Integral points on the elliptic curve $y^2=x^3+27x-62$

Abstract

We give a new proof that the elliptic curve $y^2=x^3+27x-62$ has only the integral points $(x,y)=(2,0)$ and $(x,y)=(28\text{,}844\text{,}402,\pm 15\text{,}491\text{,}585\text{,}540)$ using elementary number theory methods and some properties of generalized Fibonacci and Lucas sequences.

MSC:11B25, 11B37.

1 Introduction

Let P and Q be non-zero integers with $P^2+4Q\ne 0$. The generalized Fibonacci sequence $(U_n(P,Q))$ and the Lucas sequence $(V_n(P,Q))$ are defined by the following recurrence relations:

$$U_0(P,Q)=0,U_1(P,Q)=1,U_{n+2}(P,Q)=P{U}_{n+1}(P,Q)+Q{U}_n(P,Q)\text{for }n\ge 0$$

and

$$V_0(P,Q)=2,V_1(P,Q)=P,V_{n+2}(P,Q)=P{V}_{n+1}(P,Q)+Q{V}_n(P,Q)\text{for }n\ge 0.$$

$U_n(P,Q)$ is called the n th generalized Fibonacci number and $V_n(P,Q)$ is called the n th generalized Lucas number. Also, generalized Fibonacci and Lucas numbers for negative subscripts are defined as

$$U_{-n}(P,Q)=\frac{-U_n(P,Q)}{{(-Q)}^n}\text{and}{V}_{-n}=\frac{V_n(P,Q)}{{(-Q)}^n}\text{for }n\ge 1,$$

(1.1)

respectively. Taking $\alpha =(P+\sqrt{P^2+4Q})/2$ and $\beta =(P-\sqrt{P^2+4Q})/2$ to be the roots of the characteristic equation $x^2-Px-Q=0$, we have the well-known expressions named Binet’s formulas

$$U_n(P,Q)=({\alpha }^n-{\beta }^n)/(\alpha -\beta )\text{and}{V}_n(P,Q)={\alpha }^n+{\beta }^n$$

(1.2)

for all $n\in \mathbb{Z}$. Instead of $U_n(P,Q)$ and $V_n(P,Q)$, we use $U_n$ and $V_n$, respectively. For $P=Q=1$, the sequence $(U_n)$ is the familiar Fibonacci sequence $(F_n)$ and the sequence $(V_n)$ is the familiar Lucas sequence $(L_n)$. If $P=2$ and $Q=1$, then we have the well-known Pell sequence $(P_n)$ and Pell-Lucas sequence $(Q_n)$. For $Q=-1$, we represent $(U_n)$ and $(V_n)$ by $(u_n)$ and $(v_n)$, respectively. Thus $u_0=0$, $u_1=P$ and $u_{n+1}=P{u}_n-u_{n-1}$ and $v_0=2$, $v_1=P$ and $v_{n+1}=P{v}_n-v_{n-1}$ for all $n⩾1$. Also, it is seen from Eq. (1.1) that

$$u_{-n}=-u_n(P,-1)\text{and}{v}_{-n}=v_n(P,-1)$$

for all $n⩾1$. For more information about generalized Fibonacci and Lucas sequences, one can consult [1–5].

There has been much interest in determining the problem of the integral points on elliptic curves, and many advanced methods have been developed to solve such problems (see [6, 7] and [8]). In 1987, Don Zagier [9] proposed that the largest integral point on the elliptic curve

$$y^2=x^3+27x-62$$

(1.3)

is $(x,y)=(28\text{,}844\text{,}402,\pm 154\text{,}914\text{,}585\text{,}540)$. Then the same problem was dealt with by some authors. In [10], Zhu and Chen found all integral points on (1.3) by using algebraic number theory and p-adic analysis. In [11], Wu proved that (1.3) has only the integral points $(x,y)=(2,0)$ and $(28\text{,}844\text{,}402,\pm 154\text{,}914\text{,}585\text{,}540)$ using some results of quartic Diophantine equations with elementary number methods. After that, in [12], the authors found the integral points on (1.3) using similar methods to those given in [11]. In this paper, we determine that the largest integral point on the elliptic curve $y^2=x^3+27x-62$ is $(x,y)=(28\text{,}844\text{,}402,\pm 154\text{,}914\text{,}585\text{,}540)$ by using elementary number theory methods and some properties of generalized Fibonacci and Lucas sequences. Our proof is extremely different from the proofs of the others.

2 Preliminaries

In this section, we present two theorems and some well-known identities regarding the sequences $(u_n)$ and $(v_n)$, which will be useful during the proof of the main theorem.

We state the following theorem from [13].

Theorem 2.1 Let $P>2$. If $u_n=c{x}^2$ with $c\in \{1,2,3,6\}$ and $n>3$, then $(n,P,c)=(4,338,1)$ or $(6,3,1)$.

The following theorem is a well-known theorem (see [14]).

Theorem 2.2 Let $m\ge 1$ and $n\ge 1$. Then $(u_m,u_n)=u_{(m,n)}$.

The well-known identities for $(u_n)$ and $(v_n)$ are as follows:

$$u_{2n}=u_n{v}_n,$$

(2.1)

$$v_n=u_{n+1}-u_{n-1},$$

(2.2)

$$u_{2k+1}-1=u_k{v}_{k+1}.$$

(2.3)

Moreover, if P is even, then

$$u_n\text{ is even}\iff n\text{ is even},$$

(2.4)

$$u_n\text{ is odd}\iff n\text{ is odd}.$$

(2.5)

3 Proof of the main theorem

The main theorem we deal with here is as follows.

Theorem 3.1 The elliptic curve $y^2=x^3+27x-62$ has only the integral points $(x,y)=(2,0)$ and $(28\text{,}844\text{,}402,\pm 154\text{,}914\text{,}585\text{,}540)$.

Proof Assume that $(x,y)$ is an integral point on the elliptic curve $y^2=x^3+27x-62$. It can be easily seen that $x>0$. On the other hand, obviously, the elliptic curve $y^2=x^3+27x-62$ has only the integral point $(x,y)=(2,0)$ with $y=0$. Hence, we may assume that $y\ne 0$. Let $k=x-2$. Substituting this value of k into $y^2=x^3+27x-62$, we get

$$y^2=k(k^2+6k+39).$$

(3.1)

Since $y\ne 0$, it is obvious that $y^2>0$. On the other hand, since $k^2+6k+39={(k+3)}^2+30>0$, we conclude that $k>0$. Clearly, $d=(k,k^2+6k+39)=1,3,13\text{ or }39$. So, we get from (3.1) that

$$k=d{a}^2,k^2+6k+39=d{b}^2,y=\pm dab$$

(3.2)

for some positive integers a and b.

If $d=1$, then from (3.2) we get $a^4+6a^2+39=b^2$. Completing the square gives ${(a^2+3)}^2+30=b^2$. This implies that $[b-(a^2+3)][b+(a^2+3)]=30$. It can be easily shown that there are no integers a and b satisfying the previous equation.

If $d=3$, then from (3.2) we obtain $9a^4+18a^2+39=3b^2$. Completing the square gives

$$b^2-3{(a^2+1)}^2=10.$$

(3.3)

Working on modulo 8 shows that (3.3) is impossible.

If $d=13$, then from (3.2) we immediately have $169a^4+78a^2+39=13b^2$. Completing the square gives

$${(13a^2+3)}^2-13b^2=-30.$$

(3.4)

Working on modulo 8 shows that (3.4) is impossible.

Lastly, we consider (1.3) for the case when $d=39$. If $d=39$, then from (3.2) we get $k=39a^2$ and $k^2+6k+39=39b^2$. Substituting $k=39a^2$ into $k^2+6k+39=39b^2$ and completing the square give

$${(39a^2+3)}^2+30=39b^2.$$

(3.5)

This equation is of the form

$$u^2-39v^2=-30.$$

(3.6)

Let $x_n+y_n\sqrt{39}$ be a solution of the equation $x^2-39y^2=1$. Since the fundamental solution of this equation is $\alpha =25+4\sqrt{39}$, we get $x_n+y_n\sqrt{39}={\alpha }^n$, and therefore $x_n=({\alpha }^n+{\beta }^n)/2$ and $y_n=({\alpha }^n-{\beta }^n)/2\sqrt{39}$, where $\beta =25-4\sqrt{39}$. It can be easily seen that $x_n=v_n(50,-1)/2$ and $y_n=4u_n(50,-1)$. Equation (3.6) has exactly two solution classes and the fundamental solutions are $3+\sqrt{39}$ and $3-\sqrt{39}$. So, the general solution of (3.6) is given by

$$a_n+b_n\sqrt{39}=(3-\sqrt{39})(x_n+y_n\sqrt{39}),$$

(3.7)

$$a_n+b_n\sqrt{39}=(3+\sqrt{39})(x_n+y_n\sqrt{39}),$$

(3.8)

with $n\ge 1$, respectively [15]. Considering first Eq. (3.7), we readily obtain $a_n=3x_n-39y_n$. Since $x_n=v_n/2$ and $y_n=4u_n$, it follows that

$$a_n=(3v_n-312u_n)/2.$$

From (2.2), if we write $u_{n+1}-u_{n-1}$ instead of $v_n$ and rearrange the above equation, then we get $a_n=-81u_n-3u_{n-1}$. This means that $39a^2+3=-81u_n-3u_{n-1}$ by (3.5). Dividing both sides of the equation by 3 gives $13a^2+1=-27u_n-u_{n-1}$. However, this is impossible for $13a^2+1>0$ and $n\ge 1$. Another possibility is that $-39a^2-3=-81u_n-3u_{n-1}$, implying that

$$13a^2+1=27u_n+u_{n-1}.$$

(3.9)

It can be shown by the induction method that

$$u_n\equiv \{\begin{array}{ll}-n(mod13)& \text{if }n\text{ is even},\\ n(mod13)& \text{if }n\text{ is odd}\end{array}$$

(3.10)

and

$$u_n\equiv n(mod8).$$

(3.11)

So, working on modulo 8 and using (3.11) in Eq. (3.9) lead to a contradiction.

Now, we consider Eq. (3.8). Then we immediately have $a_n=3x_n+39y_n$. Since $x_n=v_n/2$ and $y_n=4u_n$, it follows that $a_n=(3v_n+312u_n)/2$. In view of (2.2), we readily obtain $a_n=3u_{n+1}+81u_n$. By (3.5), we get $39a^2+3=3u_{n+1}+81u_n$, implying that

$$13a^2+1=u_{n+1}+27u_n.$$

(3.12)

Assume that n is odd. By using (3.10), we get

$$u_{n+1}+27u_n\equiv -n-1+27n\equiv -1(mod13),$$

a contradiction by (3.12). So, n is even. Now, let us assume that a is odd in Eq. (3.12). Then using (3.11) gives

$$u_{n+1}+27u_n\equiv n+1+3n\equiv 4n+1\equiv 6(mod8),$$

i.e.,

$$4n\equiv 5(mod8),$$

which is impossible. So, a is even, and therefore $a=2m$ for some positive integer m. Substituting $a=2m$ into (3.12), we get

$$52m^2+1=u_{n+1}+27u_n.$$

(3.13)

In the above equation, if m is odd, then from (3.11) we get

$$u_{n+1}+27u_n\equiv n+1+3n\equiv 4n+1\equiv 5(mod8),$$

which implies that

$$n\equiv 1(mod2).$$

But this is impossible since n is even. As a consequence, m is even and therefore we conclude that $4|a$. We now return to (3.12). Since n is even, $n=2r$ for some $r>0$. Then (3.12) becomes

$$13a^2=u_{2r+1}-1+27u_{2r}.$$

By (2.3) and (2.1), it can be seen that $u_{2r+1}-1+27u_{2r}=u_r{v}_{r+1}+27u_r{v}_r=u_r(v_{r+1}+27v_r)$ and therefore

$$13a^2=u_r(v_{r+1}+27v_r).$$

By using (2.2), we get $13a^2=u_r(u_{r+2}-u_r+27u_{r+1}-27u_{r-1})$. In view of the recurrence relation of the sequence $u_r$, we immediately have

$$13a^2=u_r(3\text{,}848u_r-104u_{r-1}).$$

Dividing both sides of the above equation by 13 and rearranging the equation gives

$$a^2=8u_r(37u_r-u_{r-1}).$$

Since $4|a$, it follows that

$$2{(a/4)}^2=u_r(37u_r-u_{r-1}).$$

By Theorem 2.2, since $(u_r,u_{r-1})=1$, clearly, $(u_r,37u_r-u_{r-1})=1$. This implies that either

$$37u_r-u_{r-1}=2c^2$$

(3.14)

or

$$u_r=2c^2$$

(3.15)

for some positive integer c, where $u_r=u_r(50,-1)$. By (2.4) and (2.5), it can be seen that $37u_r-u_{r-1}$ is always odd. Therefore (3.14) is impossible. By Theorem 2.1, (3.15) is impossible for the case when $r>3$. Hence, we have $r\le 3$. On the other hand, since $u_r=2c^2$ is even, from (2.4), it follows that r is even. Since r is even and $n=2r$, we get $n=4$. Substituting this value of n into (3.12), we obtain

$$13a^2+1=u_5+27u_4.$$

Since $u_5=6\text{,}242\text{,}501$ and $u_4=124\text{,}900$, a simple computation shows that $a=860$. Moreover, since $k=39a^2$ and $x=k+2$, we get $k=28\text{,}844\text{,}400$ and therefore $x=28\text{,}844\text{,}402$. Substituting $x=28\text{,}844\text{,}402$ into $y^2=x^3+27x-62$ gives $y=\pm 15\text{,}491\text{,}585\text{,}540$. Hence, the theorem is proved, the elliptic curve $y^2=x^3+27x-62$ has only the integral points $(x,y)=(2,0)$ and $(x,y)=(28\text{,}844\text{,}402,\pm 15\text{,}491\text{,}585\text{,}540)$, which is the largest integral point on it. This completes the proof of the main theorem. □