The upper bound estimate of the number of integer points on elliptic curves $y^2=x^3+p^{2r}x$

Abstract

Let p be a fixed prime and r be a fixed positive integer. Further let $N(p^{2r})$ denote the number of pairs of integer points $(x,\pm y)$ on the elliptic curve $E:y^2=x^3+p^{2r}x$ with $y>0$. Using some properties of Diophantine equations, we give a sharper upper bound estimate for $N(p^{2r})$. That is, we prove that $N(p^{2r})\le 1$, except with $N({17}^{2(2s+1)})=2$, where s is a nonnegative integer.

MSC:11G05, 11Y50.

1 Introduction

Let ℤ, ℕ be the sets of all integers and positive integers, respectively. Let p be a fixed prime and k be a fixed positive integer. Recently, the integer points $(x,y)$ on the elliptic curve

$$y^2=x^3+p^{k}x$$

(1.1)

have been investigated in many papers (see [1, 2] and [3]). In this paper we deal with the number of integer points on (1.1) for even k. Then (1.1) can be rewritten as

$$y^2=x^3+p^{2r}x,$$

(1.2)

where r is a positive integer.

An integer point $(x,y)$ on (1.2) is called trivial or non-trivial according to whether $y=0$ or not. Obviously, (1.2) has only the trivial integer point $(x,y)=(0,0)$. Notice that if $(x,y)$ is a non-trivial integer point on (1.2), then $(x,-y)$ is also. Therefore, $(x,y)$ along with $(x,-y)$ are called by a pair of non-trivial integer points and denoted by $(x,\pm y)$, where $y>0$. For any positive integer n, let

$$u(n)=\frac{1}{2}({\alpha }^n+{\beta }^n),v(n)=\frac{1}{2\sqrt{2}}({\alpha }^n-{\beta }^n),$$

(1.3)

where

$$\alpha =3+2\sqrt{2},\beta =3-2\sqrt{2}.$$

(1.4)

Using some properties of Diophantine equations, we give a sharper upper bound estimate for $N(p^{2r})$, the number of pairs of non-trivial integer points $(x,\pm y)$ on (1.2). That is, we shall prove the following results.

Theorem 1.1 All non-trivial integer points on (1.2) are given as follows.

• (i) $p=u(2^m)$, $r=2s+1$, $(x,\pm y)=(p^{2s}{v}^2(2^m),\pm p^{3s}v(2^m)(v^2(2^m)+1))$, where m, s are nonnegative integers.

• (ii) $p\equiv 1(mod8)$, $r=2s+1$, $(x,\pm y)=(p^{2s+1}{X}^2,\pm p^{3s+2}XY)$, where s is a nonnegative integer, $(X,Y)$ is a solution of the equation

  $$X^4-p{Y}^2=1,X,Y\in \mathbb{N}.$$

  (1.5)

Theorem 1.2 Let p be an odd prime, r be a positive integer. Then for any nonnegative integer s, we have $N(p^{2r})\le 1$, except with $N({17}^{2(2s+1)})=2$. Moreover, if $p\not\equiv 1(mod8)$, then $N(p^{2r})=0$, except with $N(3^{2(2s+1)})=1$.

2 Lemmas

Lemma 2.1 ([[4], Theorem 244])

Every solution $(u,v)$ of the equation

$$u^2-2v^2=1,u,v\in \mathbb{N}$$

(2.1)

can be expressed as $(u,v)=(u(n),v(n))$, where n is a positive integer.

Lemma 2.2 If $p=u(n)$, then $n=2^m$, where m is a nonnegative integer.

Proof Assume that n has an odd divisor d with $d>1$. Then we have either $u(1)|u(n)$ and $1<u(1)<u(n)$ or $u(n/d)|u(n)$ and $1<u(n/d)<u(n)$. Therefore, since p is a prime, it is impossible. Thus, we get $n=2^m$. The lemma is proved. □

Any fixed positive integer a can be uniquely expressed as $a=b{c}^2$, where b, c are positive integers with b is square free. Then b is called the quadratfrei of a and denoted by $Q(a)$.

Lemma 2.3 For any positive integer m, we have $3|Q(v(2^m))$.

Proof By (1.3) and (1.4), we get

$$v\left(2^m\right)=2^{m+1}\prod _{i=0}^{m-1}u\left(2^i\right)$$

(2.2)

and

$$u\left(2^i\right)=2u^2\left(2^{i-1}\right)-1,i\in \mathbb{N}.$$

(2.3)

Since $(2/3)=-1$, where $(2/3)$ is the Legendre symbol, we see from (2.3) that $3\nmid u(2^i)$ for $i\ge 1$. Therefore, since $u(1)=3$, by (2.2), we obtain $3\parallel v(2^m)$. It implies that $3|Q(v(2^m))$. The lemma is proved. □

Let D be a non-square positive integer. It is a well known fact that if the equation

$$U^2-D{V}^2=-1,U,V\in \mathbb{N}$$

(2.4)

has solutions $(U,V)$, then it has a unique solution $(U_1,V_1)$ such that $U_1+V_1\sqrt{D}\le U+V\sqrt{D}$, where $(U,V)$ through all solutions of (2.4). For any odd positive integer l, let

$$U(l)+V(l)\sqrt{D}={(U_1+V_1\sqrt{D})}^l.$$

Then $(U,V)=(U(l),V(l))$ ($l=1,3,\dots$) are all solutions of (2.4).

Lemma 2.4 ([[5], Theorem 1])

The equation

$$X^4-D{Y}^2=-1,X,Y\in \mathbb{N}$$

(2.5)

has at most one solution $(X,Y)$. Moreover, if the solution $(X,Y)$ exists, then $(X^2,Y)=(U(l),V(l))$, where $l=Q(U_1)$.

Lemma 2.5 ([[5], Theorem 3])

If $3|Q(U_1)$, then (2.5) has no solutions $(X,Y)$.

Lemma 2.6 If $p=u(2^m)$, where m is a positive integer with $m>1$, then (1.5) has no solutions $(X,Y)$.

Proof Since $p=u(2^m)$ with $m>1$, by (2.3), we have

$$p=2u^2\left(2^{m-1}\right)-1=4v^2\left(2^{m-1}\right)+1.$$

(2.6)

We see from (2.6) that the equation

$$U^2-p{V}^2=-1,U,V\in \mathbb{N}$$

(2.7)

has solution $(U,V)$ and its fundamental solution is $(U_1,V_1)=(2v(2^{m-1}),1)$. Further, since $m-1\ge 1$, by Lemma 2.3, we have $3|Q(v(2^{m-1}))$. Hence, we get $3|Q(U_1)=Q(2v(2^{m-1}))$. Therefore, by Lemma 2.5, the lemma is proved. □

Lemma 2.7 ([6])

The equation

$$2X^2+1=Y^n,X,Y,n\in \mathbb{N},n>3$$

(2.8)

has no solutions $(X,Y,n)$.

Lemma 2.8 The equation

$$X^2-Y^4=p^{2n},X,Y,n\in \mathbb{N},gcd(X,Y)=1$$

(2.9)

has only the solutions $(p,X,Y,n)=(u(2^m),v^2(2^m)+1,v(2^m),1)$, where m is a nonnegative integer.

Proof Assume that $(p,X,Y,n)$ is a solution of (2.9). If $p=2$, since $gcd(X,Y)=1$, then we have $2\nmid XY$, $gcd(X+Y^2,X-Y^2)=2$, $X+Y^2=2^{2n-1}$, $X-Y^2=2$ and $Y^2=2^{2n-2}-1$. But since $Y^2+1$ is not a square, it is impossible.

If p is an odd prime, then we have $gcd(X+Y^2,X-Y^2)=1$, and by (2.9), we get $X+Y^2=p^{2n}$, $X-Y^2=1$,

$$2X=p^{2n}+1$$

(2.10)

and

$$2Y^2=p^{2n}-1.$$

(2.11)

By Lemma 2.7, we get from (2.11) that $n=1$ and

$$p^2-2Y^2=1.$$

(2.12)

Further, applying Lemma 2.1 to (2.12) yields

$$p=u(n),Y=v(n),n\in \mathbb{N}.$$

(2.13)

Further, by Lemma 2.2, we see from the first equality of (2.13) that $n=2^m$. Thus, by (2.10) and (2.13), the lemma is proved. □

3 Proof of Theorem 1.1

Assume that $(x,\pm y)$ is a pair of non-trivial integer points on (1.2). Since $y>0$, we have $x>0$ and x can be expressed as

$$x=p^{t}z,t\in \mathbb{Z},t\ge 0,z\in \mathbb{N},p\nmid z.$$

(3.1)

Substituting (3.1) into (1.2) yields

$$p^{t}z(p^{2t}{z}^2+p^{2r})=y^2.$$

(3.2)

We first consider the case that $r>t$. By (3.2), we have

$$p^{3t}z(z^2+p^{2r-2t})=y^2.$$

(3.3)

Since $p\nmid z$, we have $p\nmid z^2+p^{2r-2t}$ and $gcd(z,z^2+p^{2k-2r})=1$. Hence by (3.3), we get

$$\begin{array}{c}t=2s,z=f^2,z^2+p^{2r-2t}=g^2,y=p^{3s}fg,\hfill \\ s\in \mathbb{Z},s\ge 0,f,g\in \mathbb{N},gcd(f,g)=1,\hfill \end{array}$$

(3.4)

whence we obtain

$$g^2-f^4=p^{2r-4s}.$$

(3.5)

Applying Lemma 2.8 to (3.5) yields

$$p=u\left(2^m\right),2r-4s=2,f=v\left(2^m\right),g=v^2\left(2^m\right)+1,m\in \mathbb{Z},m\ge 0.$$

(3.6)

Therefore, by (3.1), (3.4), and (3.6), the integer points of type (i) are given.

We next consider the case that $r=t$. Then we have

$$p^{3r}z(z^2+1)=y^2.$$

(3.7)

Since $p\nmid z$, $gcd(z,z^2+1)=1$ and $z^2+1$ is not a square, we see from (3.7) that

$$\begin{array}{c}r=2s+1,z=f^2,z^2+1=p{g}^2,y=p^{3s+2}fg,\hfill \\ s\in \mathbb{Z},s\ge 0,f,g\in \mathbb{N},gcd(f,g)=1.\hfill \end{array}$$

(3.8)

By (3.8), we get

$$f^4-p{g}^2=-1.$$

(3.9)

It implies that $(X,Y)=(f,g)$ is a solution of (1.5). Therefore, by (3.1) and (3.8), we obtain the integer points of type (ii).

We finally consider the case that $r<t$. Then we have

$$p^{t+2r}z(p^{2t-2r}{z}^2+1)=y^2.$$

(3.10)

Since $p\nmid z(p^{2t-2r}{z}^2+1)$ and $gcd(z,p^{2t-2r}{z}^2+1)=1$, we see from (3.10) that $p^{2t-2r}{z}^2+1$ is a square, a contradiction.

To sum up, the theorem is proved.

4 Proof of Theorem 1.2

By (2.3), if $p=u(2^m)$ with $m\ge 1$, then $p\equiv 1(mod8)$. Therefore, by Theorem 1.1, if $p\not\equiv 1(mod8)$, then (1.2) has only the non-trivial integer point

$$p=3,r=2s+1,(x,\pm y)=(3^{2s}\cdot 4,\pm 3^{3s}\cdot 10).$$

(4.1)

It implies that the theorem is true for $p\not\equiv 1(mod8)$.

For $p\equiv 1(mod8)$, let $N_1$ and $N_2$ denote the number of pairs of non-trivial integer points of types (i) and (ii) in Theorem 1.1, respectively. Obviously, we have

$$N\left(p^{2r}\right)=N_1+N_2$$

(4.2)

and $N_1\le 1$. By Lemma 2.4, we get $N_2\le 1$. Hence, by (4.2), we have $N(p^{2r})\le 2$ for $p\equiv 1(mod8)$. Since $u(2)=17$ and (1.5) has the solution $(X,Y)=(2,1)$ for $p=17$, by Theorem 1.1, we get

$$\begin{array}{c}p=17,r=2s+1,(x,\pm y)=({17}^{2s}\cdot 144,\pm {17}^{3s}\cdot 1\text{,}740)\text{and}\hfill \\ ({17}^{2s+1}\cdot 4,\pm {17}^{3s+2}\cdot 2)\hfill \end{array}$$

(4.3)

and $N({17}^{2(2s+1)})=2$. However, by Lemma 2.6, if $p=u(2^m)$ with $m>1$, then $N_2=0$. Therefore, by (4.2), if $p\equiv 1(mod8)$, then $N(p^{2r})\le 1$, except with $N({17}^{2(2s+1)})=2$. The theorem is proved.