# Principal Primitive Ideals in Quadratic Orders and Pell’s Equations

Authors: Ahmad Issa; Hasan Sankari
Journal: Journal of Mathematics
 (2021)

---

## Abstract

In this paper, we introduce a method of determining whether the primitive ideal is principal in a real quadratic order, depending on the solvability of Pell’s equation.


---

## Body

## 1. Introduction

Letd>1 be square-free integer and n a nonzero integer. Then, the equation x2−dy2=n is known as Generalized Pell’s equation. This equation has been studied by many authors (cf. [2, 4, 6–8]). In [1, 3, 5], this equation has been applied to study the class numbers of real cyclotomic fields.In this work, we give necessary and sufficient condition to solvex2−dy2=n, depending on the ideals in orders of the real quadratic fields. We also introduce the formula of the opposite ideal for every solution of this equation, in special cases of d and n.


## 2. Preliminaries

A nonsquare integerΔ∈Z is called a quadratic discriminant if Δ≡0or1mod4, we set ωΔ=σΔ+Δ/2, where σΔ=0,ifΔ≡0mod41,ifΔ≡1mod4 and(1)OΔ=1,ωΔ=ℤωΔ=u+vΔ2|u,v∈ℤ,u≡vΔmod2.We callωΔ the basic number and OΔ the order of the quadratic discriminant Δ.Now, we investigate ideals. LetΔ be a quadratic discriminant. An ideal Ι⊂OΔis called OΔ-primitive if e−1Ι⊈OΔ for all e∈ℕ≥2. For a nonzero ideal Ι of OΔ, we call NΙ=OΔ:Ι the absolute norm of Ι. If α is a nonzero of OΔ, then Nα=Nα.


## 3. Main Results

In this section, we will prove our main results. As a start, we record the following theorem.Theorem 1.
LetΔ=4d∈ℕ be a quadratic discriminant. Suppose that c=1ifd≡3mod4,andc=2ifd≡0mod8,d=c2ab, where a,b∈ℕ, and a is odd and square-free. Then,(1)
Ι=2ca,ca+d is an OΔ-primitive ideal satisfying NΙ=2ca. In addition, the following assertions are equivalent:(a)
Ι is a principal ideal of OΔ.(b)
There existx,y∈Z such that x2−dy2=2ca and gcd c,x/cy=1.(2)
Letx,y∈Z be such that x2−dy2=2ca and gcd c,x/cy=1. Then,(2)Ι=x+yd.Proof (1).
Ι is an ideal of OΔ satisfying NΙ=2ca. Let e−1Ι⊂OΔ with e∈ℕ which implies e=1. Therefore, Ι is an OΔ-primitive ideal.
(a)⟹ (b) Let Ι be principal. Then, there exists α=x1+y1Δ/2∈OΔ, where Ι=α=αOΔ. We set u=x1/2,v=y1. Then, Ι=u+vd. Since I is an OΔ-primitive ideal, gcd u,v=1.
Ifc=1, I=2a,a+d=u+vd, then(3)2a=NI=Nu+vd=u2−dv2.
On the other hand, ifc=2, I=4a,2a+d=u+vd, since α∈I, there existm,Y∈Z such that(4)u+vd=4am+2a+dY=2a2m+Y+Yd.
IfX=2m+Y, then u=2aX and v=Y, so I=2aX+YdOΔ and gcd 2,aXY=1. On the other hand, 4a=NI=N2aX+Yd=4a2X2−dY2. Let x=2aXandy=Y; hence, x2−dy2=4aand2,x/2y=1.(5)b⟹a.
Ifc=1, let x,y∈Z such that x2−dy2=2a, which implies x=ax1, for some x1∈Z and gcd 2,xy=1; then, gcd 2,x1=1 and x1−y=2w, for some w∈Z. Now, we obtain x+yd=2aw+ya+d∈I; hence, x+yd⊂I, and equality holds since(6)Nx+yd=Nx+yd=x2−dy2=2a=NI.
Ifc=2, let x,y∈Z such that x2−dy2=4a and 2,x/2y=1; then, x/22−aby2=a, which implies x/2=ax1, for some x1∈Z; then, x1 is odd and x1−y=2w, for somew∈Z Now, we obtain x+yd=4aw+y2a+d∈I. Hence, x+yd⊂I, and equality holds since(7)Nx+yd=Nx+yd=4a=NI.(2) By (b) ⟹ (a). Now, we will be devoted to numerical examples illustrating Theorem1.Example 1.
LetΔ=364=4dandd=91=7×13=ab≡3mod4. Since 105,11 is the solution of x2−91y2=14; hence, 14,7+d=105+11d.Example 2.
LetΔ=1248=4dandd=312=4×3×26=4ab≡0mod8. Since 18,1 is the solution of x2−312y2=12; hence, 12,6+d=18+d.Theorem 2.
LetΔ=4d∈ℕ be a quadratic discriminant. Suppose that d=4ab, where a,b∈ℕ, and a is square-free. Then,(1)
I=a,d is an OΔ-primitive ideal satisfying NΙ=a. In addition, the following assertions are equivalent:(a)
Ι is a principal ideal of OΔ.(b)
There existx,y∈Z such that x2−dy2=4a and gcd x,y=2.(2)
Letx,y∈Z be such that x2−dy2=4a and gcd x,y=2.Then,(8)Ι=x2+y2d.Proof (1).
Ι is an ideal of OΔ satisfying NΙ=a. Let e−1Ι⊂OΔ with e∈ℕ which implies e=1. Therefore, Ι is OΔ-primitive.
(a)⟹ (b) Let Ι be principal. Then, there exists α=x1+y1Δ/2∈OΔ, where Ι=α=αOΔ. We set u=x1/2,v=y1. Then, Ι=u+vd. Since I is an OΔ-primitive ideal, gcd u,v=1.
On the other hand,a=NΙ=Nu+vd=u2−dv2. Let x=2u and y=2v; then, x2−dy2=4a and gcd x,y=2.
(2) (b)⟹ (a) Let x,y∈Z such that x2−dy2=4a and gcd x,y=2. Then, x/22−aby2=a, which implies x/2=ax1, for some x1∈Z. x/2+y/2d=ax1+y/2d∈Ι. Hence, x/2+y/2dOΔ⊂Ι, and equality holds since(9)Nx2+y2dOΔ=Nx2+y2d=x2−dy24=a=NΙ.(2) By (b) ⟹ (a). Now, we will be devoted to numerical examples illustrating Theorem2.Example 3.
LetΔ=1488=4dandd=372=4×3×31=4ab. Since 270,14 is the solution of x2−372y2=−12; hence, 3,d=135+7d.



---

*Source: 3124437-2021-11-08.xml*