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Article

Some Remarks on the Divisibility of the Class Numbers of Imaginary Quadratic Fields

Department of Mathematics, Chosun University, 309 Pilmundaero, Gwangju 61452, Korea
Mathematics 2022, 10(14), 2488; https://doi.org/10.3390/math10142488
Received: 8 June 2022 / Revised: 8 July 2022 / Accepted: 10 July 2022 / Published: 17 July 2022

Abstract

:
For a given integer n, we provide some families of imaginary quadratic number fields of the form Q ( 4 q 2 p n ) , whose ideal class group has a subgroup isomorphic to Z / n Z .

1. Introduction

The class number of a number field is by definition the order of the ideal class group of its ring of integers. Thus, a number field has class number one if and only if its ring of integers is a principal ideal domain. In this sense, the ideal class group measures how far R is from being a principal ideal domain, and hence from satisfying unique prime factorization. The divisibility properties of class numbers are very important to know the structure of ideal class groups of number fields. Numerous results about the divisibility of the class numbers of quadratic fields have been introduced by many authors ([1,2,3,4,5,6,7,8,9,10,11,12,13,14,15]). By their works, it was shown that there exist infinitely many imaginary quadratic number fields whose ideal class numbers are multiples of n. They proved that there exist infinitely many imaginary quadratic number fields such that the ideal class group has a cyclic subgroup of order n. Most of such families are of the type Q ( x 2 t n ) or of the type Q ( x 2 4 t n ) , where x and t are positive integers with some restrictions. (For the case of Q ( x 2 t n ) , see [1,2,6,7,9,11,12,13,15] and for the case of Q ( x 2 4 t n ) see [3,4,5,8,10,14]).
Recently, K. Chakraborty, A. Hoque, Y. Kishi and P.P. Pandey considered the family K p , q = Q ( q 2 p n ) when p and q were distinct odd prime numbers and n 3 was an odd integer (see Theorem 1.2 of [2]). However, they just dealt with the case when n was an odd integer. We want to deal with the case when n is an even integer. In this article, we treat the family K p , 2 q = Q ( 4 q 2 p n ) when p and q are distinct odd prime numbers.

2. Preliminaries

In this section, we review some previous results which we will use.

2.1. Being a pth Power

Proposition 1.
(Proposition 2.2 in [2]). Let d 5   ( mod 8 ) be an integer and ℓ be a prime. For odd integers a, b, we have
a + b d 2 Z [ d ] if   and   only   if = 3 .
Definition 1.
If L / K is a Galois extension and α is in L, then the trace of α is the sum of all the Galois conjugates of α, i.e.,
T r ( α ) = σ Gal ( L / K ) σ ( α ) ,
where Gal ( L / K ) denotes the Galois group of L / K .
Lemma 1.
(Lemma 4 in [10]). Let K be a quadratic number field and O K be its ring of algebraic integers. If α O K , then α is a square in O K if and only if there exists A Z such that N ( α ) = A 2 and such that T r ( α ) + 2 A is a square in Z . If K is imaginary, we may assume that A 0 .

2.2. Result of Y. Bugeaud and T. N. Shorey

In this section, we review a result of Y. Bugeaud and T.N. Shorey (see [16]). Let F n be the nth Fibonacci sequence and L n be the nth Lucas sequence. Let us define the sets F and G N × N × N by
F : = { ( F h 1 2 ϵ , L h 1 + ϵ , F h 1 ) | h 1 N   s . t . h 1 2 and ϵ { ± 1 } }
and
G : = { ( 1 , 4 p 1 h 2 1 , p 1 ) | p 1 is a prime number and h 2 N } .
For λ { 1 , 2 , 2 } , we define the set H λ N × N × N by
H λ : = ( D 1 , D 2 , p ) | D 1 , D 2 and p are mutually coprime positive integers with p an odd prime and there exist positive integers r , s such that D 1 s 2 + D 2 = λ 2 p r and 3 D 1 s 2 D 2 = ± λ 2
Theorem 1.
(Theorem 1 in [16]). Let D 1 , D 2 and p be mutually coprime positive integers with p a prime number. Let λ { 1 , 2 , 2 } be such that λ = 2 if p = 2 . We assume that D 2 is odd if λ { 2 , 2 } . Then, the number of positive integer solutions ( x , y ) of the equation
D 1 x 2 + D 2 = λ 2 p y
is at most one except for
( λ , D 1 , D 2 , p ) E : = ( 2 , 13 , 3 , 2 ) , ( 2 , 7 , 11 , 3 ) , ( 1 , 2 , 1 , 3 ) , ( 2 , 7 , 1 , 2 ) , ( 2 , 1 , 1 , 5 ) , ( 2 , 1 , 1 , 13 ) , ( 2 , 1 , 3 , 7 ) .
or
( D 1 , D 2 , p ) F G H λ .
We recall the result of J.H.E Cohn [17] about the appearance of squares in the Lucas sequence.
Theorem 2.
Let L n be the nth Lucas sequence. Then, the only perfect square appearing in the Lucas sequences are L 1 = 1 and L 3 = 4 .

3. Main Result

In this section, we will describe the main result. Here is the crucial theorem.
Theorem 3.
Suppose that n 3 is an integer and q is an odd prime number such that ( q , n ) = 1 and q ± 1   ( mod ) for all odd prime number 3 dividing n. Let p be an odd prime number with 4 q 2 < p n and ( q , p ) = 1 . Let d be the square-free part of 4 q 2 p n , i.e., 4 q 2 p n = m 2 d for some positive integer m. Assume that 2 q ± 1   ( mod   | d | ) ) . Moreover, we assume q 2   ( mod   3 ) when 3 | n . Then, we have the following:
(i) Assume that n is an even integer or p 1   ( mod 4 ) . Then, the class number of K p , 2 q = Q ( d ) is divisible by n.
(ii) Assume that n is an odd integer and p 3   ( mod 4 ) . Moreover, we assume p n / 3 ( 4 q + 1 ) / 3 when 3 | n . Then, the class number of K p , 2 q = Q ( d ) is divisible by n.
Remark 1.
By Dirichlet’s theorem on arithmetic progressions, we know that there exist infinitely many q such that q ± 1   ( mod ) for all odd prime number 3 dividing n.
Theorem 4.
Let n, q be as in Theorem 3. For each q, the class number of K p , 2 q is divisible by n for all but finitely many p s . Furthermore, for each q there are infinitely many fields K p , 2 q .

4. Proof of Main Theorem

4.1. Crucial Proposition

Lemma 2.
Let p , d and m be as in Theorem 3 (i) or (ii). Let ℓ be an odd prime such that
α = 2 q + m d = ( a + b d )
for some integer a and b. Then, a | 2 q if and only if a | 2 q .
Proof. 
Suppose that
α = 2 q + m d = ( a + b d ) .
If we compare the real parts, we know that
2 q = a + i = 1 ( 1 ) / 2 2 i a 2 i b 2 i d i .
This implies that a | 2 q . Since a | 2 q , we also know that a | 2 q . Similarly, a | 2 q implies that a | 2 q . □
Proposition 2.
Let n , q , p , d and m be as in Theorem 3 (i) or (ii). Then, the element α = 2 q + m d is not an ℓth power of an element in the ring of integers of K p , 2 q for any odd prime divisor ℓ of n. In addition, α and α are not a square in O K p , 2 q .
Proof.  
(i) Assume that n is an even integer or p 1   ( mod 4). Moreover, we assume p n / 3 ( q + 16 ) / 3 when 3 | n . Since n is an even integer or p 1   ( mod 4), we know that d 3   ( mod 4). Let be an odd prime divisor of n. If α = 2 q + m d is an th power, then
α = 2 q + m d = ( a + b d )
for some integer a and b. If we compare the real parts, we know that
2 q = a + i = 1 ( 1 ) / 2 2 i a 2 i b 2 i d i .
This implies that a | 2 q . By Lemma 2, we can assume that a = 2 q , a = q , a = 2 or a = 1 .
Case (i-A1): a = 2 , 3
Comparing the real parts, we have
2 q = ( ± 2 ) + i = 1 ( 1 ) / 2 2 i ( ± 2 ) 2 i b 2 i d i ± 2   ( mod ) .
From these, we have q ± 1   ( mod ) , which violates our assumption.
Case (i-A2): a = 2 , = 3
Suppose that
α = 2 q + m d = ( 2 + b d ) 3 .
Comparing the real parts, we have
2 q = 8 + 6 b 2 d .
Since d < 0 , we have q = 4 + 3 b 2 d < 0 . This is impossible.
Case (i-B1): a = q , 3
Comparing the real parts, we have
2 q = ( ± q ) + i = 1 ( 1 ) / 2 2 i ( ± q ) 2 i b 2 i d i ± q ( mod ) .
Thus, we get 3 q 0 ( mod ) or q 0 ( mod ) , which contradicts the assumption “ ( q , n ) = 1 ” and “ 3 ”.
Case (i-B2): a = q , = 3
Suppose that
α = 2 q + m d = ( q + b d ) 3 .
Comparing the real parts, we have
2 q = q 3 + 3 q b 2 d .
By (3), we have 2 = q 2 + 3 b 2 d , and hence 2 q 2 ( mod 3 ) . This is impossible.
Case (i-C): a = 2 q
We have 2 q + m d = ( 2 q + b d ) . Taking the norm on both sides, we obtain
p n = ( 4 q 2 b 2 d ) .
If we write D 1 = d > 0 , we have
D 1 b 2 + 4 q 2 = p n / .
We also obtain
D 1 m 2 + 4 q 2 = p n .
Then, we easily know that ( | b | , n / ) and ( m , n ) are distinct solutions of (1) for D 1 = d > 0 , D 2 = 4 q 2 , λ = 1 . The next thing we have to do is to show that ( 1 , D 1 , D 2 , p ) E and ( D 1 , D 2 , p ) F G H λ . Clearly, ( 1 , D 1 , D 2 , p ) E and ( D 1 , D 2 , p ) G . By Theorem 2, we know that ( D 1 , D 2 , p ) F . Finally suppose that ( D 1 , D 2 , p ) H λ . Then, there exist positive integers r , s such that
3 D 1 s 2 4 q 2 = ± 1
and
D 1 s 2 + 4 q 2 = p r .
By (4), we have q 3 , and hence we have 3 D 1 s 2 4 q 2 = 1 . From this together with (5), we obtain
16 q 2 = 3 p r + 1 ,
that is,
( 4 q 1 ) ( 4 q + 1 ) = 3 p r .
This implies that 4 q 1 = 1 or 4 q 1 = 3 . It contradicts the fact that q is an odd prime number. Hence, ( D 1 . D 2 , p ) H 1 . By Theorem 1, the equation
d x 2 + 4 q 2 = p y
has at most one integer solutions ( x , y ) . Thus, a 2 q
Case (i-D): a = 1
Comparing the real parts, we have
2 q = ( 1 ) + i = 1 ( 1 ) / 2 2 i ( 1 ) 2 i b 2 i d i 1 ( mod | d | ) .
It contradicts our assumption “ 2 q 1 ( mod | d | ) ”.
(ii) Assume that n is an odd integer and p 3   ( mod 4). Then, we know that d 1   ( mod 4). Moreover, we assume p n / 3 ( 4 q + 1 ) / 3 when 3 | n . Let be an odd prime divisor of n. If α = 2 q + m d is an th power, then
α = 2 q + m d = a + b d 2 , a b ( mod 2 ) .
for some integer a and b. In case both a and b are even, then we can proceed as in the above and obtain a contradiction. Thus, we can assume that both a and b are odd. If we take the norm on both sides we obtain
4 p n / = a 2 b 2 d .
Since a and b are odd integers and p 2 , we can get d 5   ( mod 8). By Proposition 1, we know that = 3 . Thus, we have
α = 2 q + m d = a + b d 2 3 .
Comparing the real parts, we have
16 q = a ( a 2 + 3 b 2 d ) .
Since a is an odd integer, we have a = 1 or a = q .
Case (ii-A): a = 1
By (7) and d < 0 , we have 16 q = 1 + 3 b 2 d < 0 . This is not possible.
Case (ii-B): a = q
By (6) and (7), we have
4 p n / 3 = q 2 b 2 d   and   16 = q 2 + 3 b 2 d .
From these, we have 3 p n / 3 = q 2 4 = ( q 2 ) ( q + 2 ) . This implies that q 2 = 3 or q + 2 = 3 . Since q is a prime, we have q 2 = 3 and p n / 3 = q + 2 = 7 . These violate our assumption p n / 3 ( 4 q + 1 ) / 3 .

4.2. Proof of Theorem 3

Next, we prove Theorem 3.
Proof of Theorem 3. 
Let n , q , p , d and m be as in Theorem 3 (i) or (ii). Set α = 2 q + m d . We can easily check that α and α ¯ are coprime and N ( α ) = α α ¯ = p n . This implies that ( α ) = a n for some integral ideal a of K p , 2 q . It suffices to show that the order of [ a ] in the ideal class group of K p , 2 q is n. If this is not the case, we have ( α ) = ( β ) for some integer β in O K p , 2 q and some prime divisor of n. Since K p , 2 q is an imaginary quadratic field, the only units of O K p , 2 q are ± 1 . Thus, we have α = ± β . If is an odd prime, we have α = γ where γ = ± β . This contradicts Proposition 2. Next, let us consider the case of = 2 . Then, we have α = ± β 2 . It means that α or α is a square in O K p , 2 q , which contradicts Proposition 2. Hence, the order of [ a ] in the ideal class group of K p , 2 q is n. □

4.3. Proof of Theorem 4

We are now in a position to prove the main theorem
Proof. 
Let n and q be as in Theorem 3. For any positive integer D, the curve
D X 2 + 4 q 2 = Y n
is an irreducible algebraic curve of genus > 0 (see [18]). By Siegel’s theorem (see [19]), there are only finitely many integral points ( X , Y ) on the curve (8). Thus, for each d < 0 , there are at most finitely many primes p such that
d x 2 + 4 q 2 = p n .
It means that there are infinitely many fields K p , 2 q for the fixed prime q. In addition, we have | d | > 2 q + 1 for sufficiently large p, so 2 q ± 1   ( mod   | d | ) . Further, if p is large enough, then p n / 3 ( q + 16 ) / 3 and p n / 3 ( 4 q + 1 ) / 3 . Hence, the class number of K p , 2 q is divisible by n for a sufficiently large p. □

5. Numerical Examples

In this section, we give several examples. All computations in this section are based on the Magma program. For example, Table 1 is the list of imaginary quadratic fields K p , 2 q corresponding to n = 3 and p 19 . In the below Table 2, Table 3, Table 4, Table 5, Table 6, Table 7 and Table 8, we use * in the column for class number to indicate the failure of condition “ p n / 3 ( q + 16 ) / 3 ” or “ p n / 3 ( 4 q + 1 ) / 3 ”. Furthermore, the appearance of ** in the column for a class number indicates the failure of condition “ 2 q ± 1   ( mod   | d | ) ”. Finally, the appearance of *** in the column for a class number indicates the failure of condition “ q ± 1   ( mod )” for an odd prime divisor 3 of n.

Funding

This study was supported by research funds from Chosun University 2022.

Conflicts of Interest

The author declares no conflict of interest.

References

  1. Ankeny, N.; Chowla, S. On the divisibility of the class numbers of quad-ratic fields. Pac. J. Math. 1955, 5, 321–324. [Google Scholar] [CrossRef]
  2. Chakraborty, K.; Hoque, A.; Kishi, Y.; Pandey, P.P. Divisibility of the class numbers of imaginary quadratic fields. J. Number Theory 2018, 185, 339–348. [Google Scholar] [CrossRef][Green Version]
  3. Cohn, J.H.E. On the class number of certain imaginary quadratic fields. Proc. Am. Math. Soc. 2002, 130, 1275–1277. [Google Scholar] [CrossRef][Green Version]
  4. Gross, B.H.; Rohrlich, D.E. Some results on the Mordell–Weil group of the Jacobian of the Fermat curve. Invent. Math. 1978, 44, 201–224. [Google Scholar] [CrossRef]
  5. Ishii, K. On the divisibility of the class number of imaginary quadratic fields. Proc. Jpn. Acad. Ser. A 2011, 87, 142–143. [Google Scholar] [CrossRef]
  6. Ito, A. A note on the divisibility of class numbers of imaginary quadratic fields Q ( a 2 k n ) . Proc. Jpn. Acad. Ser. A 2011, 87, 151–155. [Google Scholar] [CrossRef]
  7. Ito, A. Remarks on the divisibility of the class numbers of imaginary quad-ratic fields Q ( 2 2 k q n ) . Glasg. Math. J. 2011, 53, 379–389. [Google Scholar] [CrossRef][Green Version]
  8. Ito, A. Notes on the divisibility of the class numbers of imaginary quadratic fields Q ( 3 2 e 4 k n ) . Abh. Math. Semin. Univ. Hambg. 2015, 85, 1–21. [Google Scholar] [CrossRef]
  9. Kishi, Y. Note on the divisibility of the class number of certain imaginary quadratic fields. Glasg. Math. J. 2009, 51, 187–191. [Google Scholar] [CrossRef][Green Version]
  10. Louboutin, S.R. On the divisibility of the class number of imaginary quadratic number fields. Proc. Am. Math. Soc. 2009, 137, 4025–4028. [Google Scholar] [CrossRef][Green Version]
  11. Murty, M.R. The ABC conjecture and exponents of class groups of quadrat-ic fields. Contemp. Math. 1998, 210, 85–95. [Google Scholar]
  12. Murty, M.R. Exponents of class groups of quadratic fields. In Topics in Number Theory; Mathematics and Its Applications; Kluwer Academic Publishers: University Park, PA, USA; Dordrecht, The Netherlands, 1999; Volume 467, pp. 229–239. [Google Scholar]
  13. Soundararajan, K. Divisibility of class numbers of imaginary quad-ratic fields. J. Lond. Math. Soc. 2000, 61, 681–690. [Google Scholar] [CrossRef]
  14. Yamamoto, Y. On unramified Galois extensions of quadratic number fields. Osaka J. Math. 1970, 7, 57–76. [Google Scholar]
  15. Zhu, M.; Wang, T. The divisibility of the class number of the imaginary quadratic field Q ( 2 2 m k n ) ). Glasg. Math. J. 2012, 54, 149–154. [Google Scholar]
  16. Bugeaud, Y.; Shorey, T.N. On the number of solutions of the generalized Ra-manujan–Nagell equation. J. Reine Angew. Math. 2001, 539, 55–74. [Google Scholar]
  17. Cohn, J.H.E. Square Fibonacci numbers, etc. Fibonacci Quart. 1964, 2, 109–113. [Google Scholar]
  18. Schmidt, W.M. Equations over Finite Fields, an Elementary Approach; Lecture Notes in Math.; Springer: Berlin, Germany; New York, NY, USA, 1976; Volume 536. [Google Scholar]
  19. Evertse, J.-H.; Silverman, J.H. Uniform bounds for the number of solutions to Yn = f(X). Math. Proc. Cambridge Philos. Soc. 1986, 100, 237–248. [Google Scholar] [CrossRef]
Table 1. Numerical examples for n = 3 .
Table 1. Numerical examples for n = 3 .
pq 4 q 2 p 3 d h ( d ) pq 4 q 2 p 3 d h ( d )
75 243 3 1 *115 1231 1231 27
117 1135 1135 181113 655 655 12
1117 175 7 1 *135 2097 233 12
137 2001 2001 481311 1713 1713 36
1317 1041 1041 361319 753 753 12
175 4813 4813 30177 4717 4717 24
1711 4429 4429 601713 4237 4237 24
1719 3469 3469 301723 2797 2797 18
1729 1549 1549 181731 1069 1069 30
195 6759 751 15197 6663 6663 60
1911 6375 255 121913 6183 687 12
1917 5703 5703 541923 4743 527 18
1929 3495 3495 361931 3015 335 18
1937 1383 1383 181941 135 15 2 *
Table 2. Numerical examples for n = 4 .
Table 2. Numerical examples for n = 4 .
pq 4 q 2 p 4 d h ( d ) pq 4 q 2 p 4 d h ( d )
53 589 589 1657 429 429 16
511 141 141 873 2365 2365 32
75 2301 2301 48711 1917 213 8
713 1725 69 8717 1245 1245 32
719 957 957 16723 285 285 16
113−14,605 14 , 605 80115−14,541−14,54164
117−14,4451605161113−13,965 285 16
1117−13,485−13,4851281119−13,197−13,19748
1123−12,525 501 161129−11,277−11,27732
1131−10,797−10,797641137 9165 9165 64
1141 7917 7917 321143 7245 805 16
1147 5805 645 161153 3405 3405 48
1159 717 717 16
Table 3. Numerical examples for n = 5 .
Table 3. Numerical examples for n = 5 .
pq 4 q 2 p 5 d h ( d ) pq 4 q 2 p 5 d h ( d )
37−47−47553−3089−308940
57−2929−292940511−2641−264120
513−2449−244940517−1969−196920
519−1681−11 **523−1009−100920
73−16,771−16,7714075−16,707−16,70720
711−16,323−16,32330713−16,131−16,13140
717−15,651−173920719−15,363−170710
723−14,691−14,69140729−13,443−13,44330
731−12,963−12,96320737−11,331−125915
741−10,083−10,08320743−9411−941130
747−7971−797130753−5571−6195
759−2883−31 **761−1923−192310
Table 4. Numerical examples for n = 6 .
Table 4. Numerical examples for n = 6 .
pq 4 q 2 p 6 d h ( d ) pq 4 q 2 p 6 d h ( d )
35−629−6293637−533−53312
311−245−52 *313−53−536
53−15,589−15,5897257−15,429−15,42996
511−15,141−30912513−14,949−166148
517−14,469−14,46996519−14,181−14,18196
523−13,509−150124529−12,261−12,26172
531−11,781−130924537−10,149−10,149120
541−8901−98936543−8229−822948
547−6789−678972553−4389−438948
559−1701−214 *561−741−74124
73−117,613−117,61316875−117,549−13,061156
711−117,165−117,165240713−116,973−12,99760
717−116,493−116,493192719−116,205−116,205192
723−115,533−12,83772729−114,285−114,285240
731−113,805−140524737−112,173−112,173240
741−110,925−49312743−110,253−110,253288
747−108,813−108,813240753−106,413−106,413216
759−103,725−46130761−102,765−102,765192
767−99,693−11,07748771−97,485−97,485192
773−96,333−96,333192779−92,685−92,685288
783−90,093−90,093192789−85,965−85,965240
797−80,013−80,0131927101−76,845−76,845192
7103−75,213−8357727107−71,853−71,853144
7109−70,125−2805487113−66,573−739772
7127−53,133−53,1331207131−49,005−52 *
7137−42,573−42,5731207139−40,365−448548
7149−28,845−3205247151−26,445−26,44596
7157−19,053−2117367163−11,373−11,37372
7167−6093−67730
Table 5. Numerical examples for n = 7 .
Table 5. Numerical examples for n = 7 .
pq 4 q 2 p 7 d h ( d ) pq 4 q 2 p 7 d h ( d )
35−2087−208735311−1703−170328
313−1511−151149317−1031−103135
319−743−74321323−71−717
53−78,089−78,089280511−77,641−77,641112
513−77,449−77,449112517−76,969−76,969196
519−76,681−76,681140523−76,009−76,009224
529−74,761−74,761140531−74,281−74,281140
537−72,649−72,649168541−71,401−71,401140
543−70,729−70,729140547−69,289−69,289196
553−66,889−66,889112559−64,201−64,201112
561−63,241−63,241196567−60,169−60,169112
571−57,961−57,961112573−56,809−56,809112
579−53,161−53,161168583−50,569−50,569168
589−46,441−46,441140597−40,489−40,489140
5101−37,321−37,321845103−35,689−35,689112
5107−32,329−32,3291405109−30,601−30,601112
5113−27,049−27,049845127−13,609−13,60956
5131−9481−9481845137−3049−304928
5139−841−11 **
Table 6. Numerical examples for n = 8 .
Table 6. Numerical examples for n = 8 .
pq 4 q 2 p 8 d h ( d ) pq 4 q 2 p 8 d h ( d )
35−6461−64619637−6365−636564
311−6077−607748313−5885−588596
317−5405−540564319−5117−511764
323−4445−444564329−3197−319764
331−2717−271732337−1085−108532
Table 7. Numerical examples for n = 9 .
Table 7. Numerical examples for n = 9 .
pq 4 q 2 p 9 d h ( d ) pq 4 q 2 p 9 d h ( d )
35−19,583−19,5839937−19,487−19,487144
311−19,199−19,199162313−19,007−19,007108
317−18,527−18,527108319−18,239−18,239144
323−17,567−17,56790329−16,319−16,319153
331−15,839−15,839180337−14,207−14,20781
341−12,959−12,95999343−12,287−12,28790
347−10,847−10,84763353−8447−844799
359−5759−5759108361−4799−479963
367−1727−172736
Table 8. Numerical examples for n = 10 .
Table 8. Numerical examples for n = 10 .
pq 4 q 2 p 10 d h ( d ) pq 4 q 2 p 10 d h ( d )
37−58,853−58,853180311−58,565−58,565240
313−58,373−58,373240317−57,893−57,893280
323−56,933−19710329−55,685−55,685160
331−55,205−55,205240337−53,573−31710
341−52,325−209340343−51,653−51,653160
347−50,213−50,213120353−47,813−47,813260
359−45,125−52 ***361−44,165−36520
367−41,093−41,093240371−38,885−38,885160
373−37,733−37,733160379−34,085−34,085200
383−31,493−31,493120389−27,365−27,365120
397−21,413−437203101−18,245−18,245160
3103−16,613−16,6131003107−13,253−13,25380
3109−11,525−461303113−7973−797380
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Kim, K.-S. Some Remarks on the Divisibility of the Class Numbers of Imaginary Quadratic Fields. Mathematics 2022, 10, 2488. https://doi.org/10.3390/math10142488

AMA Style

Kim K-S. Some Remarks on the Divisibility of the Class Numbers of Imaginary Quadratic Fields. Mathematics. 2022; 10(14):2488. https://doi.org/10.3390/math10142488

Chicago/Turabian Style

Kim, Kwang-Seob. 2022. "Some Remarks on the Divisibility of the Class Numbers of Imaginary Quadratic Fields" Mathematics 10, no. 14: 2488. https://doi.org/10.3390/math10142488

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