Open Access

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

Journal of Inequalities and Applications20132013:221

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

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 + 27 x 62 has only the integral points ( x , y ) = ( 2 , 0 ) and ( x , y ) = ( 28 , 844 , 402 , ± 15 , 491 , 585 , 540 ) using elementary number theory methods and some properties of generalized Fibonacci and Lucas sequences.

MSC:11B25, 11B37.

Keywords

elliptic curves integral point generalized Fibonacci and Lucas sequences

1 Introduction

Let P and Q be non-zero integers with P 2 + 4 Q 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 ) for  n 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 ) for  n 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 ) = U n ( P , Q ) ( Q ) n and V n = V n ( P , Q ) ( Q ) n for  n 1 ,
(1.1)
respectively. Taking α = ( P + P 2 + 4 Q ) / 2 and β = ( P P 2 + 4 Q ) / 2 to be the roots of the characteristic equation x 2 P x Q = 0 , we have the well-known expressions named Binet’s formulas
U n ( P , Q ) = ( α n β n ) / ( α β ) and V n ( P , Q ) = α n + β n
(1.2)
for all n 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 ) and v n = v n ( P , 1 )

for all n 1 . For more information about generalized Fibonacci and Lucas sequences, one can consult [15].

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 + 27 x 62
(1.3)

is ( x , y ) = ( 28 , 844 , 402 , ± 154 , 914 , 585 , 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 , 844 , 402 , ± 154 , 914 , 585 , 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 + 27 x 62 is ( x , y ) = ( 28 , 844 , 402 , ± 154 , 914 , 585 , 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 { 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 1 and n 1 . Then ( u m , u n ) = u ( m , n ) .

The well-known identities for ( u n ) and ( v n ) are as follows:
u 2 n = u n v n ,
(2.1)
v n = u n + 1 u n 1 ,
(2.2)
u 2 k + 1 1 = u k v k + 1 .
(2.3)
Moreover, if P is even, then
u n  is even n  is even ,
(2.4)
u n  is odd n  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 + 27 x 62 has only the integral points ( x , y ) = ( 2 , 0 ) and ( 28 , 844 , 402 , ± 154 , 914 , 585 , 540 ) .

Proof Assume that ( x , y ) is an integral point on the elliptic curve y 2 = x 3 + 27 x 62 . It can be easily seen that x > 0 . On the other hand, obviously, the elliptic curve y 2 = x 3 + 27 x 62 has only the integral point ( x , y ) = ( 2 , 0 ) with y = 0 . Hence, we may assume that y 0 . Let k = x 2 . Substituting this value of k into y 2 = x 3 + 27 x 62 , we get
y 2 = k ( k 2 + 6 k + 39 ) .
(3.1)
Since y 0 , it is obvious that y 2 > 0 . On the other hand, since k 2 + 6 k + 39 = ( k + 3 ) 2 + 30 > 0 , we conclude that k > 0 . Clearly, d = ( k , k 2 + 6 k + 39 ) = 1 , 3 , 13  or  39 . So, we get from (3.1) that
k = d a 2 , k 2 + 6 k + 39 = d b 2 , y = ± d a b
(3.2)

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
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 169 a 4 + 78 a 2 + 39 = 13 b 2 . Completing the square gives
( 13 a 2 + 3 ) 2 13 b 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 = 39 a 2 and k 2 + 6 k + 39 = 39 b 2 . Substituting k = 39 a 2 into k 2 + 6 k + 39 = 39 b 2 and completing the square give
( 39 a 2 + 3 ) 2 + 30 = 39 b 2 .
(3.5)
This equation is of the form
u 2 39 v 2 = 30 .
(3.6)
Let x n + y n 39 be a solution of the equation x 2 39 y 2 = 1 . Since the fundamental solution of this equation is α = 25 + 4 39 , we get x n + y n 39 = α n , and therefore x n = ( α n + β n ) / 2 and y n = ( α n β n ) / 2 39 , where β = 25 4 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 + 39 and 3 39 . So, the general solution of (3.6) is given by
a n + b n 39 = ( 3 39 ) ( x n + y n 39 ) ,
(3.7)
a n + b n 39 = ( 3 + 39 ) ( x n + y n 39 ) ,
(3.8)
with n 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
a n = ( 3 v n 312 u 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 = 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 1 . Another possibility is that 39 a 2 3 = 81 u n 3 u n 1 , implying that
13 a 2 + 1 = 27 u n + u n 1 .
(3.9)
It can be shown by the induction method that
u n { n ( mod 13 ) if  n  is even , n ( mod 13 ) if  n  is odd
(3.10)
and
u n n ( mod 8 ) .
(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 = 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
13 a 2 + 1 = u n + 1 + 27 u n .
(3.12)
Assume that n is odd. By using (3.10), we get
u n + 1 + 27 u n n 1 + 27 n 1 ( mod 13 ) ,
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 + 27 u n n + 1 + 3 n 4 n + 1 6 ( mod 8 ) ,
i.e.,
4 n 5 ( mod 8 ) ,
which is impossible. So, a is even, and therefore a = 2 m for some positive integer m. Substituting a = 2 m into (3.12), we get
52 m 2 + 1 = u n + 1 + 27 u n .
(3.13)
In the above equation, if m is odd, then from (3.11) we get
u n + 1 + 27 u n n + 1 + 3 n 4 n + 1 5 ( mod 8 ) ,
which implies that
n 1 ( mod 2 ) .
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 = 2 r for some r > 0 . Then (3.12) becomes
13 a 2 = u 2 r + 1 1 + 27 u 2 r .
By (2.3) and (2.1), it can be seen that u 2 r + 1 1 + 27 u 2 r = u r v r + 1 + 27 u r v r = u r ( v r + 1 + 27 v r ) and therefore
13 a 2 = u r ( v r + 1 + 27 v r ) .
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
13 a 2 = u r ( 3 , 848 u r 104 u r 1 ) .
Dividing both sides of the above equation by 13 and rearranging the equation gives
a 2 = 8 u r ( 37 u r u r 1 ) .
Since 4 | a , it follows that
2 ( a / 4 ) 2 = u r ( 37 u r u r 1 ) .
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
37 u r u r 1 = 2 c 2
(3.14)
or
u r = 2 c 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 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 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 = 2 r , we get n = 4 . Substituting this value of n into (3.12), we obtain
13 a 2 + 1 = u 5 + 27 u 4 .

Since u 5 = 6 , 242 , 501 and u 4 = 124 , 900 , a simple computation shows that a = 860 . Moreover, since k = 39 a 2 and x = k + 2 , we get k = 28 , 844 , 400 and therefore x = 28 , 844 , 402 . Substituting x = 28 , 844 , 402 into y 2 = x 3 + 27 x 62 gives y = ± 15 , 491 , 585 , 540 . Hence, the theorem is proved, the elliptic curve y 2 = x 3 + 27 x 62 has only the integral points ( x , y ) = ( 2 , 0 ) and ( x , y ) = ( 28 , 844 , 402 , ± 15 , 491 , 585 , 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

(1)
Department of Mathematics, Sakarya University

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