Solution to Abstract Algebra by Dummit & Foote 3rd edition Chapter 7.1 Exercise 7.1.24
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.
Solution: 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 $$N(a+b\sqrt{D}) = a^2 – Db^2 = 1.$$For $D = 3$, note that $N(2 + \sqrt{3}) = 1$.
For $D = 5$, note that $N(9 + 4\sqrt{5}) = 1$.
For $D = 6$, note that $N(5 + 2 \sqrt{6}) = 1$.
For $D = 7$, note that $N(8 + 3 \sqrt{7}) = 1$.
