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Integral points on the elliptic curve y 2 = x 3 +27x62

Abstract

We give a new proof that the elliptic curve y 2 = x 3 +27x62 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.

1 Introduction

Let P and Q be non-zero integers with P 2 +4Q0. 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 n0

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 n0.

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 n1,
(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 PxQ=0, we have the well-known expressions named Binet’s formulas

U n (P,Q)= ( α n β n ) /(αβ)and V n (P,Q)= α n + β n
(1.2)

for all nZ. 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 n1. Also, it is seen from Eq. (1.1) that

u n = u n (P,1)and v n = v n (P,1)

for all n1. 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 +27x62
(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 +27x62 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 m1 and n1. 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 evenn is even,
(2.4)
u n  is oddn 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 +27x62 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 +27x62. It can be easily seen that x>0. On the other hand, obviously, the elliptic curve y 2 = x 3 +27x62 has only the integral point (x,y)=(2,0) with y=0. Hence, we may assume that y0. Let k=x2. Substituting this value of k into y 2 = x 3 +27x62, we get

y 2 =k ( k 2 + 6 k + 39 ) .
(3.1)

Since y0, 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 or 39. So, we get from (3.1) that

k=d a 2 , k 2 +6k+39=d b 2 ,y=±dab
(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 +6k+39=39 b 2 . Substituting k=39 a 2 into k 2 +6k+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=254 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 n1, 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 n1. 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(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 =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 n1+27n1(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 +27 u n n+1+3n4n+16(mod8),

i.e.,

4n5(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

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+3n4n+15(mod8),

which implies that

n1(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

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 r3. 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

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 +27x62 gives y=±15,491,585,540. Hence, the theorem is proved, the elliptic curve y 2 = x 3 +27x62 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. □

References

  1. Kalman D, Mena R: The Fibonacci numbers-exposed. Math. Mag. 2003, 76: 167–181.

    Article  MathSciNet  Google Scholar 

  2. Karaatlı O, Keskin R: On some diophantine equations related to square triangular and balancing numbers. J. Algebra, Number Theory: Adv. Appl. 2010, 4(2):71–89.

    Google Scholar 

  3. Muskat JB: Generalized Fibonacci and Lucas sequences and rootfinding methods. Math. Comput. 1993, 61: 365–372.

    Article  MathSciNet  Google Scholar 

  4. Rabinowitz S: Algorithmic manipulation of Fibonacci identities. 6. In Application of Fibonacci Numbers. Kluwer Academic, Dordrecht; 1996:389–408.

    Chapter  Google Scholar 

  5. Ribenboim P: My Numbers, My Friends. Springer, New York; 2000.

    Google Scholar 

  6. Baker A:The Diophantine equation y 2 =a x 3 +b x 2 +cx+d. J. Lond. Math. Soc. 1968, 43: 1–9.

    Article  Google Scholar 

  7. Stroeker RJ, Tzanakis N: On the elliptic logarithm method for elliptic Diophantine equations: reflections and an improvement. Exp. Math. 1999, 8: 135–149.

    Article  MathSciNet  Google Scholar 

  8. Stroeker RJ, Tzanakis N: Computing all integer solutions of a genus 1 equation. Math. Comput. 2003, 72: 1917–1933.

    Article  MathSciNet  Google Scholar 

  9. Zagier D: Large integral points on elliptic curves. Math. Comput. 1987, 48: 425–436.

    Article  MathSciNet  Google Scholar 

  10. Zhu H, Chen J:Integral points on y 2 = x 3 +27x62. J. Math. Study 2009, 42(2):117–125.

    MathSciNet  Google Scholar 

  11. Wu H:Points on the elliptic curve y 2 = x 3 +27x62. Acta Math. Sin., Chin. Ser. 2010, 53(1):205–208.

    Google Scholar 

  12. He Y, Zhang W: An elliptic curve having large integral points. Czechoslov. Math. J. 2010, 60(135):1101–1107.

    Article  Google Scholar 

  13. Mignotte M, Pethő A: Sur les carrés dans certanies suites de Lucas. J. Théor. Nr. Bordx. 1993, 5(2):333–341.

    Article  Google Scholar 

  14. Ribenboim P: An algorithm to determine the points with integral coordinates in certain elliptic curves. J. Number Theory 1999, 74: 19–38.

    Article  MathSciNet  Google Scholar 

  15. Nagell T: Introduction to Number Theory. Wiley, New York; 1981.

    Google Scholar 

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Acknowledgements

Dedicated to Professor Hari M Srivastava.

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Correspondence to Olcay Karaatlı.

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Karaatlı, O., Keskin, R. Integral points on the elliptic curve y 2 = x 3 +27x62. J Inequal Appl 2013, 221 (2013). https://doi.org/10.1186/1029-242X-2013-221

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Keywords

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