Some Criteria for Class Numbers to Be Non-One

Abstract

Let be a positive integer which is not a perfect square and be any nonzero fixed integer. Then, the equation 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 .

1. Introduction

Let be a positive square-free integer, be a real quadratic field, and 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 , , 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 only for six values of and Chowla’s conjecture which says that only for six values of .

Mollin [4] considered and proved that if , where is a prime, then . In 2007, Byeon et al. proved (see [5]) Mollin’s conjecture which says that whenever . 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 of , where is a positive integer that is not a perfect square. We denote this expansion bywhere is the period length, (the floor of ), and is a palindrome.

The th convergent of for is given bywherewith . The complete quotients are given by , where , , and for ,and . We will also need the following fact:

Also, when is even, 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 integer is called a quadratic discriminant if . The quadratic discriminant is called a fundamental discriminant if it admits no factorization such that is a quadratic discriminant and In addition, is a fundamental discriminant if and only if either is square-free and or for some square-free integer such that .

Let be a quadratic discriminant, then the order of the quadratic discriminant isand if is a fundamental discriminant, we have , where is the ring of integers of The unit group of is given by (see [15], Theorem 5.2.1, p. 122).

Now, we investigate ideals. Let be a quadratic discriminant. An ideal is called -primitive if , for all . For a nonzero ideal of , we call the absolute norm of . If is a nonzero of , then (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. Let be a fundamental discriminant.(1)Suppose that , , where , and set(a) is an -primitive ideal of satisfying . In addition, the following assertions are equivalent:(1) is a principal ideal of (2)There exist such that (b)Let be such that . Then, .(2)Suppose that and , where , and set :(a) is an -primitive ideal of satisfying . In addition, the following assertions are equivalent:(1) is a principal ideal of (2)There exist such that (b)Let be such that . Then, .(3)Suppose that and , where , and set :(a) is an - primitive ideal of satisfying . In addition, the following assertions are equivalent:(1) is a principal ideal of (2)There exist such that (b)Let be such that . Then, .

Proof. (1)(a) By [17] (Theorem 7.1.5, p. 144), we find that is an ideal of satisfying . Let with which implies . Therefore, is an -primitive ideal of  (1)(2). Now, let be principal. Then, there exists , where . Since is an -primitive ideal, gcd . If , , then On the other hand, If , , then (2)(1). Let such that , which implies for some  If , now we obtain . Hence, , and equality holds, since On the other hand, If , which implies , then and for some Now, we obtain . Hence, , and equality holds, since(b) By (2)(1).(2)(a) By [17] (Theorem 7.1.5, p. 144), we find that is an ideal of satisfying . Let with which implies . Therefore, is an -primitive ideal of  (1)(2). Let now be principal. Then, there exists , where . On the other hand, , and therefore . (2)(1). Let such that , which implies and for some Now, we obtain . Hence, , and equality holds, since(b)   By (2) (1).(3)(a) We note is an -primitive ideal of satisfying  (1)(2). Let now be principal. Then, there exist such that . Since , is odd. But is even, so the equation implies that and have the same parity. Let , hence . Since is odd and is even, then is odd and is even. Let , , hence . Let , , then there exist such that . (2) (1). Let such that . Then, , which implies for some Now, we obtain . Hence, , and equality holds, since(b)By (2)(1).Now, we will be devoted to numerical examples illustrating Theorem 1.

Example 1. Let . Since is the solution of , then .

Example 2. Let . Since is the solution of , then .

Example 3. Let . Since is the solution of , then .

Example 4. Let . Since is the solution of , then .
We need the following proposition in order to prove that .

Proposition 1. Let be a quadratic discriminant, or , , where . Then, the equation has no integer solutions for all primes , where .

Proof. We will provide the complete proof for the case , and the other case can be handled along the similar lines. Suppose this equation has integer solutions , choose from these solutions with , , and is the smallest. Let and . Then,Hence,By the definition of , we findThen, we have or . Thus, or .  Case (i): .  Case (ii): In both cases, we find a contradiction.

Theorem 2. Let be a fundamental discriminant, or , , where and is not a prime number, then

Proof. Let be a prime number and ; hence, , where is square-free integer, which implies for .
From Theorem 1 and Proposition 1, we obtain  Case (i): is not a principal ideal of if   Case (ii): is not a principal ideal of if In both cases, we find that .
The following proposition gives us a condition for solving the equation ,

Proposition 2. Let , where , be a positive integer that is not a perfect square and suppose that is square-free and , . If the equation is solvable, then .

Proof. By [16] (Proposition 3.1), it follows that is even and . We put in (5). We obtainHence, . We getSo, This completes the proof.

Example 5. Let sinceHence, , and the equation has no integer solutions.

Example 6. Let sinceHence, , and the equation has no integer solutions.
We use Theorem 1 and Proposition 2 in order to prove the following theorem, which gives a sufficient condition for in two cases.

Theorem 3. Let be a fundamental discriminant and suppose that if , and ; where and , . Then,(a)If and then (b)If then

Proof. (a)From Proposition 2, the equation is unsolvable. Since , then the equation is unsolvable. By Theorem 1, it follows that is not a principal ideal; then, .(b)From Proposition 2, the equation is unsolvable. Since , then the equation is unsolvable. By Theorem 1, it follows that is not a principal ideal; then,

Example 7. Let , , , and . Then, .

Example 8. Let , and . Then, .

Data Availability

No data were used to support this study.

Conflicts of Interest

The authors declare that they have no conflicts of interest.

Acknowledgments

This work was supported by Tishreen University.