# The Hecke GroupHλ4 Acting on Imaginary Quadratic Number Fields

Authors: Abdulaziz Deajim
Journal: Journal of Mathematics
 (2021)

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## Abstract

LetHλ4 be the Hecke group x,y:x2=y4=1 and, for a square-free positive integer n, consider the subset ℚ∗−n=a+−n/c|a,b=a2+n/c∈ℤ,c∈2ℤ of the quadratic imaginary number field ℚ−n. Following a line of research in the relevant literature, we study the properties of the action of Hλ4 on ℚ∗−n. In particular, we calculate the number of orbits arising from this action for every such n. Some illustrative examples are also given.


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## Body

## 1. Introduction

### 1.1. Preliminaries

For a positive real numberλ, Hecke introduced in [5] the group Hλ generated by the linear transformations x:z⟼−1/z and w:z⟼z+λ. He further showed that if λ=λk=2cosπ/k for an integer k≥3, then Hλk is Fuchsian (i.e., a discrete subgroup of PSL2,ℝ) (see [5, 13]). The groups Hλk are justifiably called Hecke groups. Letting y=xw:z⟼−1/z+λk, it can be shown that Hλk is generated by x and y, so Hλk=x,y:x2=yk=1, and Hλk is isomorphic to a free product C2∗Ck of the cyclic groups of orders 2 and k (see [2]). Such a presentation of Hλk is by no means unique. The actions of Hecke groups on many discrete structures play important roles in various branches of mathematics (see [2] for more details).Fork=3, it is known (see [6]) that Hλ3 is the modular group PSL2,ℤ. The action of the modular group on real quadratic number fields was extensively studied; see for instance [7, 9–11]. On the other hand, some aspects of the action of the modular group (resp., a particular subgroup of the modular group) on imaginary quadratic number fields were also studied; see [4] (resp., [1]).Fork=4, we set H=Hλ4 in this paper. Following [12], we choose the following presentation for H:(1)H=x,y:x2=y4=1withx:z⟼−12zandy:z⟼−12z+1⋅The action ofH on real quadratic number fields was also studied (e.g., see [8, 12]). In this paper, we consider the action of H on imaginary quadratic number fields. In particular, for a square-free positive integer n, we study the action of H on the following subset of ℚ−n:(2)ℚ∗−n≔a+−nc∈ℚ−n|a,b=a2+nc∈ℤ,c∈2ℤ.We observe that the setℚ∗−n is closed under complex conjugation since, for α=a+−n/c∈ℚ∗−n, we have α¯=−a+−n/−c. Furthermore, such an α and its conjugate are roots of the quadratic polynomial ct2−2at+b with discriminant 4a2−4bc=−4n. In fact, there is a two-to-one correspondence from the set ℚ∗−n to the set of such polynomials.


### 1.2. Summary of Contributions

After displaying and proving some interesting preliminary results in Sections2 and 3 on the action of H on ℚ∗−n, we utilize them to prove the main result of the paper (Theorem 1), which aims at precisely calculating the number of orbits resulting from this action. We present in Section 4 some examples to illustrate various computational aspects that arise when applying Theorem 1 to concrete cases.


## 1.1. Preliminaries

For a positive real numberλ, Hecke introduced in [5] the group Hλ generated by the linear transformations x:z⟼−1/z and w:z⟼z+λ. He further showed that if λ=λk=2cosπ/k for an integer k≥3, then Hλk is Fuchsian (i.e., a discrete subgroup of PSL2,ℝ) (see [5, 13]). The groups Hλk are justifiably called Hecke groups. Letting y=xw:z⟼−1/z+λk, it can be shown that Hλk is generated by x and y, so Hλk=x,y:x2=yk=1, and Hλk is isomorphic to a free product C2∗Ck of the cyclic groups of orders 2 and k (see [2]). Such a presentation of Hλk is by no means unique. The actions of Hecke groups on many discrete structures play important roles in various branches of mathematics (see [2] for more details).Fork=3, it is known (see [6]) that Hλ3 is the modular group PSL2,ℤ. The action of the modular group on real quadratic number fields was extensively studied; see for instance [7, 9–11]. On the other hand, some aspects of the action of the modular group (resp., a particular subgroup of the modular group) on imaginary quadratic number fields were also studied; see [4] (resp., [1]).Fork=4, we set H=Hλ4 in this paper. Following [12], we choose the following presentation for H:(1)H=x,y:x2=y4=1withx:z⟼−12zandy:z⟼−12z+1⋅The action ofH on real quadratic number fields was also studied (e.g., see [8, 12]). In this paper, we consider the action of H on imaginary quadratic number fields. In particular, for a square-free positive integer n, we study the action of H on the following subset of ℚ−n:(2)ℚ∗−n≔a+−nc∈ℚ−n|a,b=a2+nc∈ℤ,c∈2ℤ.We observe that the setℚ∗−n is closed under complex conjugation since, for α=a+−n/c∈ℚ∗−n, we have α¯=−a+−n/−c. Furthermore, such an α and its conjugate are roots of the quadratic polynomial ct2−2at+b with discriminant 4a2−4bc=−4n. In fact, there is a two-to-one correspondence from the set ℚ∗−n to the set of such polynomials.


## 1.2. Summary of Contributions

After displaying and proving some interesting preliminary results in Sections2 and 3 on the action of H on ℚ∗−n, we utilize them to prove the main result of the paper (Theorem 1), which aims at precisely calculating the number of orbits resulting from this action. We present in Section 4 some examples to illustrate various computational aspects that arise when applying Theorem 1 to concrete cases.


## 2. Lemmas

Throughout the paper, unless a particular value ofn is specified, n denotes a square-free positive integer.Lemma 1.
Fora+−n/c∈ℚ∗−n, we have the following:(i)
a is odd if and only if n is odd.(ii)
b and c have the same sign.Proof.
Clear.
Following [4], for α=a+−n/c∈ℚ∗−n, we sometimes use the notations aα for a, bα for b, and cα for c for obvious reasons (see the lemma below).Lemma 2.
Forα=a+−n/c∈ℚ∗−n, the effect of the action of t∈x,y,y2,y3 on α is summarized in the table below.Proof.
With the use of the equalitybc=a2+n, we have(3)xα=−12α=−12ca+−n=−12ca−−na2+n=−a+−n2b.
So,axα=−a, cxα=2b, and bxα=axα2+n/cxα=−a2+n/2b=c/2 as claimed. We also have(4)yα=−12α+1=−c2a+c+−n=−ca+c−−n2a2+2ac+c2+n=−ca+c−−n2bc+2ac+c2=−a−c+−n22a+b+c.
So,ayα=−a−c, cyα=22a+b+c, and(5)byα=ayα2+ncyα=−a−c2+n22a+b+c=a2+2ac+c2+n22a+b+c=bc+2ac+c222a+b+c=cb+2a+c22a+b+c=c2,as claimed. Now, the rest of the proof is direct recursive computations using the above formulae. For instance, ay2α=−ayα−cyα=−−a−c−22a+b+c=−3a−2b−c, and so on.Remark 1.
It follows from Table1 that ℚ∗−n is an H-set. To see this, we only need to show that ℚ∗−n is stable under the action of H (because H acts naturally on ℂ). For this, it suffices to show that ℚ∗−n is stable under the action of the two generators of H. By means of Table 1, direct computations show that, for α∈ℚ∗−n and t=x,y, we have atα2+n=btαctα and ctα is even, which imply that tα∈ℚ∗−n.Table 1
Effect ofx, y, y2, and y3 on α.
tatαbtαctαx−ac/22by−a−cc/222a+b+cy2−3a−2b−c2a+b+c4a+4b+cy3−a−2b4a+4b+c/22bLemma 3.
Under the action ofH on ℚ∗−n, the denominators of all elements of the same orbit are of the same sign.Proof.
This follows directly from Table1, taking into account that the signs of btα and ctα are the same for t=x,y (Lemma 1(ii)).
Recall the following classical terminology. An elementα∈ℚ∗−n is said to be purely imaginary if aα=0. We can thus have the following observation, where we use the usual notation dm for the number of positive divisors of m∈ℕ.Lemma 4.
ℚ∗−n contains purely imaginary elements if and only if n is even, in which case the number of purely imaginary elements is equal to 2dn/2.Proof.
Suppose thatℚ∗−n contains a purely imaginary element α. Then, α must be of the form −n/c with b=n/c∈ℤ. Since c is even, so must n be. Conversely, if n is even, then −n/2∈ℚ∗−n is purely imaginary. Now, let n be even with n=2n′, n′∈ℕ. The set of purely imaginary elements in ℚ∗−n is(6)−nccis even andcdividesn=−n2c′c′∈ℤandc′dividesn′.
The cardinality of the above set is obviously2dn′, taking into consideration positive and negative divisors of n′.
An elementα∈ℚ∗−n is said to be totally positive (resp., totally negative) if aαcα>0 (resp., aαcα<0). It is obvious that an element of ℚ∗−n is either purely imaginary, totally positive, or totally negative. We observe that the set of totally positive elements of ℚ∗−n lies in the open right-half of the complex plane, the set of totally negative elements lies in the open left-half of the complex plane, and the set of purely imaginary elements lies on the imaginary axis.
The following lemma studies the effects ofx,y,y2, and y3 on elements of ℚ∗−n in each of the three states.Lemma 5.
Letα=a+−n/c∈ℚ∗−n. We have the following:(1)
α is purely imaginary if and only if xα is purely imaginary.(2)
α is totally positive if and only if xα is totally negative.(3)
Ifα is totally positive or purely imaginary, then yα, y2α, and y3α are all totally negative.(4)
α,yα,y2α, and y3α are all totally negative if and only if α is totally negative, a<c, a<2b, and 3a<2b+c.Proof.
(1)
From Table1, axα=−aα.(2)
This follows from Table1, as axα and aα have opposite signs, while cxα and cα have the same sign.(3)
Recall from Lemma1 that b and c have the same sign. If α is totally positive, then either a,b,c>0 or a,b,c<0. In both cases, it follows directly from Table 1 that yα, y2α, and y3α are all totally negative. Similarly, if α is purely imaginary, then a=0 and b,c<0 or b,c>0. In either case, it follows directly from Table 1 that yα, y2α, and y3α are all totally negative in this case as well.(4)
Assume thatα is totally negative, a<c, a<2b, and 3a<2b+c. Suppose first that a>0 and b,c<0. So, a<−c, a<−2b, and 3a<−2b−c. By Table 1, we deduce that ayα=−a−c>0, ay2α=−3a−2b−c>0, and ay3α=−a−2b>0. Note also that cyα<0 as byα=c/2<0, cy2α<0 as by2α=2a+b+c<0, and cy3α=2b<0. It thus follows that yα, y2α, and y3α are all totally negative. The case when a<0 and b,c>0 is handled similarly.Conversely, assume thatα,yα,y2α, and y3α are all totally negative. Suppose that a>0 and b,c<0. Since yα is totally negative and byα=c/2<0, it follows that −a−c=ayα>0. So, a=a<−c=c. Since byα and cyα have the same sign and byα=c/2<0, 22a+b+c=cyα<0. Since y2α is totally negative and by2α=2a+b+c=cyα/2<0, −3a−2b−c=ay2α>0. Thus, 3a=3a<−2b−c=2b+c. Finally, since y3α is totally negative and cy3α=2b<0, it follows that −a−2b=ay3α>0. Thus, a=a<−2b=2b. The case when a<0 and b,c>0 is handled similarly.Example 1.
(1)
For the totally positive elementα1=2+−2/2∈ℚ∗−2, we see that yα1=−4+−2/18, y2α1=−14+−2/22, y3α1=−6+−2/6, which are all totally negative.(2)
For the purely imaginary elementα2=−2/2∈ℚ∗−2, we see that yα2=−2+−2/6, y2α2=−4+−2/6, and y3α2=−2+−2/2, which are all totally negative.(3)
For the totally negative elementα3=1+−5/−2∈ℚ∗−5, we see that aα3=1, bα3=−3, and cα3=−2, which satisfy the required inequalities of part 4 of Lemma 5. Clearly, yα3=1+−5/−6, y2α3=5+−5/−10, and y3α3=5+−5/−6 are all totally negative.Remark 2.
(1)
For any set of four elementsα,yα,y2α,y3α⊆ℚ∗−n, we see from parts 3 and 4 of Lemma 5 that either all four elements are totally negative, one is totally positive and the other three are totally negative, or one is purely imaginary and the other three are totally negative.(2)
We show here that the three conditionsa<c, a<2b, and 3a<2b+c in part 4 of Lemma 5 are all necessary, and so none of them can be dropped. The totally negative element α1=2+−2/−2∈ℚ∗−2 has the two properties a<2b and 3a<2b+c, but not a<c (in fact, a=c). So, yα1 is not totally negative. The totally negative element α2=4+−2/−6∈ℚ∗−2 has the two properties a<c and a<2b, but not 3a<2b+c (in fact, 3a=2b+c). So, y2α2 is not totally negative. The totally negative element α3=2+−2/−6∈ℚ∗−2 has the two properties a<c and 3a<2b+c, but not a<2b (in fact, a=2b). So, y3α3 is not totally negative.Lemma 6.
Under the action ofH on ℚ∗−n, every orbit contains a totally positive element.Proof.
Letα∈ℚ∗−n and αH=β|hα=βfor someh∈H be the orbit containing α. If α is totally positive, then there is nothing to prove. If α is totally negative, then xα∈αH is totally positive (Lemma 5(2)). If α is purely imaginary, then yα∈αH is totally negative and so xyα∈αH is totally positive (Lemma 5(2,3)).


## 3. Orbits inℚ∗−n under the Action of H

For anyα∈ℚ∗−n, denote the quadruplet α,yα,y2α,y3α by Qα for short. Note that Qα is stable under the action of the cyclic group y generated by y; that is, Qα=Qyiα for i=1,2,3. If all elements of Qα are totally negative, then we call Qα a totally negative quadruplet. Set(7)TN−n≔Qα⊆ℚ∗−n|Qαis a totally negative quadruplet.By part 4 of Lemma5, we have(8)TN−n=Qα⊆ℚ∗−n|aαcα<0,aα<cα,aα<2bα,3aα<2bα+cα.IfTN−n is not empty, then it follows from Lemma 3 that we can partition TN−n into the disjoint union TN−n=TNa>0−n∪TNa<0−n, where(9)TNa>0−n≔Qα⊆ℚ∗−n|aα>0,cα<0,aα<cα,aα<2bα,3aα<2bα+cα,TNa<0−n≔Qα⊆ℚ∗−n|aα<0,cα>0,aα<cα,aα<2bα,3aα<2bα+cα.Moreover, it is easy to see that there is a bijection between the two setsTNa>0−n and TNa<0−n and, therefore,(10)TN−n=TNa>0−n+TNa<0−n=2TNa>0−n.For anα with Qα∈TNa>0−n, we call aα,−bα,−cα∈ℕ3 the signature of α, and we set(11)Sa>0−n≔aα,−bα,−cα∈ℕ3|Qα∈TNa>0−n.It is clear that we can also writeSa>0−n as(12)Sa>0−n=a,−b,−c∈ℕ3|bc=a2+n,a<−c,a<−2b,3a<−2b−c.It shall be apparent shortly that in order to count the number of orbits inℚ∗−n, we will need to calculate the cardinality of Sa>0−n, and for this endeavor, it is very helpful to find the maximum value that a can possibly take in order for a,−b,−c to belong to Sa>0−n. Based on Lemma 8, we see that such an a cannot exceed n if n is odd, and it cannot exceed n/2 if n is even. Lemma 7 is needed in proving Lemma 8.Lemma 7.
Leta,−b,−c∈Sa>0−n with c=2c′.(i)
Ifn is odd and b≤n or c′≤n, then a≤n.(ii)
Ifn is even and b≤n/2 or c′≤n/2, then a≤n/2.Proof.
We will only prove part (i), as the proof of part (ii) follows a similar argument. Assume thatn is odd. Denote b, c, and c′ by B, C, and C′, respectively. Notice that as b and c are of the same sign, bc=BC. We then have a2+n=BC=2BC′, a<2C′, a<2B, and 3a<2B+2C′. Due to the apparent symmetry between B and C′, it suffices to prove that if B≤n, then a≤n (the proof of the other statement is similar). Assume that B≤n.Case 1. Assume that B=n. Since a2=nC−1, n divides a2. Since n is square-free, n divides a and so a≥n. If a>n, then a/n>1. Since n and a are both odd (Lemma 1), a/n≥3. So, a≥3n. But a<2B=2n, which is a contradiction. Thus, a=n in this case.Case 2. Assume that B≤n+1/2. Then we have a<2B≤n+1 and so a≤n.Case 3. Assume that n+1/2<B<n. For the sake of contradiction, suppose that a>n. Let a=n+s, where s is even and s≥2. Since n+1/2<B and n is odd, B=n+t/2 for some odd t with t≥3. Note that as a<2B, s<t+1. Since s is even and t is odd, s≤t−1. Now, we have(13)a2+n=BCiffn+s2+n=n+tC2,iffn2+2sn+s2+n=n+tC2,iff2n2+4sn+2s2+2n=Cn+Ct,iff2n2+4s+2−Cn+2s2−Ct=0.
Sincea2+n=BC, the last quadratic equation (in n) is solvable in ℤ. So, its discriminant Δ is a perfect square in ℤ. But(14)Δ=4s+2−C2−422s2−Ct=16s2+4+C2+16s−8sC−4C−16s2+8Ct=C2+8t−8s−4C+16s+4.
SinceΔ is a perfect square in ℤ, there is some x∈ℤ such that Δ=C+x2. So, we must have(15)2x=8t−8s−4,x2=16s+4.
Thus,(16)4t−4s−22=16s+4,16t2+16s2+4−32st−16t+16s=16s+4,16t2+16s2−32st−16t=0,t2+s2−2st−t=0,t−s2=t.
So,x=4t−4s−2. As t−1≥s, x≥2. This implies, in particular, that C+x>0 and so Δ=C+x. We thus have n=−4s+2−C±C+x/4. Considering the negative sign and substituting for x, we would have(17)n=−4s−2+C−C−4t+4s+24=−t<0,a contradiction (as n>0). So, we consider the positive sign to get(18)n=−4s−2+C+C+4t−4s−24=2C+4t−8s−44,2n=C+2t−4s−2.
So,C=2n−2t+4s+2. Since 3a<2B+C, it follows that 3n+3s<n+t+2n−2t+4s+2 and thus −t+s+2>0. However, s≤t−1 implies that −t+s+2≤1. Combining the last two inequalities yields that −t+s+2=1, and so −t+s=−1. Thus, t=t−s2=1 contradicts the inequality t≥3. Hence, a>n is false, and, therefore, a≤n in this case.Lemma 8.
Leta,−b,−c∈Sa>0−n.(i)
Ifn is odd, then a≤n.(ii)
Ifn is even, then a≤n/2.Proof.
We shall only prove part (i), as the proof of part (ii) follows a similar argument. Assume thatn is odd. Fix a,−b,−c∈Sa>0−n, let c=2c′, and denote b, c, and c′ by B, C, and C′, respectively. We proceed in cases as follows.Case 1. Suppose that a≤B and let B=a+t for some t≥0. If a>n, then let a=n+s for some s≥1. Then,(19)a2+n=BC⟹a2+a−s=a+tC=aC+tC⟹aa+1−C=s+tC>0⟹a+1−C>0⟹a+1>C⟹a≥C,a contradiction. So,a≤n in this case.Case 2. Suppose that a≤C′. If a>n, then let a=n+s for some s≥1. Since 2a≤2C′=C in this case, let C=2a+t for some t≥0. We then have(20)a2+n=BC⟹a2+a−s=B2a+t=2aB+Bt⟹aa+1−2B=s+Bt>0⟹a+1−2B>0⟹a+1>2B⟹a≥2B.a contradiction. So,a≤n in this case too.Case (3): Suppose thata>B and a>C′. Assume that a>n. It then follows from Lemma 7 that B>n and C′>n. We then have n<B<a<2B and n<C′<a<C=2C′. Let a=n+1+s, B=n+1+t, C′=n+1+u, for some s,t,u≥0. Since B<a and C′<a, t<s, u<s, and s≥1. Now,(21)3a<2B+C⟹3n+3+3s<2n+2+2t+n+1+u⟹3s<2t+u.
But2t+u<2s+s=3s. This is a contradiction. So, a≤n in this case as well.Proposition 1.
Under the action ofH on ℚ∗−n, we have the following:(i)
Ifn is odd, then every orbit contains a unique totally negative quadruplet.(ii)
Ifn is even, then every orbit contains either a unique pair of purely imaginary elements or a unique totally negative quadruplet (but not both).Proof.
We first show that for any totally positiveβ∈ℚ∗−n, aβ<ayiβ for all i=1,2,3. If aβ,cβ>0 (the case aβ,cβ<0 is similar), then (by Table 1)(22)ayβ=−aβ−cβ=aβ+cβ>aβ=aβ.
In a similar manner, we prove thatay2β>aβ and ay3β>aβ.
LetαH be an orbit in ℚ∗−n and, by Lemma 6, let α1∈αH be totally positive. So, by Lemma 5, xα1 is totally negative. If Qxα1∈TN−n, then we stop. Otherwise, for some i=1,2,3, α2=yixα1 is either purely imaginary or totally positive. If α2 is purely imaginary, then xα2 is purely imaginary too (Lemma 5) and we stop. Otherwise, α2 is totally positive, and so xα2 is totally negative. If Qxα2∈TN−n, then we stop. Else, for some i=1,2,3, α3=yixα2 is either purely imaginary or totally positive. We repeat for α3 what we did for α1 and α2 to get either a pair of purely imaginary elements, a totally negative quadruplet, or carry on to get a totally positive α4=yixα3 for some i=1,2,3. We continue in this manner and suppose that no pair of purely imaginary elements pups up in the process, so we have a sequence of totally positive elements α1,α2,α3,…, such that, for every j=1,2,3,…, αj=yixαj−1 for some i=1,2,3. By the argument at the beginning of this proof, we see that if α2=yxα1, then y3α2=xα1 and so(23)aα2<ay3α2=axα1=aα1.
In a similar manner, we haveaα2<aα1 if α2=y2xα1 or α2=y3xα1. Doing the same for each j=1,2,3,…, we have(24)aα1>aα2>aα3>⋯.
Sinceaαi is a decreasing sequence of positive integers, it must terminate at some αk, say. We notice that xαk is totally negative. If, for some i=1,2,3, yixαk is purely imaginary, then we stop. Otherwise, yixαk is totally negative for all i=1,2,3, as if any of yixαk is totally positive, then ayixαk<axαk=aαk contradicting the minimality of aαk. By this, we have reached at the totally negative quadruplet Qxαk.
To complete the proof, we need yet to tackle the uniqueness claim. Suppose that an orbit contains a totally negative quadrupletQα=α,yα,y2α,y3α. We show that as we depart from Qα, every quadruplet we get must contain a totally positive element and three totally negative elements and, thus, we can never reach another totally negative quadruplet in the orbit. The action of y on Qα obviously permutes the elements of Qα. So, the only way to depart from Qα is via the action of x. If we depart Qα from the direction of α, for instance, then we get xα, which is then totally positive. Now the quadruplet Qxα=xα,yxα,y2xα,y3xα contains only one totally positive element, namely, xα, whereas the remaining elements are totally negative. The same scenario exactly occurs if we depart Qα from the direction of either of yiα, for i=1,2,3, so we keep dealing with Qxα with no loss of generality. Now, to depart from Qxα in order to get another quadruplet, we note that if we depart from the direction of xα, then we would land back in Qα as xxα=α. So, we should consider departing from the direction of yixα, i=1,2,3, to get the totally positive xyixα. Again, the quadruplet Qxyixα contains only one totally positive element, namely, xyixα, whereas the remaining elements are totally negative. Continuing this process with no return from one quadruplet to its predecessor, we can never reach another totally negative quadruplet. This shows the uniqueness of a totally negative quadruplet in a single orbit, if it exists. The uniqueness of a pair of purely imaginary elements in a single orbit, if exists, follows a similar line of argument as above. This completes the proof.
In the following, we denote the set of purely imaginary pairsα,xα of elements of ℚ∗−n by PI−n for n even. For m∈ℕ and k∈ℕ∪0 with k≤m, we denote the number of positive divisors of m, which do not exceed k by d≤km.
Now, we state the main theorem of this paper.Theorem 1.
The number of orbits inℚ∗−n under the action of H is equal to(25)TN−n;ifnis odd,PI−n+TN−n;ifnis even,=2;ifn=1orn=212∑1≤i≤n,iodddi2+n2−2d≤i/2i2+n2−Ein;ifnis odd,n≠1,dn2+12∑2≤i≤n/2,ievendi2+n2−2d≤i/2i2+n2−Ein;ifnis even,n≠2,where Ein is the number of positive even integers C dividing i2+n and 2i2+n/C+C≤3i.Proof.
The expression,(26)TN−n;ifnis odd,PI−n+TN−n;ifnis even,for the number of orbits in ℚ∗−n follows directly from Proposition 1, based on which we shall derive the second expression.
Ifn=1, the number of orbits in ℚ∗−1 is TN−1=2TNa>0−1. If Qα∈TNa>0−1, then (as aα,−bα,−cα∈Sa>0−1); it follows from Lemma 8 that 0<aα≤1 and so aα=1. But then cα<0 is an even divisor of 12+1=2; that is, cα=−2. Thus, TNa>0−1=1 and so TN−1=2 as claimed. Note that this is an exceptional case of signatures in the sense that Sa>0−1=1 and not a multiple of 4 as expected. The reason for this is that the totally negative element α=1+−1/−2 is fixed by y and thus Qα=α. Similarly, Qα¯=α¯. In fact, α and α¯ are the only elements of ℂ fixed by y, since z∈ℂ is fixed by y if and only if −1/2z+1=z; that is, z must be a solution of the quadratic equation 2z2+2z+1=0.
For the casen=2, the number of orbits in ℚ∗−2 is PI−2+TN−2. Since −2/c∈ℚ∗−2 if and only if c is an even divisor of 2, the only purely imaginary elements of ℚ∗−2 are −2/2 and −2/−2. Note that the pair of purely imaginary elements containing −2/2 is a singleton in the sense that it contains only −2/2. The same thing is said about −2/−2. The reason for this is that these two elements are fixed by x. In fact, they are the only elements of ℂ fixed by x, since z∈ℂ is fixed by x if and only if −1/2z=z; that is, z must be a solution of the quadratic equation 2z2+1=0. Since, by Lemma 3, these two elements lie in distinct orbits, we have PI−2=2. On the other hand, if Qα∈TNa>0−2, then (as aα,−bα,−cα∈Sa>0−2); it follows from Lemma 8 that 0<aα≤1 and so aα=1. But then cα<0 is an even divisor of 12+2=3, which is absurd. Thus, TNa>0−2=∅ and, therefore, TN−2=∅. This shows that the number of orbits in ℚ∗−2 is 2.
Now, suppose thatn∉1,2. Let Qα∈TNa>0−n. Since Qα is stable under the action of y, Sa>0−n contains the signature of α if and only if it contains the signatures of yα,y2α, and y3α. This shows that we can associate to every Qα∈TNa>0−n four distinct elements of Sa>0−n. Moreover, if Qα,Qβ∈TNa>0−n are not equal, then they are disjoint indeed. Thus, Sa>0−n=4TNa>0−n. Since TN−n=2TNa>0−n, we get TN−n=1/2Sa>0−n. To ease notations in the calculations below, we denote an element a,−b,−c∈Sa>0−n by a,B,C (so, B=−b and C=−c). Considering this notation, recall that(27)Sa>0−n=a,B,C∈ℕ3|Cis even,a2+n=BC,a<2B,a<C,3a<2B+C.
Due to Lemmas1 and 8, we partition Sa>0−n as follows:(28)Sa>0−n=∪1≤i≤n,ioddSa>0i−n;ifnis odd,n≠1,∪2≤i≤n/2,ievenSa>0i−n;ifnis even,n≠2,where(29)Sa>0i−n=i,B,C∈ℕ3|Cis even,i2+n=BC,i<2B,i<C,3i<2B+C.
Since the partitioning sets are mutually disjoint, the cardinality ofSa>0−n is equal to the sum of the cardinalities of its partitioning sets. For a fixed i, we see that an element i,B,C∈Sa>0i−n can also be written in the form i,i2+n/C,C. So, we write Sa>0i−n in the form(30)Sa>0i−n=i,i2+nC,C0<Cis even,Ci2+n,i<2i2+nC,i<C,3i<2i2+nC+C.
On the contrary, we observe that(31)Sa>0i−n=Sa>0i,∗−n−Sa>0i,2B−n∪Sa>0i,C−n∪Sa>0i,2B+C−n,
where(32)Sa>0i,∗−n=i,i2+nC,C0<Cis even,Ci2+n,Sa>0i,2B−n=i,i2+nC,C0<Cis even,Ci2+n,2i2+nC≤i,Sa>0i,C−n=i,i2+nC,C0<Cis even,Ci2+n,C≤i,Sa>0i,2B+C−n=i,i2+nC,C0<Cis even,Ci2+n,2i2+nC+C≤3i.
We now claim that the setsSa>0i,2B−n, Sa>0i,C−n, and Sa>0i,2B+C−n are mutually disjoint. Indeed, if Sa>0i,2B−n∩Sa>0i,C−n≠∅, then there is an even positive integer C such that C|i2+n, 2i2+n/C≤i, and C≤i. This implies that 2i2+n/i≤i and so 2i2+n≤i2, which is absurd. Also, if Sa>0i,2B−n∩Sa>0i,2B+C−n≠∅, then there is an even positive integer C such that C|i2+n, 2i2+n/C≤i, and 2i2+n/C+C≤3i. Let 0≤t<i be such that 2i2+n/C+t=i. Then, 2i2+n+Ct=iC and so C=2i2+n/i−t. Substituting this in the inequality 2i2+n/C+C≤3i yields(33)i−t2+2i2+n≤3ii−t,i2−2it+t2+2i2+2n≤3i2−3it,2n+t2≤it≤0,which is absurd too. Similarly, if Sa>0i,C−n∩Sa>0i,2B+C−n≠∅, then there is an even positive integer C such that C|i2+n, C≤i, and 2i2+n/C+C≤3i. Let 0≤t<i be such that C+t=i. Substituting the value of C in the inequality 2i2+n/C+C≤3i yields(34)2i2+n+i−t2≤3ii−t,2i2+2n+i2+t2−2it≤3i2−3it,2n+t2≤it≤0,which is absurd as well. By this, the claim is settled. Hence, we have(35)|Sa>0i−n=Sa>0i,∗−n−Sa>0i,2B−n−Sa>0i,C−n−Sa>0i,2B+C−n.
It is obvious thatSa>0i,∗−n is equal to the number of positive even divisors of i2+n, which is, by elementary number theory, equal to the number of positive divisors of i2+n/2. So, Sa>0i,∗−n=di2+n/2. As for the cardinality Sa>0i,C−n, we observe that it is equal to the number of positive even divisors of i2+n, which do not exceed i; that is, d≤ii2+n. Again, by elementary number theory, d≤ii2+n=d≤i/2i2+n/2. On the contrary, it can be easily checked that the map Sa>0i,2B−n⟶Sa>0i,C−n defined by i,i2+n/C,C⟼i,C/2,2i2+n/C is bijective and, thus, Sa>0i,2B−n=Sa>0i,C−n=d≤i/2i2+n/2. Finally, it is obvious that Sa>0i,2B+C−n=Ein, and by this, the proof is complete.Corollary 1.
The action ofH on ℚ∗−n is intransitive for every n.Proof.
Forn=1 and n=2, the number of orbits is 2 and so the action is intransitive in these two cases. For an even n different from 2, dn/2≥2 and so the action is intransitive in this case as well. If n is odd and different from 1, we can check that(36)1,n+12,2,1,1,n+1,n,n+12,2n,n,n,n+1⊆Sa>0−n.
Thus,TN−2=1/2Sa>0−n≥1/2⋅4=2 and, therefore, the action is also intransitive in this case.


## 4. Examples

To compute the number of orbits based on Theorem1, we remark that the first formula (for n≠1,2) requires exhaustive calculation of the sets PI−n and Sa>0−n, whereas the second formula requires the calculation of elements leading to Ein. We shall follow both approaches below.We know that, for an evenn, purely imaginary elements always exist. In the examples below, however, we chose to consider n=10 and n=14 to illustrate a case for an even n in the former case where totally negative quadruplets do not exist in contrast to the latter case where they exist. On the other hand, we know that for an odd n, purely imaginary elements do not exist while totally negative quadruplets always exist. However, we chose to consider below n=11 and n=15 to illustrate a case for an odd n in the former case where all Ein are zero in contrast to the latter case where not all Ein are zero.Example 2.
n=10. By Theorem 1, the number of orbits under the action of H on ℚ∗−10 is(37)d5+12Sa>0−10=2+12∑i=2,even5Sa>0i−10.
Fori=2, i2+n=14 and the only possible values of C are 2 and 14. But, if C=2 then i≮C and if C=14 then B=1 and i≮2B. So, Sa>02−10=∅. For i=4, i2+n=26 and the only possible values of C are 2 and 26. Similar to the argument above, we get Sa>04−10=∅. Thus, Sa>0−10=0 and the number of orbits in ℚ∗−10 is 2.
On the contrary, the other formula given by Theorem1 gives(38)d5+12∑i=2,even5di2+n2−2d≤i/2i2+n2−Ein.
As above, fori=2, if C=2, then B=7 and 2⋅7+2=16≰3⋅i=6, and if C=14, then B=1 and 2⋅1+14=16≰3⋅i=6. So, E210=0. Similarly, E410=0. Thus, the number of orbits in ℚ∗−10 is equal to(39)2+12d7−2d≤17−0+d13−2d≤213−0=2+122−2+2−2=2.Example 3.
n=14. Similar to the case n=10, it can be verified that Sa>02−14=2,3,6, Sa>04−14=4,5,6,4,3,10, and Sa>06−14=6,5,10. So, the number of orbits in ℚ∗−14 under the action of H is equal to(40)d7+12Sa>0−14=2+12∑i=2,even7Sa>0i−14=2+121+2+1=2+2=4.
On the contrary, using the other formula given by Theorem1, it can be checked that Ei14=0 for all i∈2,4,6. Thus, the number of orbits in ℚ∗−14 is equal to(41)d7+12∑i=2,even7di2+n2−2d≤i/2i2+n2−Ein=2+12d9−2d≤19−0+d15−2d≤215−0+d25−2d≤325−0=2+123−2⋅1+4−2⋅1+3−2⋅1=2+2=4.Example 4.
n=11. By Theorem 1, the number of orbits under the action of H on ℚ∗−11 is(42)12Sa>0−11=12∑i=1,odd11Sa>0i−11.
Fori=1, i2+n=12 and the only possible values of C are 2,4,6, and 12. The corresponding values of B are, respectively, 6,3,2, and 1. It can be checked that for every such triplet i,C,B, the conditions i<C, i<2B, and 3i<2B+C are satisfied. Thus,(43)Sa>01−11=1,6,2,1,3,4,1,2,6,1,1,12.
Similarly, the following can be verified:(44)Sa>03−11=3,5,4,3,2,10,Sa>05−11=5,6,6,5,3,12,Sa>07−11=7,6,10,7,5,12,Sa>09−11=∅,Sa>011−11=11,11,12,11,6,22.
Thus, the number of orbits inℚ∗−11 is 1/24+2+2+2+0+2=6.
On the contrary, the other formula given by Theorem1 gives(45)12∑i=1,odd11di2+n2−2d≤i/2i2+n2−Ein.
As above, fori=1, it can be checked that 2B+C≰3⋅i=3 for every possible pair B,C. So, E111=0. It can also be similarly checked that Ei11=0 for every i∈3,5,7,9,11. Thus, the number of orbits in ℚ∗−11 is equal to(46)12d6−2d≤06−0+d10−2d≤110−0+d18−2d≤218−0+d30−2d≤330−0+d46−2d≤446−0+d66−2d≤566−0=124−0+4−2+6−4+8−6+4−4+8−6=12⋅12=6.Example 5.
n=15. Similar to the case n=11, it can be verified that(47)Sa>01−15=1,8,2,1,4,4,1,2,8,1,1,16,Sa>03−15=3,6,4,3,4,6,3,3,8,3,2,12,Sa>05−15=5,5,8,5,4,10,Sa>07−15=7,8,8,7,4,16,Sa>09−15=9,8,12,9,6,16,Sa>011−15=Sa>013−15=∅,Sa>015−15=15,15,16,15,8,30.
Thus, by Theorem1, the number of orbits in ℚ∗−15 is(48)12Sa>0−15=12∑i=1,odd15Sa>0i−15=124+4+2+2+2+0+0+2=8.
On the contrary, using the other formula given in Theorem1, it can be verified that Ei15=0 for all i∈1,3,5,7,9,11,13. However, for i=15, each of the two triplets, and none else, 15,10,24 and 15,12,20 satisfy the condition 2B+C≤3i. So, E1515=2. Thus, the number of orbits in ℚ∗−15 is equal to(49)12d8−2d≤08−0+d12−2d≤112−0+d20−2di≤220−0+d32−2di≤332−0+d48−2di≤448−0=d68−2di≤568−0+d92−2di≤668−0+d120−2di≤7120−2=124−0+6−2+6−4+6−4+10−8+6−6+6−6+16−12−2=8.



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