# A Necessary and Sufficient Condition fork=ℚ4n2+1 to Have Class Number ωn+c

Authors: Ahmad Issa; Hasan Sankari
Journal: International Journal of Mathematics and Mathematical Sciences
 (2022)

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## Abstract

In this paper, we give a necessary and sufficient condition for real quadratic fieldk=ℚ4n2+1 to have class number ωn+c where c∈1,2 and ωn is the number of odd prime divisors of n.


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## Body

## 1. Introduction

Letd be a positive square-free integer and let hd and Ck denote the class number and the class group of a real quadratic field k=ℚd, respectively.The class number problem of quadratic fields is one of the most intriguing unsolved problems in algebraic number theory, and it has been the object of attention for many years of research.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. On the other hand, it was conjectured by Gauss that there are infinitely many real quadratic fields with class number one. This conjecture is still open.Many fruitful research studies have been conducted in this direction (see [2–7]). In particular, we mention some results showing finiteness of class number one real quadratic fields in special families of real quadratic fields. 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.Another line of work, in these contexts, is finding bounds for class number of a number field. Hasse [8] and Yokoi [9, 10] studied lower bounds for class numbers of certain real quadratic fields. Mollin [11, 12] generalized their results for certain real quadratic and biquadratic fields. Recently, Chakraborty et al. [13] derived a lower bound for ℚn2+1. Also, Mishra [14] gave a lower bound for ℚn2+r, where r=1,4. For a given fixed number h, it is interesting to find necessary and sufficient conditions for a real quadratic field to have class number h, see [15–18]. In particular, Yokoi [18] showed that the class number of k=ℚ4n2+1 is 1 if and only if n2−tt+1 (with 1≤t≤n−1) is a prime. In this work, we use the lower bound for h4n2+1 to give a necessary and sufficient condition for k=ℚ4n2+1 to have class number ωn+c where c∈1,2.


## 2. Preliminaries

Letk be a real quadratic field and ζks be its Dedekind zeta function. Siegel [19] developed a method of computing ζk1−2n, where n is a positive integer. By specializing Siegel’s formula for a real quadratic field, we obtain the following result.Theorem 1 (Zagier, [20]).
Letk be a real quadratic field with discriminant D. Then,(1)ζk−1=160∑t<Dt2≡Dmod4σ1D−t24,where σ1r denotes the sum of divisors of r.However, there is another method, due to Lang, of computing special values ofζks if k is a real quadratic field.Letk=ℚd be a real quadratic field of discriminant D and Η an ideal class of k. Let Ι be any integral ideal belonging to Η−1 with an integral basis {r1, r2}. We put(2)δΙ=r1r2′−r1′r2,where r1´ and r2´ are the conjugates of r1 and r2, respectively.Letε be the fundamental unit of k. Then, {εr1, εr2} is also integral basis of Ι, and thus we can find a matrix M=abcd with integer entries satisfying(3)εr1r2=M.r1r2.The entries ofM are given by(4)a=trr1r2′εδΙ,b=trr1r1′ε′δΙ,c=trr2r2′εδΙ,d=trr1r2′ε′δΙ.Moreover, detM=Nε and bc≠0. See ([13, 14, 21]). Now, we can state Lang’s formula.Theorem 2 (Lang, [22]).
By keeping the abovementioned notation, we have(5)ζk−1,Η=sgnδΙr2r2′360NΙc3a+d3−6a+dNε−240c3sgncS3a,c+180ac3sgncS2a,c−240c3sgncS3d,c+180dc3sgncS2d,c,where NΙ denotes the norm of an ideal Ι, Nε is the norm of ε, and Si−,− denotes the generalized Dedekind sum as defined in [23].Now, we will introduce some basics about quadratic number fieldk=ℚ4n2+1. Let n be a positive integer and let d=4n2+1 be a square-free integer. Clearly d≡1or5mod8. In this case, the fundamental unit of k is ε=2n+d and Nε=−1. We also know that if p∣n, where p is an odd prime, then p splits in k=ℚd as(6)p=p,1+d2p,1−d2.By ([21], Theorem 2.4), we also know that(7)ζk−1,A=4n3+14n180,where A will always denote the principal ideal class in k.By ([13], Proposition 3.1), we also know that(8)ζk−1,C=2n3+n2p4+5p290p2,where C is the ideal class containing p,1+d/2 or p,1−d/2.


## 3. Main Results

In this section, we will prove our main results. As a start, we record the following theorem that we derive from ([13], Proposition 3.1).Theorem 3.
Letn be a positive integer and let d=4n2+1 be a square-free integer and suppose that(i)
ωn≥3(ii)
c=1 if d≡5mod8 and c=2 if d≡1mod8
Then,(9)hd≥ωn+c.Now, we find necessary and sufficient condition fork=ℚ4n2+1 to have class number ωn+c.Theorem 4.
By keeping the abovementioned notation, we have
hd=ωn+c if and only if(10)∑m=0n−1σ1n2−mm+1=4n3+14n6+∑i=1ωn2n3+n2pi4+5pi23pi2ifc=1,5n3+40n6+∑i=1ωn2n3+n2pi4+5pi23pi2ifc=2,where pi are odd primes, pi|n, and ωn≥3.Proof.
Now, by Theorem1,(11)ζk−1=160∑t<4n2+1t2≡4n2+1mod4σ14n2+1−t24,=160∑1≤t≤2n−1ti soddσ14n2+1−t24,=130∑1≤t≤2n−1ti soddσ14n2+1−t24,=130∑m=0n−1σ14n2+1−2m+124,so that(12)ζk−1=130∑m=0n−1σ1n2−mm+1.
LetCi1≤i≤ωn be the ideal class in k such that pi,1+d/2∈Ci, then by (8), we obtain(13)ζk−1,Ci=2n3+n2pi4+5pi290pi2.
Ifc=1, then 2 remains prime in k. We note C1,…Cωn are distinct nonprincipal ideal classes in k.
Ifc=2, then 2 splits in k, that is,(14)2=2,1+d22,1−d2.
We assume thatB denotes the ideal class in k such that 2,1+d/2∈B, then by ([21], Theorem 2.5), we obtain(15)ζk−1,B=n3+26n180.
We noteB,C1,…Cωn are distinct nonprincipal ideal classes in k.
Necessary: lethd=ωn+c. Now, by definition, we have(16)ζk−1=∑J∈Ckζk−1,J.
Ifc=1, we obtain Ck=A,C1,…Cωn, then(17)ζk−1=ζk−1,A+∑i=1ωnζk−1,Ci.
This implies(18)ζk−1=4n3+14n180+∑i=1ωn2n3+n2pi4+5pi290pi2.
Hence, by (12), we find(19)∑m=0n−1σ1n2−mm+1=4n3+14n6+∑i=1ωn2n3+n2pi4+5pi23pi2.
Ifc=2, we obtain Ck=A,B,C1,…Cωn, then(20)ζk−1=ζk−1,A+ζk−1,B+∑i=1ωnζk−1,Ci.
This implies(21)ζk−1=5n3+40n180+∑i=1ωn2n3+n2pi4+5pi290pi2.
Hence, by (12), we find(22)∑m=0n−1σ1n2−mm+1=5n3+40n6+∑i=1ωn2n3+n2pi4+5pi23pi2.
Sufficiency: let(23)∑m=0n−1σ1n2−mm+1=4n3+14n6+∑i=1ωn2n3+n2pi4+5pi23pi2ifc=1,5n3+40n6+∑i=1ωn2n3+n2pi4+5pi23pi2ifc=2.
Then, by (12), we find(24)ζk−1=4n3+14n180+∑i=1ωn2n3+n2pi4+5pi290pi2ifc=1,5n3+40n180+∑i=1ωn2n3+n2pi4+5pi290pi2ifc=2.
By Theorem3, we get hd≥ωn+c.
Supposehd>ωn+c. Then, there exist at least ωn+c+1 ideal classes in k. Since for any ideal class E, ζk−1,E>0, thus(25)ζk−1>ζk−1,A+c−1ζk−1,B+∑i=1ωnζk−1,Ci.
It is a contradiction.



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