining the rank of a peculiar quadratic number ring, which should involve some cohomological calculations and some genus theory(or not?). In any case, I see no reason not to tag the number-theory tag... –  awllower Jan 17 '13 at 18:08
en.wikipedia.org/wiki/Continued_fraction See its theorem 3 –  awllower Jan 17 '13 at 18:17
@awllower Actually, it is equivalent to finding solutions to$x^2-2y^2=\pm 1$. Indeed, $1+\sqrt{2}$ corresponds to such a solution for $x^2-2y^2=-1$ –  Thomas Andrews Jan 17 '13 at 18:21

First, note that $a+b\sqrt{2}$ is a unit if and only if $a^2-2b^2=\pm1$. Use that to show that if $b\neq 0$ then $|b|\leq |a|< 2|b|$.

Now we first restrict ourselves to $a,b\geq 0$, and prove by induction on $b$.

If $b=0$ then $a=\pm 1$, and $u>0$ implies $a=1$, so $u=(1+\sqrt 2)^0$.

If $a,b>0$ and $a+b\sqrt{2}$ is a unit with then $$(a+b\sqrt{2})(\sqrt{2}-1) = (2b-a)+(a-b)\sqrt{2}$$ is also a unit.

Since we know that $b\leq a< 2b$, we have that $2b-a>0$ and $a-b<b$, so by induction:

$$(a+b\sqrt{2})(\sqrt{2}-1) =(1+\sqrt{2})^n$$

But multiplying both sides by $1+\sqrt{2}$ you get:

$$a+b\sqrt{2}=(1+\sqrt{2})^{n+1}$$

Then you have to deal with the case where one of $a,b$ is negative...

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$a+b\sqrt(2)$ is a unit if and only if $a^2−2b^2|a$ and $a^2−2b^2|b$; why does this imply $a^2−2b^2=±1$? –  Temitope.A Jan 18 '13 at 23:17
thomas Whhy is it so? –  Temitope.A Jan 19 '13 at 10:34
@Temitope.A If $a+b\sqrt{2}$ is a unit, show that $a-b\sqrt{2}$ is also a unit, so $a^2-2b^2=(a+b\sqrt{2})(a-b\sqrt{2})$ is a unit. But the only elements of $\mathbb Z$ that are units in $\mathbb Z[\sqrt{2}]$ are $\pm 1$ –  Thomas Andrews Jan 19 '13 at 12:49
@Temitope.A Alternatively, you could show directly that if $a,b$ are not relatively prime, then $a+b\sqrt{2}$ cannot be a unit. –  Thomas Andrews Jan 19 '13 at 12:52
Let $a+b \sqrt{2}$ be a unit. To show $a^2-2b^2=\pm 1$, define $N(a+b \sqrt{2}):=a^2-2b^2$. It can be verified that $N(\alpha \beta)=N(\alpha)N(\beta), \forall \alpha, \beta \in \mathbb{Z}[\sqrt{2}]$. If $\alpha=a+b \sqrt{2}$ is a unit, then $\alpha \beta=1$ for some $\beta$, whence $N(\alpha)N(\beta)=1$ in $\mathbb{Z}$. Thus, $N(\alpha)=\pm 1$. –  AG. Jul 13 '13 at 0:07

Here is a simpler proof that relies only on basic bounding arguments. Note that if $u$ is a unit then $u, 1/u, -u, -1/u$ are all units also. Now, since once of these is greater than $1$, it suffices to show if $u<1$ then $u$ is a power of $1+\sqrt{2}$. Clearly the nonnegative powers of $1+\sqrt{2}$ monotonically tend to $\infty$ starting from $1$, so we can write $$(1+\sqrt{2})^k\le u <(1+\sqrt{2})^{k+1}$$ for some $k\in\mathbf{Z}^{+}\cup\{0\}$. Dividing by $(1+\sqrt{2})^k$ yields $$1\le u(1+\sqrt{2})^{-k}<1+\sqrt{2}.$$ Note that $u(1+\sqrt{2})^{-k}\in\mathbf{Z}[\sqrt{2}]^{\times}$, and since $1+\sqrt{2}$ is the smallest unit greater than $1$, we must have $u(1+\sqrt{2})^{-k}=1\implies u=(1+\sqrt{2})^k$. Due to norm being multiplicative, all powers of $1+\sqrt{2}$ are units, so we are done.

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