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

*Journal of Inequalities and Applications*
**volume 2013**, Article number: 221 (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.

## 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:

and

{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

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

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\u2a7e1. Also, it is seen from Eq. (1.1) that

for all n\u2a7e1. 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

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:

Moreover, if *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

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

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.

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

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

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

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=39{a}^{2} and {k}^{2}+6k+39=39{b}^{2}. Substituting k=39{a}^{2} into {k}^{2}+6k+39=39{b}^{2} and completing the square give

This equation is of the form

Let {x}_{n}+{y}_{n}\sqrt{39} be a solution of the equation {x}^{2}-39{y}^{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}=4{u}_{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

with n\ge 1, respectively [15]. Considering first Eq. (3.7), we readily obtain {a}_{n}=3{x}_{n}-39{y}_{n}. Since {x}_{n}={v}_{n}/2 and {y}_{n}=4{u}_{n}, it follows that

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}=-81{u}_{n}-3{u}_{n-1}. This means that 39{a}^{2}+3=-81{u}_{n}-3{u}_{n-1} by (3.5). Dividing both sides of the equation by 3 gives 13{a}^{2}+1=-27{u}_{n}-{u}_{n-1}. However, this is impossible for 13{a}^{2}+1>0 and n\ge 1. Another possibility is that -39{a}^{2}-3=-81{u}_{n}-3{u}_{n-1}, implying that

It can be shown by the induction method that

and

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}=3{x}_{n}+39{y}_{n}. Since {x}_{n}={v}_{n}/2 and {y}_{n}=4{u}_{n}, it follows that {a}_{n}=(3{v}_{n}+312{u}_{n})/2. In view of (2.2), we readily obtain {a}_{n}=3{u}_{n+1}+81{u}_{n}. By (3.5), we get 39{a}^{2}+3=3{u}_{n+1}+81{u}_{n}, implying that

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

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

*i.e.*,

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

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

which implies that

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

By (2.3) and (2.1), it can be seen that {u}_{2r+1}-1+27{u}_{2r}={u}_{r}{v}_{r+1}+27{u}_{r}{v}_{r}={u}_{r}({v}_{r+1}+27{v}_{r}) and therefore

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

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

Since 4|a, it follows that

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

or

for some positive integer *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. □

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## Acknowledgements

Dedicated to Professor Hari M Srivastava.

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Karaatlı, O., Keskin, R. Integral points on the elliptic curve {y}^{2}={x}^{3}+27x-62.
*J Inequal Appl* **2013**, 221 (2013). https://doi.org/10.1186/1029-242X-2013-221

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

### Keywords

- elliptic curves
- integral point
- generalized Fibonacci and Lucas sequences