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

- Olcay Karaatlı
^{1}Email author and - Refik Keskin
^{1}

**2013**:221

https://doi.org/10.1186/1029-242X-2013-221

© Karaatlıand Keskin; licensee Springer 2013

**Received: **9 January 2013

**Accepted: **6 March 2013

**Published: **2 May 2013

## 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.

## Keywords

## 1 Introduction

*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:

*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

for all $n\u2a7e1$. For more information about generalized Fibonacci and Lucas sequences, one can consult [1–5].

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* $({u}_{m},{u}_{n})={u}_{(m,n)}$.

*P*is even, then

## 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

for some positive integers *a* and *b*.

If $d=1$, then from (3.2) we get ${a}^{4}+6{a}^{2}+39={b}^{2}$. 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.

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

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

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

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

*n*is even. Now, let us assume that

*a*is odd in Eq. (3.12). Then using (3.11) gives

*i.e.*,

*a*is even, and therefore $a=2m$ for some positive integer

*m*. Substituting $a=2m$ into (3.12), we get

*m*is odd, then from (3.11) we get

*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

*c*, where ${u}_{r}={u}_{r}(50,-1)$. By (2.4) and (2.5), it can be seen that $37{u}_{r}-{u}_{r-1}$ 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 ${u}_{r}=2{c}^{2}$ 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

Since ${u}_{5}=6\text{,}242\text{,}501$ and ${u}_{4}=124\text{,}900$, a simple computation shows that $a=860$. Moreover, since $k=39{a}^{2}$ 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. □

## Declarations

### Acknowledgements

Dedicated to Professor Hari M Srivastava.

## Authors’ Affiliations

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