Let $K=\mathbb{Q}(\sqrt{d})$, the quadratic field over $\mathbb{Q}$, where $\mathbb{Q}$ is the rational number field and $d$ is a square-free integer other than $0$ and $1$. The ring of algebraic integers of $K$ is denoted by $R_d$, and it is very important for the study of dynamical systems, e.g., see [1, 2]. We call $R_d$ an imaginary quadratic number ring if $d < 0$. From the work of Stark [3], we know that there are only finite negative integers $d$ such that the complex quadratic ring $R_d$ is a unique-factorization domain, namely, $d=-1, -2, - 3, - 7, -11, - 19, - 43, - 67, - 163$. For an arbitrary prime element $\vartheta\in R_d$, and a positive integer $n$, the unit groups of $R_d/\langle{\vartheta^n}\rangle$ were determined for the cases $d=-1, -2, -3$ in [4-6], respectively. Moreover, the square mapping graphs for the Gaussian integer ring modulo $n$ is studied in paper [7]. In this paper, we investigate the unit groups of $R_d/\langle{\vartheta^n}\rangle$ for the cases $d=-3, -7, -11, -19, -43, -67, -163$, and we make some corrections to the case of $d=-3$ in paper [6].

Throughout this paper, we denote by $\mathbb{Z}$ the set of rational integers, $\mathbb{Z}_n$ is the additive cyclic group of order $n$, $\mathbb{Z}/\langle n\rangle$ is the ring of integers modulo $n$, and $o(\theta)$ is the order of $\theta$ in a group. For a given ring $R$, let $U(R)$ denote the unit group of $R$, let $\langle \gamma\rangle$ denote the ideal of $R$ generated by $\gamma\in R$. If $\gamma$ is an element of a given group $G$, we also use $\langle \gamma\rangle$ to denote the subgroup of $G$ generated by $\gamma\in G$. The Legendre symbol $(\frac{a}{p})$, where $a$ is an integer, $p$ is a prime and $p\nmid a$, is defined as follows: if there exists an integer $x$ such that $x^{2}\equiv a\ ({\rm mod}\ p)$, then $(\frac{a}{p})=1$, otherwise, $(\frac{a}{p})=-1$.

Lemma 1.1 [8, Lemma 2.4.2] The ring $R_d$ of algebraic integers of $K=\mathbb{Q}(\sqrt{d})$ is

(1) $R_d=\{a+b\sqrt{d}:\ a, b\in\mathbb{Z}\}$, if $d\equiv\, 2, 3\ ({\rm mod}\ 4)$.

(2) $R_d=\{\frac{1}{2}(a+b\sqrt{d}):\ a, b\in\mathbb{Z}\ \rm are\ of\ the\ same\ parity\}$, if $d\equiv\, 1\ ({\rm mod}\ 4)$.

By Lemma 1.1, for $d=-3, -7, -11, -19, -43, -67, -163$, the elements of $R_d$ are all of the form $\frac{1}{2}(a+b\sqrt{d})$, where $a, b\in\mathbb{Z}$ are of the same parity. Moreover, we know that $U(R_d)=\{\pm1\}$ for all $d=-3, -7, -11, -19, -43, -67, -163$.

Now, we need to identify all primes in the ring $R_d$. The following theorem is obtained from [9, Theorem 9.29].

Theorem 1.2  For $d=-3, -7, -11, -19, -43, -67, -163$, up to multiplication by units, the primes of $R_d$ are the following three types ($D=-d$):

(1) $p$, where $p\in\mathbb{Z}$ is a prime satisfying the Legendre symbol $(\frac{p}{D})=-1$;

(2) $\pi$ or $\overline\pi$, where $q=\pi\overline\pi\in\mathbb{Z}$ is a prime satisfying the Legendre symbol $(\frac{q}{D})=1$;

(3) $\delta=\sqrt{d}$.

Throughout this section, $d=-3, -7, -11, -19, -43, -67, -163$. For conveniences, we denote by $D=-d$. Let $n$ be a positive integer, and $\vartheta$ is a prime in $R_d$. We determine the structure of unit groups of $R_d/\langle\vartheta^n\rangle$.

First, we characterize the equivalence classes of $R_{d}/\langle\vartheta^{n}\rangle$, where $\vartheta$ is prime in $R_d$. For $\alpha\in R_d$, we denote by $[\alpha]\in R_{d}/\langle\vartheta^{n}\rangle$ the equivalence class which $\alpha$ belongs to. Simultaneously, we make corrections to the equivalence classes which are given in [6, Theorem 3.2] for the case $d=-3$.

Theorem 2.3  Let $\vartheta$ denote a prime of $R_d$, $\delta=\sqrt{d}$, $D=-d$. For an arbitrary positive integer $n$, the equivalence classes of $R_d/\langle{\vartheta^n}\rangle$ are of the following types:

(1) $R_d/\langle{\delta^{2m}}\rangle=\left\{[r_1+r_2\sqrt{d}\, ]:\ 0\leqslant r_i\leqslant D^{m}-1, r_i\in \mathbb{Z}, i=1, 2\right\}$, $m\geqslant1$;

(2) $R_d/\langle{\delta^{2m+1}}\rangle=\left\{[r_1+r_2\sqrt{d}\, ]:\ 0\leqslant r_1\leqslant D^{m+1}-1, 0\leqslant r_2\leqslant D^{m}-1, r_1, r_2\in \mathbb{Z}\right\}$, $m\geqslant0$;

(3) $R_d/\langle{p^n}\rangle=\left\{[r_1+r_2\sqrt{d}\, ]:\ 0\leqslant r_i\leqslant p^n-1, r_i\in \mathbb{Z}, i=1, 2\right\}$, where $p$ is an odd prime in $\mathbb{Z}$ satisfying the Legendre symbol $(\frac{p}{D})=-1$;

(4) $R_d/\langle{\pi^n}\rangle=\{[a]:\ 0\leqslant a\leqslant q^n-1, a\in \mathbb{Z}\}$, where $q=\pi\overline{\pi}$ is a prime in $\mathbb{Z}$ satisfying the Legendre symbol $(\frac{q}{D})=1$;

(5) Suppose that $d\neq-7$. Then

(a) $R_d/\langle{2}\rangle=\left\{[0], [1], [\frac{\, 1\, }{2}{+}\frac{\, 1\, }{2}\sqrt{d}\, ], [\frac{\, 1\, }{2}{-}\frac{\, 1\, }{2}\sqrt{d}\, ]\right\};$

(b) For $n\geqslant2$, $R_d/\langle{2^n}\rangle=R_1\cup R_2\cup R_3$, where

$ \begin{eqnarray*} R_1&=&\left\{[r_1+r_2\sqrt d\, ]:\ 0\leqslant r_i\leqslant 2^{n-1}-1, \ \ r_i\in \mathbb{Z}, \ \ i=1, 2\right\}, \\ R_2&=&\left\{[r_1-r_2\sqrt d\, ]:\ 0\leqslant r_1\leqslant 2^{n-1}-1, \ \ 1\leqslant r_2\leqslant 2^{n-1}, \ \ r_1, r_2\in \mathbb{Z}\right\}, \\ R_3&=&\left\{[\frac{r_1}{2}\pm\frac{r_2}{2}\sqrt d\, ]:\ 1\leqslant r_i\leqslant 2^{n}-1, \ \ r_i\in \mathbb{Z}, \ \ 2\nmid r_i, \ \ i=1, 2 \right\}. \end{eqnarray*} $

Proof  (1) As $\delta^{2m}=d^m$, we get that $\langle{\delta^{2m}}\rangle=\langle{D^m}\rangle$. Suppose $\alpha=a_1+a_2\sqrt{d}\in R_d$, where $a_1, a_2\in \mathbb{Z}$. Let $a_i=D^m k_i+r_i$ with $0\leqslant r_i\leqslant D^m-1$, $k_i\in \mathbb{Z}$, $i=1, 2$. Then $\alpha=(r_1+r_2\sqrt d)+D^m(k_1+k_2\sqrt d)$. So $\alpha$ and $r_1+r_2\sqrt d$ belong to the same equivalence class of $R_d/\langle{\delta^{2m}}\rangle$.

On the other hand, let $\beta= \frac{\, 1\, }{2}(b_1+b_2\sqrt{d})\in R_d$, where $b_1$ and $b_2$ are odd integers. Since $D$ is odd for $i=1, 2$, there exists a unique integer $g_i\in\{0, 1, \cdots, D^m-1\}$ satisfying the congruence $2g_i\equiv\, b_i\ ({\rm mod}\ D^m)$. Hence, there exists an odd integer $x_i$ such that $b_i=D^m x_i+2g_i$, $i=1, 2$. Therefore, $\gamma=\frac{x_1}{2}+\frac{x_2}{2}\sqrt{d}\in R_d$, and $\beta=(g_1+g_2\sqrt d)+D^m\gamma$, which implies that $\beta$ and $g_1+g_2\sqrt d$ belong to the same equivalence class of $R_d/\langle{\delta^{2m}}\rangle$. Finally, it is easy to verify that the classes of $(1)$ are distinct.

(2) As $\delta^{2m+1}=d^m\delta$, we get that $\langle{\delta^{2m+1}}\rangle=\langle D^m\sqrt{d}\, \rangle$. Suppose $\alpha=a_1+a_2\sqrt{d}\in R_d$, where $a_1, a_2\in \mathbb{Z}$. Let $a_1=D^{m+1} k_1+r_1$ with $0\leqslant r_1\leqslant D^{m+1}-1$. Let $a_2=D^m k_2+r_2$ with $0\leqslant r_2\leqslant D^m-1$. Then $\alpha=(r_1+r_2\sqrt d)+D^m\sqrt{d}(k_2-k_1\sqrt d)$. So $\alpha$ and $r_1+r_2\sqrt d$ belong to the same equivalence class of $R_d/\langle{\delta^{2m+1}}\rangle$.

On the other hand, let $\beta=\frac{\, 1\, }{2}(b_1+b_2\sqrt{d})\in R_d$, where $b_1$ and $b_2$ are odd integers. Since $D$ is odd, there exists a unique integer $g_1\in\{0, 1, \cdots, D^{m+1}-1\}$ satisfying congruence $2g_1\equiv\, b_1\ ({\rm mod}\ D^{m+1})$. Analogously, there exists a unique integer $g_2\in\{0, 1, \cdots, D^{m}-1\}$ satisfying congruence $2g_2\equiv\, b_2\ ({\rm mod}\ D^{m})$. Therefore, there exist odd integers $x_1, x_2$ such that $b_1=D^{m+1} x_1+2g_1$, and $b_2=D^{m} x_2+2g_2$. Hence, $\gamma=\frac{x_2}{2}-\frac{x_1}{2}\sqrt{d}\in R_d$, and $\beta=(g_1+g_2\sqrt d)+D^m\sqrt d(\frac{x_2}{2}-\frac{x_1}{2}\sqrt d)$, which implies that $\beta$ and $g_1+g_2\sqrt d$ belong to the same equivalence class of $R_d/\langle{\delta^{2m+1}}\rangle$.

Finally, it is easy to verify that the classes of $(2)$ are distinct.

(3) It can be proved with the similar method to $(1)$. Suppose $\alpha=a_1+a_2\sqrt{d}\in R_d$, where $a_1, a_2\in \mathbb{Z}$. Let $a_i=p^n k_i+r_i$ with $0\leqslant r_i\leqslant p^n-1$, $k_i\in \mathbb{Z}$, $i=1, 2$. Then $\alpha=(r_1+r_2\sqrt d)+p^n(k_1+k_2\sqrt d)$. So $\alpha$ and $r_1+r_2\sqrt d$ belong to the same equivalence class of $R_d/\langle{p^n}\rangle$.

On the other hand, let $\beta= \frac{\, 1\, }{2}(b_1+b_2\sqrt{d})\in R_d$, where $b_1$ and $b_2$ are odd integers. Since $p$ is odd for $i=1, 2$, there exists a unique integer $g_i\in\{0, 1, \cdots, p^n-1\}$ satisfying the congruence $2g_i\equiv\, b_i\ ({\rm mod}\ p^n)$. Hence, there exists an odd integer $x_i$ such that $b_i=p^n x_i+2g_i$, $i=1, 2$. Therefore, $\gamma=\frac{x_1}{2}+\frac{x_2}{2}\sqrt{d}\in R_d$, and $\beta=(g_1+g_2\sqrt d)+p^n\gamma$, which implies that $\beta$ and $g_1+g_2\sqrt d$ belong to the same equivalence class of $R_d/\langle{p^n}\rangle$. Finally, it is easy to verify that the classes of $(3)$ are distinct.

(4) Let $q=\pi\overline{\pi}$ be a prime in $\mathbb{Z}$ satisfying the Legendre symbol $(\frac{q}{D})=1$. Let $ \pi^n= \frac{\, 1\, }{2}(s+t\sqrt{d})$, where $s, t\in\mathbb{Z}$ are of the same parity. Then it is clear that $q\nmid st$. Suppose that $\beta=\frac{\, 1\, }{2}(b_1+b_2\sqrt{d})\in R_d$, where $b_1, b_2\in \mathbb{Z}$ are of the same parity. We show that in the quotient ring $R_d/\langle{\pi^n}\rangle$, $\beta$ belongs to the equivalence class $[a]$ for some $a\in\{0, 1, \cdots, q^n-1\}$. Indeed, Let $\gamma=\frac{\, 1\, }{2}(x+y\sqrt{d})\in R_d$, where $x, y\in \mathbb{Z}$ are of the same parity, such that $\beta=a+\pi^n\gamma$. Then the following equations hold

$ a+\frac{\, 1\, }{4}xs+\frac{\, 1\, }{4}dyt=\frac{\, 1\, }{2}b_1, $

(2.1)

$ \frac{\, 1\, }{4}ys+\frac{\, 1\, }{4}xt=\frac{\, 1\, }{2}b_2. $

(2.2)

Now we solve the integer $a$ from the above equations. By equation (2.1), we obtain

$ 4as+xs^2+dyts=2b_1s. $

(2.3)

And by equation (2.2), we get $-dyts-dt^2x=-2b_2dt.$ Eliminating $dyts$ between this equation and (2.3), we obtain

$ 4as+x(s^2-dt^2)=2(b_1s-db_2t). $

(2.4)

Note that $q=\pi\overline{\pi}$ and $\pi^n= \frac{\, 1\, }{2}(s+t\sqrt{d})$, we have $s^2-dt^2=4q^n$. Substituting this into (2.4), it follows that

$ 4as+4q^nx=2(b_1s-db_2t). $

(2.5)

Moreover, since $s, t\in\mathbb{Z}$ are of the same parity and $b_1, b_2\in \mathbb{Z}$ are of the same parity and note that $d$ is odd, we derive $b_1s-db_2t$ is even. Hence, equation (2.5) can be written as $as+q^nx=\frac{\, 1\, }{2}(b_1s-db_2t), $ which implies that

$ as\equiv\, \frac{\, 1\, }{2}(b_1s-db_2t)\ ({\rm mod}\ q^n). $

(2.6)

Because $q\nmid s$, the last congruence (2.6) in $a$ has a unique solution $a\in\{0, 1, \cdots, q^n-1\}$. Therefore, $\beta$ belongs to the equivalence class $[a]$, as desired.

Finally, it is easy to verify that the classes of $(4)$ are distinct.

(5) Suppose $d\neq-7$.

(a) We first determine the structure of the quotient ring $R_d/\langle{2}\rangle$. Suppose $\alpha_1=a\in\mathbb{Z}$. If $a$ is even, then $\frac{\, a\, }{2}\in R_d$. It follows from $\alpha_1=0+2\times\frac{\, a\, }{2}$ that $\alpha_1$ belongs to the equivalence class $[0]$ in the quotient ring $R_d/\langle{2}\rangle$. If $a$ is odd, then $a=1+2k$ for some $k\in \mathbb{Z}$. Then clearly $\alpha_1$ belongs to the equivalence class $[1]$.

Suppose $\alpha_2=b\sqrt d$, where $b\in\mathbb{Z}$. If $b$ is even, then $\frac{\, b\, }{2}\sqrt d\in R_d$. We have

$ \alpha_2=b\sqrt d=0+2\times \frac{\, b\, }{2}\sqrt d. $

So clearly $\alpha_2$ belongs to the equivalence class $[0]$. If $b$ is odd, then

$ \alpha_2=b\sqrt d=1+2(-\frac{\, 1\, }{2}+\frac{\, b\, }{2}\sqrt d). $

Therefore, $\alpha_2$ belongs to the equivalence class $[1]$.

Suppose $\alpha_3=s+t\sqrt d\in R_d$, where $s, t\in \mathbb{Z}$. If $s$ and $t$ are of the same parity, then $\frac{\, s\, }{2}+\frac{\, t\, }{2}\sqrt d\in R_d$. Moreover, we have $s+t\sqrt d=0+2(\frac{\, s\, }{2}+\frac{\, t\, }{2}\sqrt d)$. Hence, $\alpha_3$ belongs to the equivalence class $[0]$. If $s$ and $t$ are not of the same parity, then $\frac{s-1}{2}+\frac{\, t\, }{2}\sqrt d\in R_d$. Since $s+t\sqrt d=1+2(\frac{s-1}{2}+\frac{\, t\, }{2}\sqrt d)$, we obtain that $\alpha_3$ belongs to the equivalence class $[1]$.

Now, suppose $\alpha_4=\frac{\, x\, }{2}+\frac{\, y\, }{2}\sqrt d$, where $x=2k_1+1$, $y=2k_2+1$, $k_1, k_2\in \mathbb{Z}$. If $k_1$ and $k_2$ are of the same parity, then $\frac{k_1}{2}+\frac{k_2}{2}\sqrt d\in R_d$. Moreover, since $\alpha_4=(\frac{\, 1\, }{2}+\frac{\, 1\, }{2}\sqrt d)+2(\frac{k_1}{2}+\frac{k_2}{2}\sqrt d)$, we obtain that $\alpha_4$ belongs to the equivalence class $[\frac{\, 1\, }{2}+\frac{\, 1\, }{2}\sqrt d\, ]$. If $k_1$ and $k_2$ are not of the same parity, then $\frac{k_1}{2}+\frac{k_2+1}{2}\sqrt d\in R_d$. Furthermore, $\alpha_4=(\frac{\, 1\, }{2}-\frac{\, 1\, }{2}\sqrt d)+2(\frac{k_1}{2}+\frac{k_2+1}{2}\sqrt d)$. Thus, $\alpha_4$ belongs to the equivalence class $[\frac{\, 1\, }{2}-\frac{\, 1\, }{2}\sqrt d\, ]$.

Finally, we show that the classes of $(5)$ (a) are distinct. Clearly

$ [0]\neq [1]\neq [\frac{\, 1\, }{2}\pm\frac{\, 1\, }{2}\sqrt d\, ]\neq[0]. $

If $[\frac{\, 1\, }{2}+\frac{\, 1\, }{2}\sqrt d\, ]=[\frac{\, 1\, }{2}-\frac{\, 1\, }{2}\sqrt d\, ]$, then there exits $\gamma=\frac{x_1}{2}+\frac{x_2}{2}\sqrt d\in R_d$, where $x_1, x_2\in\mathbb{Z}$ are of the same parity, such that

$ \frac{\, 1\, }{2}+\frac{\, 1\, }{2}\sqrt d=(\frac{\, 1\, }{2}-\frac{\, 1\, }{2}\sqrt d)+2(\frac{x_1}{2}+\frac{x_2}{2}\sqrt d). $

Clearly, the above equation holds if and only if $x_1=0$ and $x_2=1$, which is impossible, since $x_1, x_2\in\mathbb{Z}$ must be of the same parity. Hence, we conclude that $[\frac{\, 1\, }{2}+\frac{\, 1\, }{2}\sqrt d\, ]\neq[\frac{\, 1\, }{2}-\frac{\, 1\, }{2}\sqrt d\, ]$. Therefore, the classes of (5) (a) are distinct.

(b) Now, let $n\geqslant2$. We determine the structure of the quotient ring $R_d/\langle{2^n}\rangle$. Suppose $\beta_1=a_1+a_2\sqrt d\in R_d$, where $a_1, a_2\in \mathbb{Z}$. Let $a_i=2^{n-1}k_i+r_i$, $k_i, r_i\in\mathbb{Z}$, and $0\leqslant r_i\leqslant 2^{n-1}-1$ for $i=1, 2$. First, if $k_1$ and $k_2$ are of the same parity, then $\frac{k_1}{2}+\frac{k_2}{2}\sqrt d\in R_d$. Moreover, since $\beta_1=(r_1+r_2\sqrt d)+2^n(\frac{k_1}{2}+\frac{k_2}{2}\sqrt d)$, we conclude that $\beta_1$ and $r_1+r_2\sqrt d$ belong to the same equivalence class in the quotient ring $R_d/\langle{2^n}\rangle$. Secondly, if $k_1$ and $k_2$ are not of the same parity, then $\frac{k_1}{2}+\frac{k_2+1}{2}\sqrt d\in R_d$. Since $\beta_1=[r_1-(2^{n-1}-r_2)\sqrt d\, ]+2^n(\frac{k_1}{2}+\frac{k_2+1}{2}\sqrt d)$, we obtain that $\beta_1$ and $r_1-(2^{n-1}-r_2)\sqrt d$ belong to the same equivalence class. Furthermore, since $0\leqslant r_2\leqslant 2^{n-1}-1$, we derive that $ 1\leqslant2^{n-1}-r_2\leqslant 2^{n-1}$. So in the second case, i.e., $k_1$ and $k_2$ are not of the same parity, we get that $\beta_1$ and $r_1-r\, '_2\sqrt d$ belong to the same equivalence class, where $1\leqslant r\, '_2\leqslant 2^{n-1}$ and $r\, '_2=2^{n-1}-r_2$.

Next, suppose that $\beta_2=\frac{b_1}{2}+\frac{b_2}{2}\sqrt d$, where $b_1$ and $b_2$ are odd integers. Let $b_i=2^n k_i+r_i$, where $k_i, r_i\in\mathbb{Z}$, $1\leqslant r_i\leqslant 2^{n}-1$ and $2\nmid r_i$ for $i=1, 2$. First, if $k_1$ and $k_2$ are of the same parity, then $\frac{k_1}{2}+\frac{k_2}{2}\sqrt d\in R_d$. Moreover, since $\beta_2=(\frac{r_1}{2}+\frac{r_2}{2}\sqrt d)+2^n(\frac{k_1}{2}+\frac{k_2}{2}\sqrt d)$, we obtain that $\beta_2$ and $\frac{r_1}{2}+\frac{r_2}{2}\sqrt d$ belong to the same equivalence class. Secondly, if $k_1$ and $k_2$ are not of the same parity, then $\frac{k_1}{2}+\frac{k_2+1}{2}\sqrt d\in R_d$. Since $\beta_2=(\frac{r_1}{2}-\frac{2^n-r_2}{2}\sqrt d)+2^n(\frac{k_1}{2}+\frac{k_2+1}{2}\sqrt d)$, it follows that $\beta_2$ and $\frac{r_1}{2}-\frac{2^n-r_2}{2}\sqrt d$ belong to the same equivalence class. Furthermore, according to $1\leqslant r_2\leqslant 2^n-1$, we have $1\leqslant2^n-r_2\leqslant 2^n-1$. So, in the second case, i.e., $k_1$ and $k_2$ are not of the same parity, we obtain that $\beta_2$ and $\frac{r_1}{2}-\frac{r\, '_2}{2}\sqrt d$ belong to the same equivalence class, where $1\leqslant r\, '_2\leqslant 2^{n}-1$ and $r\, '_2=2^{n}-r_2$.

Finally, we claim that the classes of (5) (b) are distinct. We only show that

$ [\frac{r_1}{2}+\frac{r_2}{2}\sqrt d\, ]\neq [\frac{x_1}{2}-\frac{x_2}{2}\sqrt d\, ], $

where $r_i, x_i\in \{1, 3, \cdots, 2^n-1\}$ with $2\nmid r_ix_i$ for $i=1, 2$. Indeed, if $[\frac{r_1}{2}+\frac{r_2}{2}\sqrt d\, ]= [\frac{x_1}{2}-\frac{x_2}{2}\sqrt d\, ]$, then there exit $t_1, t_2\in \mathbb{Z}$ of the same parity such that

$ \frac{r_1}{2}+\frac{r_2}{2}\sqrt d=(\frac{x_1}{2}-\frac{x_2}{2}\sqrt d)+2^n(\frac{t_1}{2}+\frac{t_2}{2}\sqrt d). $

So we obtain $r_1=x_1+2^nt_1$ and $r_2=-x_2+2^nt_2$. It is easy to show that $t_1=0$ and $t_2=1$, which is a contradiction.

Example 2.4  To illustrate the case $d=-19$, $q=23=\pi\, \overline\pi$ and $n=2$, let $\gamma=\frac{1}{2}(b_1+b_2\sqrt{-19})\in R_d$, where $b_1=3$ and $b_2=1$. We give the equivalence class in $R_d/\langle{\pi^2}\rangle$ which $\gamma$ belongs to. Since $\pi=2-\sqrt{-19}$ is a proper factor of $q$ in $R_{d}$, $\pi^{2}=-15-4\sqrt{-19}=\frac{-30}{2}-\frac{8}{2}\sqrt{-19}$. Denoted by $s=-30$, $t=-8$. Substituting the values for $s, t, b_1, b_2, d, q$ and $n$ into congruence (2.6), we get that $a=198$ is a solution to congruence (2.6). Moreover, substituting the values for $a, s, t, b_1, b_2$ and $d$ into equations (2.1) and (2.2), we have $x=11$ and $y=-3$. Therefore,

$ \gamma=\frac{\, 3\, }{2}+\frac{\, 1\, }{2}\sqrt{-19}=198+\pi^2(\frac{\, 11\, }{2}-\frac{\, 3\, }{2}\sqrt{-19}), $

which implies that $\gamma$ belongs to the class $[198]$.

As an easy consequence of Theorem 2.1 (5), we have

Corollary 2.5  Suppose that $2$ is prime in $R_d$. Let $\alpha=[a+b\sqrt d]\in R_d/\langle{2^n}\rangle$, where $0\leqslant a, b\leqslant 2^{n-1}-1$, $a, b\in \mathbb{Z}$. Then

(1) $\alpha=[1]$ if and only if $a=2^{n-1}k_1+1$, $b=2^{n-1}k_2$, where $k_1, k_2\in\mathbb{Z} $ are of the same parity.

(2) If $a=2^nk_1+1$, $b=2^nk_2$, $k_1, k_2\in\mathbb{Z}$, then $\alpha=[1]$.

Now, we determine the structure of unit groups of $R_{d}/\langle\vartheta^{n}\rangle$ for an arbitrary prime $\vartheta$ of $R_d$. First of all, we consider the case of $\vartheta=\delta=\sqrt{d}$. Let $\overline R=R_d/\langle{\delta^n}\rangle$. For $\alpha=[a+b\sqrt{d}\, ]\in\overline R$, it is easy to show that $\alpha\in U(\overline R)$ if and only if $d\nmid (a^2-db^2)$, if and only if $d\nmid a$, if and only if $D\nmid a$.

Theorem 2.6  Let $\overline R=R_d/\langle{(\sqrt d)^{n}}\rangle$, $n$ is an arbitrary positive integer. Let $D=-d$. Then the unit groups $U(\overline R)$ of $\overline R$ are as the follows:

(1) Let $n=1$. Then $U(\overline R)\cong\mathbb{Z}_{D-1}$.

(2) Let $n=2$. Then $U(\overline R)\cong\mathbb{Z}_{D-1}\times\mathbb{Z}_{D}$.

(3) Let $n=2m$ with $m\geqslant2$.

(a) If $d\neq-3$, then $U(\overline R)\cong\mathbb{Z}_{D-1}\times\mathbb{Z}_{D^{m-1}}\times\mathbb{Z}_{D^{m}}$;

(b) If $d=-3$, then $U(\overline R)\cong\mathbb{Z}_2\times\mathbb{Z}_3\times\mathbb{Z}_{3^{m-1}}\times\mathbb{Z}_{3^{m-1}}.$

(4) If $n=2m+1$ with $m\geqslant1$, then $U(\overline R)\cong\mathbb{Z}_{D-1}\times\mathbb{Z}_{D^{m}}\times\mathbb{Z}_{D^{m}}$.

Proof  (1) If $n=1$, by Theorem 2.1 (2), $\overline R$ is a field of order $D=-d$, so $|U(\overline R)|=D-1$. Therefore, $U(\overline R)$ is a cyclic group of order $D-1$ and hence $U(\overline R)\cong \mathbb{Z}_{D-1}$.

(2) If $n=2$, then $|U(\overline R)|=-d(-d-1)=D(D-1)$. Note that $D$ is a prime, moreover $D$ and $D-1$ are relatively prime, we get that $U(\overline R)\cong H\times\mathbb{Z}_{D}$, where $H$ is a subgroup of order $D-1$. Moreover, we can easily show that $D-1$ is square-free for $D=3, 7, 11, 43$ and $67$.

On the other hand, if $D=19$, then $D-1=2\times3^2$, clearly $[4]\in U(\overline R)$ is of order $3^2$. If $D=163$, then $D-1=2\times3^4$, clearly $[4]\in U(\overline R)$ is of order $3^4$. Therefore $H\cong\mathbb{Z}_{D-1}$. So $U(\overline R)\cong\mathbb{Z}_{D-1}\times\mathbb{Z}_{D}$.

(3) (a) Suppose that $d\neq-3$. Let $n=2m$ with $m\geqslant2$. Let $\alpha=[a+b\sqrt{d}\, ]\in\overline R$, where $a, b\in \{0, 1, \cdots, D^m-1\}$. Since $\alpha\in U(\overline R)$ if and only if $D\nmid a$, $|U(\overline R)|=(D-1) D^{2m-1}$, and we can write $U(\overline R)=P\times H$, where $P$, $H$ are finite groups, and $|P|=D-1$, $|H|=D^{2m-1}$.

We determine the structure of $H$. Let $\alpha=[a+b\sqrt{d}\, ]\in\overline R$ with $D\nmid a$. By Theorem 2.1 (1), for an arbitrary odd integer $W>1$, $\alpha^W$ equals to the equivalence class $[1]$, i.e., $\alpha^W=[1]$ if and only if the following congruences hold

$ a^W+d\left(\begin{smallmatrix}W\\2\end{smallmatrix}\right)a^{W-2}b^2+\cdots+d^{\frac{W-1}{2}}\left(\begin{smallmatrix}W\\W-1\end{smallmatrix}\right)ab^{W-1} \equiv\, 1\ ({\rm mod}\ D^m), $

(2.7)

$ \left(\begin{smallmatrix}W\\1\end{smallmatrix}\right)a^{W-1}b+d\left(\begin{smallmatrix}W\\3\end{smallmatrix}\right)a^{W-3}b^3+\cdots+d^{\frac{W-1}{2}}b^W \equiv\, 0\ ({\rm mod}\ D^m). $

(2.8)

First, we claim that for any $\alpha\in H$, $\alpha^{D^{m}}=[1]$. Let $W=D^{m}$. Since $d^m\, |\, d^j\left(\begin{smallmatrix}W\\2j\end{smallmatrix}\right)$ for $j\geqslant1$, the congruence (2.7) is equivalent to $a^{D^m}\equiv\, 1\ ({\rm mod}\ D^m)$. It is well known that the unit group of the ring $\mathbb{Z}/\langle D^m\rangle$ is isomorphic to $\mathbb{Z}_{D^{m-1}}\times \mathbb{Z}_{D-1}$. Hence, we obtain that $a^{D^m}\equiv\, 1\, ({\rm mod}\ D^m)$ if and only if $a\in\mathbb{Z}_{D^{m-1}}$. So in the set $\{0, 1, \cdots, D^m-1\}$, there are precisely $D^{m-1}$ elements $a$ such that $a^{D^m}\equiv\, 1\, ({\rm mod}\ D^m)$.

On the other hand, since $d^m\, |\, d^j\left(\begin{smallmatrix}W\\2j+1\end{smallmatrix}\right)$ for $j\geqslant0$, congruence (2.8) holds for any positive integer $b$. Therefore, we can conclude that $\alpha^W=[1]$ if and only if $a\in\mathbb{Z}_{D^{m-1}}$ and $b\in\{0, 1, \cdots, D^m-1\}$. Hence, the number of $\alpha\in U(\overline R)$ satisfying $\alpha^{D^{m}}=[1]$ is

$ D^{m-1}\times D^m=D^{2m-1}. $

Recall that $U(\overline R)=P\times H$ with $|P|=D-1$ and $|H|=D^{2m-1}$, we get that $\alpha^{D^{m}}=[1]$ for $\alpha\in H$.

Second, we consider the number of $\alpha\in U(\overline R)$ satisfying $\alpha^{D^{m-1}}=[1]$. Let $W=D^{m-1}$. Since $d^m\, |\, d^j\left(\begin{smallmatrix}W\\2j\end{smallmatrix}\right)$ for $j\geqslant1$, congruence (2.7) holds if and only if $a^{D^{m-1}}\equiv\, 1\ ({\rm mod}\ D^m), $ if and only if $a\in\mathbb{Z}_{D^{m-1}}$.

On the other hand, note that $d\neq-3$ and $d^m\, |\, d^j\left(\begin{smallmatrix}W\\2j+1\end{smallmatrix}\right)$ for $1\leqslant j\leqslant\frac{W-1}{2}$, congruence (2.8) is equivalent to $D^{m-1}a^{D^{m-1}-1}b\equiv\, 0\ ({\rm mod}\ D^m)$. That is, $D^{m-1}b\equiv\, 0\ ({\rm mod}\ D^m)$, since $D\nmid a$. Hence, we obtain $d\, |\, b$. So the solutions to congruence (2.8) are $b=D\cdot l$ with $l=0, 1, \cdots, D^{m-1}-1$. Thus the number of $\alpha\in U(\overline R)$ satisfying $\alpha^{D^{m-1}}=[1]$ is $D^{m-1}\times D^{m-1}=D^{2m-2}$. Then the number of elements of order $D^{m}$ in $U(\overline R)$ is

$ D^{2m-1}-D^{2m-2}= d\, ^{2m-2}(-d-1). $

Finally, let we calculate the number of $\alpha\in H$ satisfying $\alpha^{D^{m-2}}\equiv\, 1\ ({\rm mod}\ D^m)$. Let $W=D^{m-2}$. Since $d^m\, |\, d^j\left(\begin{smallmatrix}W\\2j+1\end{smallmatrix}\right)$ for $2\leqslant j\leqslant \frac{W-1}{2}$, congruence (2.8) holds if and only if

$ Wa^{W-3}b\, [6a^2+d(W-1)(W-2)b^2\, ]\equiv\, 0\ ({\rm mod}\ D^m). $

(2.9)

Since $D\nmid a$ and $d\neq-3$, we derive that $D\nmid [6a^2+d(W-1)(W-2)b^2]$. So congruence (2.9) holds if and only if $d^2\, |\, b$, i.e., congruence (2.8) holds if and only if $d^2\, |\, b$. Furthermore, in the case of $d^2\, |\, b$, we have $d^m \mid d^j\left(\begin{smallmatrix}W\\2j\\ \end{smallmatrix}\right)b^{2j}$ for $j\geqslant1$. Hence, in the case of $d^2\, |\, b$ congruence (2.7) holds if and only if $a^{W}\equiv\, 1\ ({\rm mod}\ D^m)$. Clearly, the number of solutions of the last congruence is $D^{m-2}$. Thus the number of $\alpha\in H$ such that $\alpha^{D^{m-2}}=1$ is $D^{m-2}\times D^{m-2}=d\, ^{2m-4}$. So we derive that the number of elements of order $D^{m-1}$ in $U(\overline R)$ is

$ D^{2m-2}-D^{2m-4}=d\, ^{2m-4}(d^2-1). $

(2.10)

Now, let $\beta=[1+\sqrt{d}\, ]\in\overline R$. Then by the above argument, we know that $\beta$ is of order $D^{m}$. Since $m\geqslant2$, clearly $\beta\in H$. Therefore $\mathbb{Z}_{D^{m}}$ is a subgroup of $H$ and we can suppose $H\cong\mathbb{Z}_{D^{m}}\times\mathbb{Z}_{D^{l_{1}}}\times\cdots\times\mathbb{Z}_{D^{l_{h}}}$, where $l_{1}+\cdots+l_{h}=m-1$. If $h\geqslant2$, then $1\leqslant l_{i}\leqslant m-2$ for $i=1, \cdots, h$ and hence there are exactly $(D-1)\cdot D^{2m-3}$ elements in $H$ of order $D^{m-1}$, which contradicts the above result (2.10). If $h=1$, then $H\cong\mathbb{Z}_{D^{m}}\times\mathbb{Z}_{D^{m-1}}$. Therefore, the number of elements of order $D^{m-1}$ in $H$ is $D^{m-1}\times D^{m-1}-D^{m-2}\times D^{m-2}=d\, ^{2m-4}(d^2-1)$, which is the same as (2.10). So we can conclude that $h=1$ and $H\cong\mathbb{Z}_{D^{m}}\times\mathbb{Z}_{D^{m-1}}$.

In the following, we determine the structure of the subgroup $P$ of $U(\overline R)$, where $|P|=-d-1$. Clearly, $-d-1$ is square-free for $d= -7, -11, -43, -67$ and hence $P\cong\mathbb{Z}_{D-1}$ in these cases. If $d=-19$, then $|P|=18=2\times3^{2}$.

On the other hand, let $a < 19^m$ be a positive integer. If $a^{19^t}\equiv1\ ({\rm mod}\ 19^m)$ for some integers $t>1$, then clearly $a=1+19x$ for some non-negative integers $x$. Hence, $4^{19^t}\not\equiv1\ ({\rm mod}\ 19^m)$ and $(4^3)^{19^t}\not\equiv1\ ({\rm mod}\ 19^m)$ for any $t>1$. Furthermore, we have

$ \begin{eqnarray*} 4^{9\times 19^{m-1}}&=&262144^{19^{m-1}}\\ &=&(19\times 13797+1)^{19^{m-1}}\\ &=&19^{19^{m-1}}\times 13797^{19^{m-1}}+\cdots+19^{m-1}\times 19\times 13797+1\\ &\equiv&1\ ({\rm mod}\ 19^m). \end{eqnarray*} $

Thus, if $d=-19$, the class $[4]\in \overline R$ is of order $3^{2}\cdot19^{m-1}$, so $P\cong\mathbb{Z}_{2}\times \mathbb{Z}_{3^{2}}\cong\mathbb{Z}_{18}$. Analogously, if $d=-163$, we have

$ \begin{eqnarray*} 4^{81\times 163^{m-1}}&=&(4^{81}-1+1)^{163^{m-1}}\\ &=&(4^{81}-1)^{163^{m-1}}+163^{m-1}(4^{81}-1)^{163^{m-1}-1}+\cdots+163^{m-1}(4^{81}-1)+1\\ &\equiv&1\ ({\rm mod}\ 163^m). \end{eqnarray*} $

Since $163\parallel (4^{81}-1)$, the element $[4]\in\overline R$ in the case of $d=-163$ is of order $3^{4}\times163^{m-1}$, so $P\cong\mathbb{Z}_{2}\times \mathbb{Z}_{3^{4}}\cong\mathbb{Z}_{162}$. Therefore, we can conclude that $P\cong\mathbb{Z}_{D-1}$ for $d=-7, -11, -19, -43, -67, -163$. Accordingly, $U(\overline R)\cong P\times H\cong \mathbb{Z}_{D^{m}}\times\mathbb{Z}_{D^{m-1}}\times \mathbb{Z}_{D-1}$, as desired.

(b) Suppose that $d=-3$, $n=2m$, $m\geqslant1$. Let $\alpha=[a+b\sqrt{d}\, ]\in U(\overline R)$, where $a, b\in \{0, 1, \cdots, 3^m-1\}$ and $3\nmid a$. Since $|U(\overline R)|=2\times 3^{2m-1}$, we can write $U(\overline R)\cong\mathbb{Z}_2\times Q$, where $|Q|=3^{2m-1}$. We claim that $\alpha^{3^{m-1}}=[1]$ for $\alpha\in Q$. Let $W=3^{m-1}$. Since $3^m\, |\, 3^j\left(\begin{smallmatrix}W\\2j\end{smallmatrix}\right)$ for $j\geqslant1$, congruence (2.7) holds if and only if $a^{3^{m-1}}\equiv\, 1\ ({\rm mod}\ 3^m), $ if and only if $a\in\mathbb{Z}_{3^{m-1}}$.

On the other hand, note that $3^m\, |\, 3^j\left(\begin{smallmatrix}W\\2j+1\end{smallmatrix}\right)$ for $2\leqslant j\leqslant\frac{W-1}{2}$, congruence (2.8) is equivalent to

$ b\left[a^2-\frac{(3^{m-1}-1)(3^{m-1}-2)}{2}b^2\right]\equiv 0\ ({\rm mod}\ 3). $

(2.11)

If $3\, |\, b$, then clearly congruence (2.11) holds. If $3\nmid b$, we show that congruence (2.11) holds, too. Indeed, since $3\nmid b$, it follows from congruence (2.11) that

$ 2a^2-(3^{m-1}-1)(3^{m-1}-2)b^2\equiv 0\ ({\rm mod}\ 3). $

(2.12)

Moreover, we have $2a^2\equiv 2\ ({\rm mod}\ 3)$ for $3\nmid a$. Thus congruence (2.12) reduces to $2-2b^2\equiv 0\ ({\rm mod}\ 3)$. The last congruence holds for $3\nmid b$. Hence, congruence (2.12) holds for any integers $b$. So we can conclude that $\alpha^{3^{m-1}}=[1]$ if and only if

$ a\in\mathbb{Z}_{3^{m-1}}, \ \ b\in\{0, 1, \cdots, 3^m-1\}. $

(2.13)

Thus there are precisely $3^{m-1}\times 3^m=3^{2m-1}$ elements $\alpha\in U(\overline R)$ such that $\alpha^{3^{m-1}}=[1]$. Recall that $|Q|=3^{2m-1}$, we obtain $\alpha^{3^{m-1}}=[1]$ for $\alpha\in Q$.

Next, we show that there exist elements in $Q$ with order $3^{m-1}$. Indeed, putting $W=3^{m-2}$. Substituting the value for $W$ into congruence (2.7). Note that $3^m\, |\, 3^j\left(\begin{smallmatrix}3^{m-2}\\2j\end{smallmatrix}\right)$ for $j\geqslant2$, we derive that congruence (2.7) holds if and only if

$ 2a^{3^{m-2}}-3^{m-1}(3^{m-2}-1)a^{3^{m-2}-2}b^2\equiv2\ ({\rm mod}\ 3^m). $

(2.14)

If we substitute $a=b=1$ into congruence (2.14), we have $3^{m-1}(3^{m-2}-1)\equiv0\ ({\rm mod}\ 3^m)$, which is impossible for $m\geqslant2$. Accordingly, congruence (2.7) does not hold for $a=b=1$, which implies that $(1+\sqrt{-3}\, )^{3^{m-2}}\neq[1]$. Moreover, by the condition (2.13), $(1+\sqrt{-3}\, )^{3^{m-1}}=[1]$. So $\beta=[1+\sqrt{-3}\, ]\in Q$. Hence $\beta$ is of order $3^{m-1}$. So $\langle 1+\sqrt{-3} \rangle\cong\mathbb{Z}_{3^{m-1}}$. Thus $Q\cong\mathbb{Z}_{3^{m-1}}\times J$, where $J$ is a subgroup of $Q$ with order $3^m$.

Now, we claim that there are elements in $J$ with order $3^{m-1}$. We first note that $(1+\sqrt{-3})^3=-8$, thus $(1+\sqrt{-3})^{3t}\in \mathbb{Z}$ for $t\geqslant1$. Moreover, since $(1+\sqrt{-3})^2=-2+2\sqrt{-3}$, we conclude that $(1+\sqrt{-3})^s=x+y\sqrt{-3}$, where $3\nmid y$ and $3\nmid s$. Let $\gamma=[1+3\sqrt{-3}\, ]$. By condition (2.13), $\gamma\in Q$. Thus $\gamma^{3^{m-1}}=[1]$ but $\gamma\not\in \langle 1+\sqrt{-3}\, \rangle$. Hence, $\gamma\in J$. Substituting $a=1$, $b=3$ and $W=3^{m-2}$ into congruence (2.8), and note that $3^m\, |\, 3^j\left(\begin{smallmatrix}3^{m-2}\\2j+1\end{smallmatrix}\right)$ for $j\geqslant2$, we derive that congruence (2.8) holds if and only if

$ 3^{m-1} -\frac{3^{m+1}(3^{m-2}-1)(3^{m-2}-2)}{2}\equiv0\ ({\rm mod}\ 3^m). $

The above congruence does not hold for $m\geqslant2$. It follows that $(1+3\sqrt{-3}\, )^{3^{m-2}}\neq[1]$. Thus, $\gamma\in J$ is of order $3^{m-1}$. Hence, $\mathbb{Z}_{3^{m-1}}$ is a subgroup of $J$, and $J\cong \mathbb{Z}_{3^{m-1}}\times \mathbb{Z}_3$. Accordingly, if $d=-3$, then $U(\overline R)\cong\mathbb{Z}_2\times\mathbb{Z}_3\times\mathbb{Z}_{3^{m-1}}\times\mathbb{Z}_{3^{m-1}}$, as desired.

(4) (a) Suppose that $d\neq-3$. Let $n=2m+1$ with $m\geqslant1$. For $\alpha=[a+b\sqrt{d}\, ]\in\overline R$, we know that $\alpha\in U(\overline{R})$ if and only if $D\nmid a$. Then, for $n=2m+1$, we have $|U(\overline R)|=(D-1)\cdot D^{2m}$. So $U(\overline R)=K\times G$, where $K$, $G$ are finite groups, and $|K|=D-1$, $|G|=D^{2m}$.

We now determine the structure of $G$. Consider the polynomial expansions of $\alpha^X$, where $X$ is an arbitrary integer. By Theorem 2.1 (2), $\alpha^X$ equals to the equivalence class $[1]$ if and only if the following congruences hold

$ a^{X}+d\left(\begin{smallmatrix}X\\2\end{smallmatrix}\right)a^{X-2}b^2+\cdots+d^{\frac{X-1}{2}}\left(\begin{smallmatrix}X\\X-1\end{smallmatrix}\right)ab^{X-1} \equiv\, 1\ ({\rm mod}\ D^{m+1}), $

(2.15)

$ \left(\begin{smallmatrix}X\\1\end{smallmatrix}\right)a^{X-1}b+d\left(\begin{smallmatrix}X\\3\end{smallmatrix}\right)a^{X-3}b^3+\cdots+d^{\frac{X-1}{2}}b^{X} \equiv\, 0\ ({\rm mod}\ D^m). $

(2.16)

Firstly, putting $X=D^m$, and noting that $D^{m+1}\, |\, d^j\left(\begin{smallmatrix}D^m\\2j\end{smallmatrix}\right)$ for $j\geqslant 1$, we derive that congruence (2.15) holds if and only if $a^{D^m}\equiv\, 1\ ({\rm mod}\ D^{m+1})$, if and only if $a\in\{1, 2, \cdots, D^{m+1}-1\}$ with $a\in \mathbb{Z}_{D^m}$. Therefore, congruence $a^{D^m}\equiv\, 1\ ({\rm mod}\ D^{m+1})$ has precisely $D^m$ solutions.

On the other hand, congruence (2.16) holds for $b\in\{1, 2, \cdots, D^{m}-1\}$. Hence, the number of elements in $U(\overline R)$ satisfying $\alpha^{D^m}=[1]$ is $D^m\times D^m=D^{2m}$. Recall that $|G|=D^{2m}$, we derive that $\alpha^{D^m}=[1]$ if and only if $\alpha\in G$.

Secondly, substituting $X=D^{m-1}$ into congruence (2.163). If $\alpha^{D^{m-1}}=[1]$, clearly $\alpha\in G$. Since $d\neq-3$, we have $D^{m}\, |\, d^j\left(\begin{smallmatrix}D^{m-1}\\2j+1\end{smallmatrix}\right)$ for $j\geqslant 1$. Therefore, congruence (2.16) holds if and only if $D\, |\, b$. In the case of $D\, |\, b$, congruence (2.15) holds if and only $a^{D^{m-1}}\equiv\, 1\ ({\rm mod}\ D^{m+1})$, if and only if $a\in \mathbb{Z}_{D^{m-1}}$. Therefore, the number of elements in $G$ satisfying $\alpha^{D^{m-1}}=[1]$ is $D^{m-1}\times D^{m-1}=D^{2m-2}$. Hence, there are precisely

$ D^{2m}-D^{2m-2}=(d^{2}-1)\cdot d^{2m-2} $

(2.17)

elements of order $D^{m}$ in $\overline R$.

Now, let $\beta=[1+\sqrt{d}\, ]$. Then $\beta^{D^{m}}=[1]$. However, by the above argument, we know that $\beta^{D^{m-1}}\neq[1]$. So the order of $\beta$ is $D^{m}$. Therefore $\mathbb{Z}_{D^{m}}$ is a subgroup of $G$, and $G\cong \mathbb{Z}_{D^{m}}\times G_{2}$, where $\langle1+\sqrt{d}\, \rangle\cong\mathbb{Z}_{D^{m}}$ and $|G_{2}|=D^{m}$.

Suppose $G_2\cong\mathbb{Z}_{D^{s_{1}}}\times\cdots\times\mathbb{Z}_{D^{s_{h}}}$, where $s_{1}+\cdots+s_{h}=m$. If $h\geqslant2$, then $1\leqslant s_{j}\leqslant m-1$ for $j=1, \cdots, h$. Hence, there are precisely $(D-1)\cdot D^{2m-1}$ elements of order $D^{m}$ in $\overline R$, which contradicts the above result (2.17). If $h=1$, then $G_2\cong\mathbb{Z}_{D^{m}}$ and hence $G\cong\mathbb{Z}_{D^{m}}\times \mathbb{Z}_{D^{m}}$. Thus the number of elements in $\overline R$ of order $D^m$ is $(d^{2}-1)\cdot d^{2m-2}$, which is the same as (2.17). Hence, we conclude that $h=1$ and $G_{2}\cong\mathbb{Z}_{D^{m}}$. Therefore, if $n=2m+1$ with $m\geqslant1$, then $U(\overline R)\cong K\times\mathbb{Z}_{D^{m}}\times\mathbb{Z}_{D^{m}}$.

Finally, we determine the structure of the subgroup $K$ for each case. Recall that $|K|=D-1$. If $d=-7$, then $|K|=6=2\times 3$, we have $K\cong\mathbb{Z}_{2}\times \mathbb{Z}_{3}\cong\mathbb{Z}_{D-1}$. If $d=-11$, then $|K|=10=2\times 5$, thus $K\cong\mathbb{Z}_{2}\times \mathbb{Z}_{5}\cong\mathbb{Z}_{D-1}$. If $d=-19$, then $|K|=18=2\times 3^{2}$, and by the similar argument to $(3)$ above, the element $[4]\in\overline R$ is of order $3^{2}\times 19^{m}$. So $K\cong\mathbb{Z}_{2}\times \mathbb{Z}_{3^{2}}\cong\mathbb{Z}_{D-1}$. If $d=-43$, then $|K|=42=6\times 7$, so $K\cong\mathbb{Z}_{6}\times \mathbb{Z}_{7}\cong\mathbb{Z}_{D-1}$. If $d=-67$, then $|K|=66=6\times 11$, thus $K\cong\mathbb{Z}_{6}\times \mathbb{Z}_{11}\cong\mathbb{Z}_{D-1}$. If $d=-163$, then $|K|=162=2\times 3^{4}$, and by the similar argument to $(3)$ above, the element $[4]\in\overline R$ is of order $3^{4}\times 163^{m}$. So $K\cong\mathbb{Z}_{2}\times \mathbb{Z}_{3^{4}}\cong\mathbb{Z}_{D-1}$. Hence $K\cong \mathbb{Z}_{D-1}$ for each case. Thus $U(\overline R)\cong\mathbb{Z}_{D-1}\times\mathbb{Z}_{D^{m}}\times\mathbb{Z}_{D^{m}}$, as desired.

(b) Suppose $d=-3$. Let $\alpha=[a+b\sqrt{-3}\, ]\in\overline R$, where $3\nmid a$. Then $|U(\overline R)|=2\times 3^{2m}$. So $U(\overline R)=\mathbb{Z}_2\times G$, where $|G|=3^{2m}$. Applying the similar argument of above (a) for the case $d\neq-3$, we get that $\alpha^{D^m}=[1]$ if and only if $a\in \mathbb{Z}_{3^m}$ and $b\in\{0, 1, \cdots, 3^m-1\}$, if and only if $\alpha\in G$.

Now, substituting $X=3^{m-1}$ into congruence (2.16). We obtain that congruence (2.16) holds if and only if $2a^2b-(3^{m-1}-1)(3^{m-1}-2)b^3\equiv\, 0\ ({\rm mod}\ 3)$. We can verify that the last congruence holds for any integers $b$.

On the other hand, congruence (2.15) holds if and only if

$ 2a^{3^{m-1}}-3^m(3^{m-1}-1)a^{3^{m-1}-2}b^2\equiv\, 2\ ({\rm mod}\ 3^{m+1}). $

(2.18)

Clearly, the above congruence (2.18) does not hold, if $a=b=1$. So $(1+\sqrt{-3}\, )^{3^{m}}=[1]$, but $(1+\sqrt{-3}\, )^{3^{m-1}}\neq[1]$. Hence, $\beta=[1+\sqrt{-3}\, ]\in G$ is of order $3^m$. Then $G\cong\mathbb{Z}_{3^m}\times E$, where $\langle 1+\sqrt{-3}\, \rangle\cong\mathbb{Z}_{3^m}$, $|E|=3^m$.

Furthermore, if we substitute $a=2$, $b=3$ into above congruence (2.18), we have

$ 2^{3^{m-1}}-1\equiv\, 0\ ({\rm mod}\ 3^{m+1}). $

(2.19)

However,

$ \begin{eqnarray*} 2^{3^{m-1}}{-}1&=&(3-1)^{3^{m-1}}{-}1\\ &=&3^{3^{m-1}}-\left(\begin{smallmatrix}3^{m-1}\\1\end{smallmatrix}\right)3^{3^{m-1}-1}+\cdots- \left(\begin{smallmatrix}3^{m-1}\\2\end{smallmatrix}\right)\times3^2+\left(\begin{smallmatrix}3^{m-1}\\1\end{smallmatrix}\right)\times3-2 \\ &\equiv&\, 3^m-2\ ({\rm mod}\ 3^{m+1}). \end{eqnarray*} $

Therefore, congruence (2.19) does not hold for $m\geqslant1$. Hence, if we let $\gamma=[2+3\sqrt{-3}\, ]$, then by the above argument, we have $\gamma^{3^m}=[1]$, but $\gamma^{3^{m-1}}\neq[1]$. Thus, $\gamma$ is of order $3^m$. It leads to $\gamma\in G$. Moreover, $(1+\sqrt{-3})^{3t}\in \mathbb{Z}$ for $t\geqslant1$, $(1+\sqrt{-3})^s=x+y\sqrt{-3}$, where $3\nmid y$ and $3\nmid s$. So we get that $\gamma\not\in\langle 1+\sqrt{-3}\, \rangle$, which implies that $\gamma\in E$. Recall that $|E|=3^m$, therefore we have $E\cong\langle 2+3\sqrt{-3}\, \rangle\cong \mathbb{Z}_{3^m}$.

Hence, if $d=-3$, then $U(\overline R)\cong\mathbb{Z}_{2}\times\mathbb{Z}_{3^{m}}\times\mathbb{Z}_{3^{m}}$, as desired.

Theorem 2.7  Let $p\in\mathbb{Z}$ be an odd prime satisfying the Legendre symbol $(\frac{p}{-d})=-1$. Let $\overline R=R_d/\langle{p^n}\rangle$, $n\geqslant1$. Then $U(\overline R)\cong\mathbb{Z}_{p^2-1}\times\mathbb{Z}_{p^{n-1}}\times\mathbb{Z}_{p^{n-1}}$.

Proof  For $\alpha=[a+b\sqrt{d}\, ]\in R_d/\langle{p^n}\rangle$, where $0\leqslant a, b\leqslant p^n-1$, it is easy to prove that $\alpha$ is a unit of $\overline R$ if and only if $p\nmid (a^2-db^2)$. So $|U(\overline R)|=(p^2-1)p^{2n-2}$.

If $n=1$, as $p$ is prime in $\overline R$, then $R_d/\langle{p}\rangle$ is a field with $p^2$ elements. Therefore $U(\overline R)\cong \mathbb{Z}_{p^2-1}$.

If $n\geqslant2$, then $U(\overline R)\cong G_1\times G_2$, where $G_1$ and $G_2$ are finite groups, and $|G_{1}|=p^2-1$, $|G_{2}|=p^{2n-2}$. First, we prove that $G_1\cong \mathbb{Z}_{p^2-1}$. Clearly, there is an epimorphism of rings

$ \mathbb{\phi}:\ \ \ R_d/\langle{p^n}\rangle\rightarrow\ R_d/\langle{p}\rangle. $

So there exists an epimorphism of groups

$ \varphi: U(R_d/\langle{p^n}\rangle)\rightarrow\ U(R_d/\langle{p}\rangle). $

That is $\varphi:\ U(\overline R)\rightarrow\ \mathbb{Z}_{p^2-1}$. Clearly, the kernel ${\rm ker}(\varphi)$ of $\varphi$ is $G_2$. If $\mathbb{Z}_{p^2-1}=\langle\eta\rangle$, then there exists $\theta\in U(\overline R)$ such that $\varphi(\theta)=\eta$. Suppose the order of $\theta\in U(\overline R)$ is $t$, then $\varphi(\theta^t)=1$. Since the order of $\eta\in\mathbb{Z}_{p^2-1}$ is $p^2-1$, we have $\varphi(\theta^{p^2-1})=\eta^{p^2-1}=1$. Therefore, $\varphi(\theta^t)=\varphi(\theta^{p^2-1})$, i.e., $\eta^t=\eta^{p^2-1}=1$. Thus we easily find that $(p^2-1)\, |\, t$, that is $(p^2-1)\, |\, o(\theta)$. Recall that ${\rm ker}(\varphi)=G_2$, and $\varphi(\theta)=\eta\neq1$, so $\theta\not\in\ker(\varphi)=G_2$. Thus $\theta\in G_1$, and $o(\theta)\, |\, (p^2-1)$. Therefore, $o(\theta)=p^2-1$. So we may conclude that $G_1\cong \mathbb{Z}_{p^2-1}$.

In the following, we investigate the structure of $G_2$. For $\alpha=[a+b\sqrt{d}\, ]\in G_2$. It is obvious that either $p\nmid a$ or $p\nmid b$. Consider the polynomial expansions of $\alpha^N$, where $N>1$ is an arbitrary odd integer. It is evident that $\alpha^N=[1]$ if and only if the following congruences hold

$ a^{N}+d\left(\begin{smallmatrix}N\\2\end{smallmatrix}\right)a^{N-2}b^2+\cdots+d^{\frac{N-1}{2}}\left(\begin{smallmatrix}N\\N-1\end{smallmatrix}\right)ab^{N-1} \equiv\, 1\ ({\rm mod}\ p^{n}), $

(2.20)

$ \left(\begin{smallmatrix}N\\1\end{smallmatrix}\right)a^{N-1}b+d\left(\begin{smallmatrix}N\\3\end{smallmatrix}\right)a^{N-3}b^3+\cdots+d^{\frac{N-1}{2}}b^{N} \equiv\, 0\ ({\rm mod}\ p^n). $

(2.21)

By the similar argument to Theorem 2.6 (3), we know that $\alpha^{p^{n-1}}=1$ for all $\alpha\in G_2$, and there are precisely $p^{2n-4}$ elements $\gamma\in G_2$ satisfying $\gamma^{p^{n-2}}=[1]$.

Moreover, let $\beta=[c+e\sqrt{d}\, ]\in G_2$ with $p\nmid c$ and $p\parallel e$. By the polynomial expansions of $\beta^{p^{n-2}}$, we know that $\beta^{p^{n-2}}\neq 1$, which implies $o(\beta)=p^{n-1}$. So $G_{2}\cong H\times P$, where $H=\langle\beta\rangle\cong\mathbb{Z}_{p^{n-1}}$ and $|P|=p^{n-1}$.

Suppose $G_2\cong \mathbb{Z}_{p^{n-1}}\times \mathbb{Z}_{p^{h_1}}\times\cdots\times \mathbb{Z}_{p^{h_r}}$, where $h_1+\cdots+h_r=n-1$. If $r\geqslant2$, then $1\leqslant h_i\leqslant n-2$ for $i=1, \cdots, r$. Thus there are $p^{n-2}p^{h_1}\cdots p^{h_r}=p^{2n-3}$ elements $\gamma\in G_2$ satisfying $\gamma^{p^{n-2}}=[1]$, which contradicts the above result. If $r=1$, then $G_2\cong \mathbb{Z}_{p^{n-1}}\times \mathbb{Z}_{p^{n-1}}$. So there are exactly $p^{n-2}p^{n-2}=p^{2n-4}$ elements $\gamma\in G_2$ satisfying $\gamma^{p^{n-2}}=[1]$, which is the same as above result. So we derive that $r=1$ and this leads to $G_2\cong \mathbb{Z}_{p^{n-1}}\times \mathbb{Z}_{p^{n-1}}$. This completes the proof.

Theorem 2.8  Let $q\in\mathbb{Z}$ be a prime satisfying the Legendre symbol $(\frac{q}{-d})=1$. Suppose that $\pi$ is a proper factor of $q$. Let $\overline R=R_d/\langle{\pi^n}\rangle$, $n\geqslant1$.

(1) Suppose $q=2$. Then $U(\overline R)\cong \mathbb{Z}_1$ if $n=1$, $U(\overline R)\cong \mathbb{Z}_2$ if $n=2$, $U(\overline R)\cong \mathbb{Z}_2\times\mathbb{Z}_{2^{n-2}}$ if $n>2$.

(2) Suppose $q\neq2$. Then $U(\overline R)\cong \mathbb{Z}_{q^{n-1}}\times\mathbb{Z}_{q-1}$.

Proof  Applying Theorem 2.1 (4), we derive that $\overline R\cong \mathbb{Z}/\langle{q^n}\rangle$. So the theorem follows.

We obtain from the proof of Theorem 1.2 that $2$ is not a prime in $R_d$ if $d=-7$. So we may assume $d\neq -7$ in the following theorems. We investigate the unit groups of $R_d/\langle{2^n}\rangle$ for $d=-3$, $-11$, $-19$, $-43$, $-67$, $-163$.

Theorem 2.9  Suppose $d=-3, -11, -19, -43, -67, -163$. Let $\overline R=R_d/\langle{2^n}\rangle$, $n\geqslant2$. Then

(1) $U(\overline R)=\overline R_1\cup \overline R_2\cup\overline R_3$, where

$\overline R_1=\left\{[r_1+r_2\sqrt d\, ]:\ 0\leqslant r_1, r_2\leqslant 2^{n-1}-1, \ r_1, r_2\in \mathbb{Z}\ {\rm are\ not\ of\ the\ same\ parity}\right\}$,

$\overline R_2=\left\{[r_1-r_2\sqrt d\, ]: \ 0\leqslant r_1\leqslant 2^{n-1}-1, \ 1\leqslant r_2\leqslant 2^{n-1}, r_1, r_2\in \mathbb{Z}\ {\rm are\ not\ of\ the\ same\ parity}\right\}$,

$\overline R_3=\left\{[\frac{r_1}{2}\pm\frac{r_2}{2}\sqrt d\, ]:\ 1\leqslant r_i\leqslant 2^{n}-1, \ r_i\in \mathbb{Z}, \ 2\nmid r_i, \ i=1, 2\right\}$.

(2) Suppose $n\geqslant4$. Then there are exactly $8$ elements $\alpha\in \overline R_1\cup\overline R_2$ satisfying $\alpha^2=[1]$.

(3) Suppose $n\geqslant5$. Then there are exactly $32$ elements $\alpha\in \overline R_1\cup\overline R_2$ satisfying $\alpha^4=[1]$.

Proof  (1) If $\alpha=[r_1\pm r_2\sqrt d\, ]\in \overline R$, where $r_1, r_2\in\mathbb{Z}$, it is easy to show that $\alpha\in U(\overline R)$ if and only if $2\nmid N(\alpha)$, i.e., $2\nmid (r_1^2-dr_2^2)$, if and only if $r_1$ and $r_2$ are not of the same parity.

If $\alpha=[\frac{r_1}{2}\pm\frac{r_2}{2}\sqrt d\, ]\in \overline R$, where $r_1, r_2\in\mathbb{Z}$ with $2\nmid r_1r_2$, then $\alpha\in U(\overline R)$ if and only if $2\nmid N(\alpha)$, i.e., $2\nmid \frac{1}{4}(r_1^2-dr_2^2)$, if and only if $8\nmid (r_1^2-dr_2^2)$. Let $r_i=2k_i+1$, $i=1, 2$. Then

$ r_1^2-dr_2^2=4(k_1^2+k_1-dk_2^2-dk_2)+(1-d). $

Clearly, $2\, |\, (k_1^2+k_1-dk_2^2-dk_2)$. However, $4\parallel (1-d)$ for $d=-3, -11, -19, -43, -67, -163$. Therefore, $8\nmid (r_1^2-dr_2^2)$. Hence, $\alpha \in U(\overline R)$.

(2) First, let $\alpha=a\in \mathbb{Z}$, where $1\leqslant a\leqslant2^{n-1}-1$. Then $\alpha\in U(\overline R)$ if and only if $2\nmid a$. By Corollary 2.5, $\alpha^2=[1]$ if and only if $a^2\equiv 1\ ({\rm mod}\ 2^n)$. The last congruence has precisely $2$ solutions.

Second, let $\alpha=\pm b\sqrt d$, where $1\leqslant b\leqslant2^{n-1}-1$. Then $\alpha\in U(\overline R)$ if and only if $2\nmid b$. Let $b=2k+1$. By Corollary 2.5, $\alpha^2=[1]$ if and only if $d(4k^2+4k+1)\equiv 1\ ({\rm mod}\ 2^n)$. Since $d-1=-4x$, where $x=1, 3, 5, 11, 17, 41$, we obtain that $d(4k^2+4k+1)-1=4(k^2d+kd-x)$. Note that $2\nmid (k^2d+kd-x)$, we derive that $d(4k^2+4k+1)\not\equiv 1\ ({\rm mod}\ 2^n)$. Therefore $\alpha^2\neq[1]$.

Thirdly, let $\alpha=a+b\sqrt d$, where $1\leqslant a, b\leqslant2^{n-1}-1$, $a, b\in\mathbb{Z}$ are not of the same parity. By Corollary 2.5, $\alpha^2=[1]$ if and only if the following congruences hold

$ a^2+b^2d=2^{n-1}k_1+1, $

(2.22)

$ 2ab=2^{n-1}k_2, $

(2.23)

where $k_1$ and $k_2$ are of the same parity. If $2\nmid a$ while $2\, |\, b$, then (2.23) reduces to $b\equiv0\ ({\rm mod}\ 2^{n-2})$. Recall that $1\leqslant b\leqslant2^{n-1}-1$, so the last congruence has exactly one solution $b=2^{n-2}$. Hence, the left hand of (2.23) is $2ab=2^{n-1}a$ with $2\nmid a$. The left hand of (2.22) is $a^2+b^2d=a^2+ 2^{2n-4}d=a^2+2^{n-1}\times 2^{n-3}d$. Because $n\geqslant4$, so $2^{n-3}$ is even. Then equality (2.22) holds for some odd integers $k_1$ if and only if $a^2=2^{n-1}k+1$ for some odd integers $k$, if and only if $a=2^{n-2}\pm1$. So we can conclude that in the case of $2\nmid a$ and $2\, |\, b$, there are exactly $2$ elements $\alpha$ satisfying $\alpha^2=[1]$.

On the other hand, suppose that $2\, |\, a$ while $2\nmid b$. Then (2.23) reduces to $a\equiv0\ ({\rm mod}\ 2^{n-2})$. Recall that $1\leqslant a\leqslant2^{n-1}-1$, so the last congruence has exactly one solution $a=2^{n-2}$. Hence, the left hand of (2.23) is $2ab=2^{n-1}b$ with $2\nmid b$. The left hand of (2.22) is $a^2+b^2d=2^{2n-4}+b^2d=2^{n-1}\times2^{n-3}+b^2d$. So equality (2.22) holds for some odd integers $k_1$ if and only if $b^2d=2^{n-1}h+1$ for some odd integers $h$. Putting $b=2s+1$, then $b^2d-1=4d(s^2+s)+(d-1)$. Because $s^2+s$ is even and $4\parallel (d-1)$ for $d=-3, -11, -19, -43, -67, -163$, we obtain that $4\parallel (b^2d-1)$. Therefore, for $n\geqslant4$, $b^2d\neq2^{n-1}h+1$ for any integers $h$. So we can conclude that in the case of $2\, |\, a$ and $2\nmid b$, there does not exist any element $\alpha$ satisfying $\alpha^2=[1]$.

Finally, let $\alpha=a-b\sqrt d$, where $1\leqslant a\leqslant2^{n-1}-1$, $1\leqslant b\leqslant2^{n-1}$, $a, b\in\mathbb{Z}$ are not of the same parity. If $2\nmid a$ while $2\, |\, b$, then (2.23) reduces to $b\equiv0\ ({\rm mod}\ 2^{n-2})$. Thus $b=2^{n-2}$ or $2^{n-1}$. In the case of $b=2^{n-2}$, applying the similar argument of above, we get that $\alpha^2=[1]$ if and only if $a=2^{n-2}\pm1$. For the other case $b=2^{n-1}$, equality (2.23) reduces to $2ab=2^na$, and the left hand of equality (2.22) is $a^2+b^2d=a^2+2^{2n-2}d$. By Corollary 2.5, $\alpha^2=[1]$ if and only if $a^2\equiv 1\ ({\rm mod}\ 2^{n})$, if and only if $a=1, 2^{n-1}-1$. Therefore, there are exactly $4$ elements $\alpha$ satisfying $\alpha^2=[1]$, if $2\nmid a$ and $2\, |\, b$.

On the other hand, if $2\, |\, a$ while $2\nmid b$, by the similar above argument, we obtain that $\alpha^2\neq[1]$.

Thus, there are exactly $8$ elements $\alpha\in \overline R_1\cup\overline R_2$ satisfying $\alpha^2=[1]$, as desired.

(3) Firstly, let $\alpha=a\in \mathbb{Z}$, where $1\leqslant a\leqslant2^{n-1}-1$ with $2\nmid a$, $a\in\mathbb{Z}$. By Corollary 2.5, $\alpha^4=[1]$ if and only if $a^4\equiv 1\ ({\rm mod}\ 2^n)$. The last congruence has precisely $4$ solutions.

Secondly, let $\alpha=\pm b\sqrt d$, where $1\leqslant b\leqslant2^{n-1}-1$ with $2\nmid b$, $b\in\mathbb{Z}$. Let $b=2k+1$. By Corollary 2.5, $\alpha^4=[1]$ if and only if $b^4d^2-1\equiv 0\ ({\rm mod}\ 2^n)$, i.e.,

$ 8d^2(2k^4+4k^3+3k^2+k)+(d^2-1)\equiv 0\ ({\rm mod}\ 2^n). $

(2.24)

It is evident that $2^4\nmid (d^2-1)$ for $d=-3, -11, -19, -43, -67, -163$. So $b^4d^2-1\not\equiv 0\ ({\rm mod}\ 2^n)$ for $n\geqslant5$. Thus, $\alpha^4\neq[1]$.

Thirdly, let $\alpha=a+b\sqrt d$, where $1\leqslant a, b\leqslant2^{n-1}-1$, $a$ and $b$ are not of the same parity. By Corollary 2.5, $\alpha^4=[1]$ if and only if the following congruences hold

$ a^4+b^2(6a^2d+b^2d^2)=2^{n-1}k_1+1, $

(2.25)

$ 4b(a^3+ab^2d)=2^{n-1}k_2, $

(2.26)

where $k_1$ and $k_2$ are of the same parity. If $2\nmid a$ while $2\, |\, b$, then (2.26) reduces to $b\equiv0\ ({\rm mod}\ 2^{n-3})$. The last congruence has exactly three solutions $b=2^{n-3}x$, where $x=1, 2, 3$. Suppose first that $b=2^{n-3}x$, $x=1, 3$. Then the left hand of equation (2.26) equals $4b(a^3+ab^2d)=2^{n-1}k_2$, where $k_2=x(a^3+ab^2d)$ is odd.

On the other hand, the left hand of equation (2.25) equals $a^4+2^{n-1}(3\times2^{n-4}a^2d+2^{3n-11}d^2x^2)x^2.$ Since $n\geqslant5$, we get that $(3\times2^{n-4}a^2d+2^{3n-11}d^2x^2)x^2$ is even. Therefore, $\alpha^4=[1]$ if and only if $a^4=2^{n-1}s+1$ for some odd integers $s$. Since $1\leqslant a\leqslant2^{n-1}-1$, clearly there are exactly $4$ elements $a$ satisfying $a^4=2^{n-1}s+1$ for some odd integers $s$. Now suppose $b=2^{n-3}x$, where $x=2$. Then the left hand of equation (2.26) equals $4b(a^3+ab^2d)=2^{n}(a^3+ab^2d)$. Therefore, by equation (2.25), we obtain that $\alpha^4=[1]$ if and only if $a^4\equiv1\ ({\rm mod}\ 2^n)$. The last congruence has exactly $4$ solutions $a\in\{1, \cdots, 2^{n-1}-1\}$. Hence, there are totally $12$ elements $\alpha$ satisfying $\alpha^4=[1]$, in the case of $2\nmid a$ and $2\, |\, b$. For another case of $2\, |\, a$ and $2\nmid b$, we reduce from equation (2.25) that $2^{n-3}\, |\, a$. Hence, $a=2^{n-3}y$, where $y=1, 2, 3$. Suppose $a=2^{n-3}y$, where $y=1, 3$. Then by equations (2.25) and (2.26), $\alpha^4=[1]$ if and only if $b^4d^2=2^{n-1}s+1$ for some odd integers $s$. Let $b=2k+1$, then $b^4d^2-1$ is equal to the left side of congruence (2.24). Since $2^4\nmid (d^2-1)$ for $d=-3, -11, -19, -43, -67, -163$. So $b^4d^2-1\not\equiv 0\ ({\rm mod}\ 2^{n-1})$ for $n\geqslant5$. Thus, $\alpha^4\neq[1]$. Next, we assume that $a=2^{n-3}y$, where $y=2$. Then by equations (2.25) and (2.26), $\alpha^4=[1]$ if and only if $b^4d^2\equiv 1\ ({\rm mod}\ 2^n)$, if and only if congruence (2.24) holds for any integers $k$ and $n$. However, this congruence does not hold for $n\geqslant5$. Therefore, we can conclude that in the case of $2\, |\, a$ and $2\nmid b$, there does not exist any element $\alpha$ satisfying $\alpha^4=[1]$. Hence, there are totally $12$ elements $\alpha=[a+b\sqrt d]\in \overline R_1$ satisfying $\alpha^4=[1]$, where $a\neq0$ and $b\neq0$.

Finally, let $\alpha=a-b\sqrt d$, where $1\leqslant a\leqslant2^{n-1}-1$, $1\leqslant b\leqslant2^{n-1}$, $a$ and $b$ are not of the same parity. If $2\nmid a$ while $2\, |\, b$, then (2.26) reduces to $b\equiv0\ ({\rm mod}\ 2^{n-3})$. The last congruence has exactly four solutions, namely $b=2^{n-3}x$, where $x=1, 2, 3, 4$. Applying the similar argument above, we obtain that there are exactly $16$ elements $\alpha\in \overline R_2$ satisfying $\alpha^4=[1]$, where $a\neq0$. On the other hand, if $2\, |\, a$ and $2\nmid b$, there does not exist any element $\alpha\in \overline R_2$ satisfying $\alpha^4=[1]$.

Thus, there are exactly $32$ elements $\alpha\in \overline R_1\cup\overline R_2$ satisfying $\alpha^4=[1]$, as desired.

In the sequel, we assume that $2$ is prime in the ring $R_d$. If $n=1$, by Theorem 2.1 (5) and Theorem 2.9, $R_d/\langle{2}\rangle$ is a field with $4$ elements. Therefore, $U(R_d/\langle{2^n}\rangle)\cong \mathbb{Z}_3$.

If $n=2$, then $|U(R_d/\langle{2^n}\rangle)|=3\times2^2$. The unit group of $R_d/\langle{2^n}\rangle$ is

$ \left\{1, \ \pm\sqrt d, \ 1-2\sqrt d, \ \frac{\, 1\, }{2}\pm \frac{\, 1\, }{2}\sqrt d, \ \frac{\, 1\, }{2}\pm \frac{\, 3\, }{2}\sqrt d, \ \frac{\, 3\, }{2}\pm \frac{\, 1\, }{2}\sqrt d, \ \frac{\, 3\, }{2}\pm \frac{\, 3\, }{2}\sqrt d\right\}. $

By calculation, we obtain that for $d=-3, -11, -19, -43, -67, -163$, $(\pm\sqrt d)^2=4k+1$ for some integers $k$. So by Corollary 2.5, $\pm\sqrt d$ is of order $2$. Similarly, $(\frac{\, 3\, }{2}\pm \frac{\, 3\, }{2}\sqrt d)^3=-27=[1]$. So the order of $\frac{\, 3\, }{2}\pm \frac{\, 3\, }{2}\sqrt d$ is $3$. Moreover, we show that $o(1-2\sqrt d)=2$, $o(\frac{\, 1\, }{2}\pm \frac{\, 1\, }{2}\sqrt d)=o(\frac{\, 1\, }{2}\pm \frac{\, 3\, }{2}\sqrt d)=o(\frac{\, 3\, }{2}\pm \frac{\, 1\, }{2}\sqrt d)=6$. Hence, $U(R_d/\langle{2^2}\rangle)\cong \mathbb{Z}_3\times \mathbb{Z}_2\times \mathbb{Z}_2$.

Analogously, if $n=3$, then $|U(R_d/\langle{2^n}\rangle)|=3\times2^4$. The unit group of $R_d/\langle{2^n}\rangle$ is

$ \begin{eqnarray*}&&\left\{1, \ 3, \ \pm\sqrt d, \ \pm3\sqrt d, \ 1\pm2\sqrt d, \ 2\pm\sqrt d, \ 2\pm3\sqrt d, \ 3\pm2\sqrt d, \ 1-4\sqrt d, \ 3-4\sqrt d\right\}\\ &\bigcup&\left\{\frac{a}{2}\pm\frac{b}{2}\sqrt d:\ \ a, b=1, 3, 5, 7\right\}.\end{eqnarray*} $

By calculation, we obtain that $o(3)=o(1\pm2\sqrt d)=o(3\pm2\sqrt d)=o(1-4\sqrt d)=o(3-4\sqrt d)=2$, and $o(\pm\sqrt d)=o(\pm3\sqrt d)=o(2+\sqrt d)=o (2+3\sqrt d)=o(2-\sqrt d)=o(2-3\sqrt d)=4$. Moreover, $o(\frac{a}{2}\pm\frac{b}{2}\sqrt d)\neq2, 4$ for $a, b=1, 3, 5, 7$. Therefore, $U(R_d/\langle{2^3}\rangle)\cong \mathbb{Z}_3\times\mathbb{Z}_{2^2} \times\mathbb{Z}_2\times \mathbb{Z}_2$.

Theorem 2.10  Suppose that $d=-3$, $-11$, $-19$, $-43$, $-67$ or $-163$. Then

(1) $U(R_d/\langle{2}\rangle)\cong \mathbb{Z}_3$.

(2) $U(R_d/\langle{2^n}\rangle)\cong \mathbb{Z}_3\times\mathbb{Z}_{2^{n-1}} \times\mathbb{Z}_{2^{n-2}}\times \mathbb{Z}_2$ for $n\geqslant2$.

Proof  The unit groups for the cases of $n=1, 2, 3$ have been stated above. So we assume $n\geqslant4$ in the following. By Theorem 2.9, we get $|U(R_d/\langle{2^n}\rangle)|=3\times 2^{2n-2}$. Thus $U(R_d/\langle{2^n}\rangle)\cong \mathbb{Z}_3\times H$, where $H$ is a subgroup with order $2^{2n-2}$.

Firstly, we claim that $\alpha^{2^{n-1}}=[1]$ for $\alpha\in\overline R_1\cup \overline R_2$, where $\overline R_1$ and $\overline R_2$ are stated in Theorem 2.9. Indeed, if we put $\alpha=a+b\sqrt d\in \overline R_1$, $\alpha^M=A+B\sqrt d$, $M$ is even, then

$ \begin{eqnarray*} &&A=a^M+d\left(\begin{smallmatrix}M\\2\end{smallmatrix}\right)a^{M-2}b^2+d^2\left(\begin{smallmatrix}M\\4\end{smallmatrix}\right)a^{M-4}b^4+\cdots+ d^{\frac{M-2}{2}}\left(\begin{smallmatrix}M\\M-2\end{smallmatrix}\right)a^2b^{M-2}+d^{\frac{M}{2}}b^{M}, \\ &&B=\left(\begin{smallmatrix}M\\1\end{smallmatrix}\right)a^{M-1}b+d\left(\begin{smallmatrix}M\\3\end{smallmatrix}\right)a^{M-3}b^3+\cdots+ d^{\frac{M-4}{2}}\left(\begin{smallmatrix}M\\M-3\end{smallmatrix}\right)a^3b^{M-3} +d^{\frac{M-2}{2}}\left(\begin{smallmatrix}M\\M-1\end{smallmatrix}\right)ab^{M-1}. \end{eqnarray*} $

Let $M=2^{n-1}$. If $2\nmid a$ while $2\, |\, b$, then $2^n\, |\, \left(\begin{smallmatrix}2^{n-1}\\s\end{smallmatrix}\right)b^s$ for $1\leqslant s\leqslant 2^{n-1}$. So we derive $2^n\, |\, (A-a^{2^{n-1}})$ and $2^n\, |\, B$. Hence, $A=2^nt+a^{2^{n-1}}$ and $B=2^{n}k$ for some integers $t, k$. By Corollary 2.5, $\alpha^{2^{n-1}}=[1]$ if and only if $a^{2^{n-1}}\equiv\, 1\ ({\rm mod}\ 2^n)$. Because $U(\mathbb{Z}/\langle{2^n}\rangle)\cong\mathbb{Z}_2\times \mathbb{Z}_{2^{n-2}}$ for $n\geqslant3$, we derive that $a^{2^{n-1}}\equiv\, 1\ ({\rm mod}\ 2^n)$ for $2\nmid a$ and $n\geqslant3$. Thus $\alpha^{2^{n-1}}=[1]$ in the case of $2\nmid a$ and $2\, |\, b$.

On the other hand, suppose $2\, |\, a$ while $2\nmid b$. Since $2^n\, |\, \left(\begin{smallmatrix}2^{n-1}\\s\end{smallmatrix}\right)a^{2^{n-1}-s}$ for $0\leqslant s\leqslant2^{n-1}-1$, it is obvious that $2^n\, |\, (A-d^{2^{n-2}}b^{2^{n-1}})$ and $2^n\, |\, B$. Since $d, b\in U(\mathbb{Z}/\langle{2^n}\rangle)$, we must have $d^{2^{n-2}}\equiv\, 1\ ({\rm mod}\ 2^n)$ and $b^{2^{n-1}}\equiv\, 1\ ({\rm mod}\ 2^n)$. Hence, $d^{2^{n-2}}b^{2^{n-1}} \equiv\, 1\ ({\rm mod}\ 2^n)$. Therefore, $\alpha^{2^{n-1}}=[1]$ in the case of $2\, |\, a$ and $2\nmid b$. So we conclude that $\alpha^{2^{n-1}}=[1]$ for $\alpha\in\overline R_1$. Similarly, we have $\alpha^{2^{n-1}}=[1]$ for $\alpha\in\overline R_2$. Thus, our claim follows.

Secondly, we prove that $\mathbb{Z}_{2^{n-1}}$ is a subgroup of $H$. Since the number of the set $\overline R_1\cup\overline R_2$ is precisely $2^{2n-2}$ and note that the subgroup $H$ is of order $2^{2n-2}$, we can conclude that $\alpha\in H$ if and only if $\alpha\in\overline R_1\cup\overline R_2$. So $H=\overline R_1\cup\overline R_2$. Furthermore, let $\alpha_0=[2+\sqrt d]\in H$. We prove that $\alpha_0^{2^{n-2}}\neq[1]$. Setting $a=2, b=1, M=2^{n-2}$. Substituting these values into the expressions for $A$ and $B$. Since $2^n\, |\, \left(\begin{smallmatrix}2^{n-2}\\s\end{smallmatrix}\right)a^s$ for $3\leqslant s\leqslant 2^{n-2}$, and $2^{n-1}\parallel\left(\begin{smallmatrix}2^{n-2}\\s\end{smallmatrix}\right)a^s$ for $s=1, 2$, we derive that $2^{n-1}\parallel (A-d^{2^{n-3}})$ and $2^{n-1}\parallel B$. So $A=2^{n-1}k+d^{2^{n-3}}$ for some odd integers $k$. Moreover, owing to Corollary 2.5, $\alpha_0^{2^{n-2}}=[1]$ if and only if $A=2^{n-1}t+1$ for some odd integers $t$, i.e., $A=2^{n-1}k+d^{2^{n-3}}=2^{n-1}t+1$, if and only if $d^{2^{n-3}}=2^{n-1}(t-k)+1$. Since $2\nmid kt$, we have $t-k$ is even. Therefore, $\alpha_0^{2^{n-2}}=[1]$ if and only if $d^{2^{n-3}}\equiv 1\ ({\rm mod}\ 2^n)$. In the following, we show that $d^{2^{n-3}}\not\equiv 1\ ({\rm mod}\ 2^n)$ for $d=-3$, $-11$, $-19$, $-43$, $-67$ or $-163$. Indeed, we have $-d=4e-1$ for some odd integers $e$. Then

$ d^{2^{n-3}}-1=(4e-1)^{2^{n-3}}-1=(4e)^{2^{n-3}}-\left(\begin{smallmatrix}2^{n-3}\\1\end{smallmatrix}\right)(4e)^{2^{n-3}-1}+\cdots+\\\left(\begin{smallmatrix}2^{n-3}\\2\end{smallmatrix}\right)(4e)^2- \left(\begin{smallmatrix}2^{n-3}\\1\end{smallmatrix}\right)4e. $

It is evident that $2^n\, |\, \left(\begin{smallmatrix}2^{n-3}\\s\end{smallmatrix}\right)(4e)^s$ for $2\leqslant s\leqslant2^{n-3}$. However, $\left(\begin{smallmatrix}2^{n-3}\\1\end{smallmatrix}\right)4e=2^{n-1}e$ is not divisible by $2^n$. Thus $d^{2^{n-3}}\not\equiv 1\ ({\rm mod}\ 2^n)$. Hence, $\alpha_0^{2^{n-2}}\neq[1]$, which implies that $\alpha_0$ is of order $2^{n-1}$. Therefore, $\mathbb{Z}_{2^{n-1}}$ is a subgroup of $H$, as desired.

Now, owing to Theorem 2.9 (2), we obtain that $H\cong\mathbb{Z}_{2^{n-1}}\times\mathbb{Z}_{2^i}\times\mathbb{Z}_{2^j}$, where $i, j\geqslant1$ and $i+j=n-1$. If $n=4$, then $i+j=3$. Hence, $H\cong\mathbb{Z}_{2^{n-1}}\times\mathbb{Z}_{2^2}\times\mathbb{Z}_{2}$ for the case $n=4$. Next, we assume that $n>4$. If $i, j\geqslant2$, then there are precisely $64$ elements $\alpha\in \mathbb{Z}_{2^{n-1}}\times\mathbb{Z}_{2^i}\times\mathbb{Z}_{2^j}$ satisfying $\alpha^4=[1]$, which contradicts Theorem 2.9 (3). If $i=n-2$ and $j=1$, then there are precisely $32$ elements $\alpha\in \mathbb{Z}_{2^{n-1}}\times\mathbb{Z}_{2^{n-2}}\times\mathbb{Z}_{2}$ satisfying $\alpha^4=[1]$, which is the same as Theorem 2.9 (3). Therefore, we conclude that $H\cong \mathbb{Z}_{2^{n-1}}\times\mathbb{Z}_{2^{n-2}}\times\mathbb{Z}_{2}$. This completes the proof of the theorem.