# Some Criteria for Class Numbers to Be Non-One

Authors: Ahmad Issa; Hasan Sankari
Journal: Journal of Mathematics
 (2020)

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## Abstract

Letd be a positive integer which is not a perfect square and n be any nonzero fixed integer. Then, the equation x2−dy2=n is known as the general Pell equation. In this paper, we give some criteria for class numbers of certain real quadratic fields to be greater than one, depending on the solvability of the general Pell equation, ideals in quadratic orders, and the period length of the simple continued fraction expansions of d.


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## Body

## 1. Introduction

Letd be a positive square-free integer, k=ℚd be a real quadratic field, and hd be the class number of this field.The class number problem of quadratic fields is one of the most intriguing unsolved problems in the algebraic number theory, and it has been the object of attention for many years of researchers.In [1], Gauss had conjectured that there exist exactly nine imaginary quadratic fields of class number 1. Later, this was solved after diverse works of Stark, Heegner, and Baker. And more, he conjectured that there exist infinitely many real quadratic fields of the form ℚp, p≡1mod4, of class number 1.Many fruitful research studies have been done in this direction. Biro’ (see [2, 3]) proved in 2003 two important results: Yokoi’s conjecture which asserts that hm2+4=1 only for six values of m=1,3,5,7,13,17 and Chowla’s conjecture which says that h4m2+1=1 only for six values of m=1,2,3,5,7,13.Mollin [4] considered m2+1 and proved that if m≠1, 2q, where q is a prime, then hm2+1>1. In 2007, Byeon et al. proved (see [5]) Mollin’s conjecture which says that hm2−4>1 whenever m>21. Numerous recent research studies have shown that the class number for a given family of quadratic fields is greater than one (see [6–14]).In this work, we give some possible generators of the principal primitive ideal by special cases of general Pell equations’ solutions. Since it is not always easy to check whether the class number for a given family of fields is greater than one or equals one, we give some results about that.


## 2. Preliminaries

Herein, we will be concerned with the simple continued fraction expansions ofd, where d is a positive integer that is not a perfect square. We denote this expansion by(1)α=d=q0;q1,q2,…,qℓ−1,2q0¯,where ℓ=ℓd is the period length, q0=d (the floor of d), and q1,q2,…,qℓ−1 is a palindrome.Thek–th convergent of α for k≥0 is given by(2)AkBk=q0;q1,q2,…,qk,where(3)Ak=qkAk−1+Ak−2,Bk=qkBk−1+Bk−2,with A−2=0,A−1=1,B−2=1,andB−1=0. The complete quotients are given by Pk+d/Qk, where P0=0, Q0=1, and for k≥1,(4)Pk+1=qkQk−Pk,qk=Pk+dQk,and d=Pk+12+QkQk+1. We will also need the following fact:(5)Ak−12−dBk−12=−1kQk.Also, whenℓ=ℓd is even, Qℓ/2 is called the central norm (for more information, see [15, 16]).Now, we will introduce some basic about quadratic order in the quadratic number field. A nonsquare integerΔ∈ℤ is called a quadratic discriminant if Δ≡0or1mod4. The quadratic discriminant Δ is called a fundamental discriminant if it admits no factorization Δ=Δ1m2 such that Δ1 is a quadratic discriminant and m∈ℕ≥2. In addition, Δ is a fundamental discriminant if and only if either Δ is square-free and Δ≡1mod4 or Δ=4d for some square-free integer d such that d≡2or3mod4.LetΔ be a quadratic discriminant, then the order of the quadratic discriminant Δ is(6)OΔ=u+vΔ2|u,v∈ℤ,u≡vΔmod2,and if Δ is a fundamental discriminant, we have OΔ=Ok, where Ok is the ring of integers of k=ℚΔ. The unit group OΔ× of OΔ is given by OΔ×=ε∈OΔ|Nε=1=u+vΔ/2|u,v∈ℤ,u2−Δv2=4 (see [15], Theorem 5.2.1, p. 122).Now, we investigate ideals. LetΔ be a quadratic discriminant. An ideal I⊂OΔ is called OΔ-primitive if e−1I⊈OΔ, for all e∈ℕ≥2. For a nonzero ideal I of OΔ, we call NI=OΔ:I the absolute norm of I. If α is a nonzero of OΔ, then N<α>=Nα (see [15], Definition 5.4.1, p. 132).


## 3. Main Results

In this section, we will prove our main results. As a start, we record the following theorem.Theorem 1.
LetΔ∈ℕ be a fundamental discriminant.(1)
Suppose thatΔ=4d, c∈1,2ifd≡3mod4,andc=1ifd≡2mod4d=ab, where a,b∈ℕ, and set(7)I=a,d,ifc=1,2a,a+d,ifc=2.(a)
I is an Ok-primitive ideal of Ok satisfying NI=ca. In addition, the following assertions are equivalent:(1)
I is a principal ideal of Ok(2)
There existx,y∈ℤ such that x2−dy2=ca(b)
Letx,y∈ℤ be such that x2−dy2=ca. Then, I=x+yd.(2)
Suppose thatΔ≡1mod4 and Δ=ab, where a,b∈ℕ, and set I=a,a+Δ/2:(a)
I is an Ok-primitive ideal of Ok satisfying NI=a. In addition, the following assertions are equivalent:(1)
I is a principal ideal of Ok(2)
There existx,y∈ℤ such that x2−Δy2=4a(b)
Letx,y∈ℤ be such that x2−Δy2=4a. Then, I=x+yΔ/2.(3)
Suppose thatΔ≡1mod8 and Δ=ab, where a,b∈ℕ, and set I=a,a+Δ/2:(a)
I is an Ok- primitive ideal of Ok satisfying NI=a. In addition, the following assertions are equivalent:(1)
I is a principal ideal of Ok(2)
There existx,y∈ℤ such that x2−Δy2=a(b)
Letx,y∈ℤ be such that x2−Δy2=a. Then, I=x+yΔ.Proof.
(1)
(a) By [17] (Theorem 7.1.5, p. 144), we find that I⊂Ok is an ideal of Ok satisfying NI=ca. Let e−1I⊂Ok with e∈ℕ which implies e=1. Therefore, I is an Ok-primitive ideal of Ok.(1)⟹(2). Now, let I be principal. Then, there exists α=x+yd∈Ok, where I=α. Since I is an Ok-primitive ideal, gcd x,y=1.Ifc=1, I=a,d=x+yd, then(8)a=NI=Nx+yd=x2−dy2.On the other hand,Ifc=2, I=2a,a+d=x+yd, then(9)2a=NI=Nx+yd=x2−dy2.(2)⟹(1). Let x,y∈ℤ such that x2−dy2=ca, which implies x=ax1 for some x1∈ℤ.Ifc=1, now we obtain x+yd=ax1+yd∈I. Hence, x+yd⊂Ι, and equality holds, since(10)N<x+yd>=Nx+yd=x2−dy2=a=NI.On the other hand,Ifc=2, which implies gcd2,xy=1, then gcd2,x1=1 and x1−y=2w for some w∈ℤ. Now, we obtain x+yd=2wa+ya+d∈I. Hence, x+yd⊂I, and equality holds, since(11)N<x+yd>=Nx+yd=x2−dy2=2a=NI.(b)
By (2)⟹(1).(2)
(a) By [17] (Theorem 7.1.5, p. 144), we find that Ι⊂Ok is an ideal of Ok satisfying NI=a. Let e−1I⊂Ok with e∈ℕ which implies e=1. Therefore, I is an Ok-primitive ideal of Ok.(1)⟹(2). Let now I be principal. Then, there exists α=x+yΔ/2∈Ok, where I=α. On the other hand, a=NI=Nα=x2−Δy2/4, and therefore x2−Δy2=4a.(2)⟹(1). Let x,y∈ℤ such that x2−Δy2=4a, which implies 2|x−y and x=ax1 for some x1∈ℤ. Now, we obtain x+yΔ/2=ax1+yΔ/2=ax1−y/2+ya+Δ/2∈I.Hence,x+yΔ/2⊂Ι, and equality holds, since(12)Nx+yΔ2=Nx+yΔ2=a=NI.(b)
By (2)⟹(1).(3)
(a) We noteI⊂Ok is an Ok-primitive ideal of Ok satisfying NI=a.(1)⟹(2). Let now I be principal. Then, there exist u,v∈ℤ such that u2−Δv2=4a. Since Δ≡1mod8, Δ is odd. But 4a is even, so the equation u2−Δv2=4a implies that u and v have the same parity. Let u1=u−v/2,v1=v, hence u12+u1v1+1−Δ/4v12=4u12+4u1v1+1−Δv12/4=u2−Δv2/4=4a/4=a. Since a is odd and 1−Δ/4 is even, then u1 is odd and v1 is even. Let m=2u1+v1/2, n=v1/2, hence m2−Δn2=u12+u1v1+1−Δ/4v12=a. Let x=m, y=n, then there exist x,y∈ℤ such that x2−Δy2=a.(2)⟹(1). Let x,y∈ℤ such that x2−Δy2=a. Then, 2x2−Δ2y2=4a, which implies 2x=ax1 for some x1∈ℤ. Now, we obtain x+yΔ=ax1/2+yΔ=ax1/2−y+2ya+Δ/2∈I.Hence,x+yΔ⊂I, and equality holds, since(13)N<x+yΔ>=Nx+yΔ=a=NI.(b)
By (2)⟹(1).
Now, we will be devoted to numerical examples illustrating Theorem1.Example 1.
LetΔ=312=4d;d=78=3×26=ab≡2mod4. Since 9,1 is the solution of x2−78y2=3, then 3,d=9+d.Example 2.
LetΔ=140=4d;d=35=5×7=ab≡3mod4. Since 5,1 is the solution of x2−35y2=−10, then 10,5+d=5+d.Example 3.
LetΔ=805=5×161=ab≡1mod4. Since 85,3 is the solution of x2−805y2=−20, then 5,5+Δ/2=85+3Δ/2.Example 4.
LetΔ=129=3×43=ab≡1mod8. Since 159,14 is the solution of x2−129y2=−3, then 3,3+Δ/2=159+14Δ.
We need the following proposition in order to prove thatham2+a>1.Proposition 1.
LetΔ∈ℕ be a quadratic discriminant,Δ=4d or Δ=d≡1mod8, d=am2+a, where a,m∈ℕ,m≥1,a>2. Then, the equation x2−dy2=p has no integer solutions for all primes p, where a>p.Proof.
We will provide the complete proof for the casex2−dy2=−p, and the other case x2−dy2=p can be handled along the similar lines. Suppose this equation x2−dy2=−p has integer solutions u,v, choose from these solutions u0,v0 with u0≥0, v0>0, and v0 is the smallest. Let x=u0−v0d∈OΔ and ε=1+2am2+2md∈OΔ×. Then,(14)Nεx=NεNx=u02−dv02=−p.
Hence,(15)N1+2am2+2mdu0−v0d=−p1+2am2u0−2mv0d2−d2mu0−v01+2am22=−p.
By the definition ofv0, we find(16)2mu0−v01+2am2≥v0.
Then, we have2mu0−v01+2am2≥v0 or −2mu0+v01+2am2≥v0. Thus, mu0≥am2+1v0 or mu0≤av0m2.Case (i):mu0≥am2+1v0.(17)u02−dv02=−p,mu02−dmv02=−pm2,am2+1v02−dmv02≤−pm2,v02a2m4+v02+2am2v02−v02a2m4−am2v02≤−pm2,v02+am2v02≤−pm2.Case (ii):mu0≤av0m2(18)u02−dv02=−p,mu02−dmv02=−pm2,av0m22−dmv02≥−pm2,−am2v02≥−pm2⟹p≥av02≥a.
In both cases, we find a contradiction.Theorem 2.
LetΔ∈ℕ be a fundamental discriminant,Δ=4d or Δ=d≡1mod8, d=am2+a, where a,m∈ℕ,m≥1,a>2 and a is not a prime number, then hd>1Proof.
Letp be a prime number and p|a; hence, a=ep, where e is square-free integer, which implies d=pr for r=pe2m2+e.
From Theorem1 and Proposition 1, we obtainCase (i):p,d is not a principal ideal of Ok if Δ=4dCase (ii):p,p+d/2 is not a principal ideal of Ok if Δ=d≡1mod8
In both cases, we find thathd>1.
The following proposition gives us a condition for solving the equationx2−dy2=ca, c∈−1,1Proposition 2.
Letd=ab, where a,b∈ℕ, be a positive integer that is not a perfect square and suppose that a is square-free and a<b, c∈−1,1. If the equation x2−dy2=ca is solvable, then ℓ=ℓd≡2c+2mod4.Proof.
By [16] (Proposition 3.1), it follows that ℓ=ℓd is even and Qℓ/2=a. We put k=ℓ/2 in (5). We obtain(19)Aℓ/2−12−dBℓ/2−12=−1ℓ/2a.
Hence,−1ℓ/2=c. We get(20)ℓ2=2r,ifc=1,2r+1,ifc=−1.
So,(21)ℓ≡0mod4,ifc=1,2mod4,ifc=−1.
This completes the proof.Example 5.
Letd=46=2×23=ab since(22)46=6;1,3,1,1,2,6,2,1,1,3,1,12¯.
Hence,ℓ=ℓ46=12, and the equation x2−46y2=−2 has no integer solutions.Example 6.
Letd=11046=21×526=ab since(23)11046=105;10,210¯.
Hence,ℓ=ℓ11046=2, and the equation x2−11046y2=21 has no integer solutions.
We use Theorem1 and Proposition 2 in order to prove the following theorem, which gives a sufficient condition for hd>1 in two cases.Theorem 3.
LetΔ=4d∈ℕ be a fundamental discriminant and suppose that c=1 if d≡2mod4, and c=2,ifd≡3mod4; d=ab, where a,b∈ℕ,a<b and a≡3mod4, ℓd≡2cmod4. Then,(a)
Ifc=1 and 8∤a+1 then hd>1(b)
Ifc=2 then hd>1Proof.
(a)
From Proposition2, the equation x2−dy2=a is unsolvable. Since a≡3mod4and8∤a+1, then the equation x2−dy2=−a is unsolvable. By Theorem 1, it follows that a,d is not a principal ideal; then, hd>1.(b)
From Proposition2, the equation x2−dy2=−a is unsolvable. Since a≡3mod4, then the equation x2−dy2=a is unsolvable. By Theorem 1, it follows that a,d is not a principal ideal; then, hd>1Example 7.
LetΔ=44184=4d;d=11046=3×3682=ab≡2mod4, a≡3mod4, 8∤a+1, and ℓd=2. Then, hd>1.Example 8.
LetΔ=7564=4d;d=1891=31×61=ab≡3mod4,a≡3mod4, and ℓd=36. Then, hd>1.



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