## Construct elements of infinite multiplicative order in some quadratic integer rings

Show that for $D \in \{ 3,5,6,7 \}$ the group of units in $\mathbb{Z}[\omega]$ is infinite by exhibiting an explicit unit of infinite multiplicative order.

We begin with a lemma.

Lemma: If $a + b\sqrt{D} \in \mathbb{Q}(\sqrt{D})$, where $D > 1$ is squarefree, $a > 1$, and $b \geq 1$, then $a + b \sqrt{D}$ has infinite multiplicative order in $\mathbb{Q}(\sqrt{D})$. Proof: We prove by induction that if $(a+b\sqrt{D})^n = p + q \sqrt{D}$, then $p > 1$ and $q \geq 1$. The base case $n = 1$ holds by hypothesis. Now suppose $(a+b\sqrt{D})^n = p + q \sqrt{D}$, and that $p > 1$ and $q \geq 1$. Then $(a + b\sqrt{D})^{n+1} = (ap + bqD) + (aq + bp)\sqrt{D}$, and we have $1 < p < ap + bqD$ and $1 \leq q < aq + bp$. By induction, $(a + b\sqrt{D})^n = 1 + 0\sqrt{D}$ has no solution $n$. $\square$

Using the lemma, it suffices to find, for each $D$, an element $a + b \sqrt{D}$ such that $a$ and $b$ are integers, $a > 1$, $b \geq 1$, and $N(a+b\sqrt{D}) = a^2 - Db^2 = 1$.

1. For $D = 3$, note that $N(2 + \sqrt{3}) = 1$.
2. For $D = 5$, note that $N(9 + 4\sqrt{5}) = 1$.
3. For $D = 6$, note that $N(5 + 2 \sqrt{6}) = 1$.
4. For $D = 7$, note that $N(8 + 3 \sqrt{7}) = 1$.

• Kreizhn  On June 15, 2011 at 3:59 pm

I think there might be a typo here. In particular, was is d? Should this be D?

• nbloomf  On June 15, 2011 at 9:03 pm

That was a typo. Thanks!

• Joseph  On July 4, 2011 at 10:46 am

Is the norm of a+bw not different in the case when D=5 as it is congruent to 1 mod 4.

• nbloomf  On July 4, 2011 at 4:36 pm

When we write the elements as $a+b\sqrt{D}$, the norm is $N(a+b\sqrt{D})$ is $a^2 - Db^2$. When $D$ is 1 mod 4, the ring of integers is enlarged to include elements of the form $a+b\omega$ where $\omega = \frac{1+\sqrt{D}}{2}$. Now $N(a+b\omega)$ looks different.