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Integral points on the elliptic curve
Journal of Inequalities and Applications volume 2013, Article number: 221 (2013)
We give a new proof that the elliptic curve has only the integral points and using elementary number theory methods and some properties of generalized Fibonacci and Lucas sequences.
Let P and Q be non-zero integers with . The generalized Fibonacci sequence and the Lucas sequence are defined by the following recurrence relations:
is called the n th generalized Fibonacci number and is called the n th generalized Lucas number. Also, generalized Fibonacci and Lucas numbers for negative subscripts are defined as
respectively. Taking and to be the roots of the characteristic equation , we have the well-known expressions named Binet’s formulas
for all . Instead of and , we use and , respectively. For , the sequence is the familiar Fibonacci sequence and the sequence is the familiar Lucas sequence . If and , then we have the well-known Pell sequence and Pell-Lucas sequence . For , we represent and by and , respectively. Thus , and and , and for all . Also, it is seen from Eq. (1.1) that
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 ). In 1987, Don Zagier  proposed that the largest integral point on the elliptic curve
is . Then the same problem was dealt with by some authors. In , Zhu and Chen found all integral points on (1.3) by using algebraic number theory and p-adic analysis. In , Wu proved that (1.3) has only the integral points and using some results of quartic Diophantine equations with elementary number methods. After that, in , the authors found the integral points on (1.3) using similar methods to those given in . In this paper, we determine that the largest integral point on the elliptic curve is 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.
In this section, we present two theorems and some well-known identities regarding the sequences and , which will be useful during the proof of the main theorem.
We state the following theorem from .
Theorem 2.1 Let . If with and , then or .
The following theorem is a well-known theorem (see ).
Theorem 2.2 Let and . Then .
The well-known identities for and 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 has only the integral points and .
Proof Assume that is an integral point on the elliptic curve . It can be easily seen that . On the other hand, obviously, the elliptic curve has only the integral point with . Hence, we may assume that . Let . Substituting this value of k into , we get
Since , it is obvious that . On the other hand, since , we conclude that . Clearly, . So, we get from (3.1) that
for some positive integers a and b.
If , then from (3.2) we get . Completing the square gives . This implies that . It can be easily shown that there are no integers a and b satisfying the previous equation.
If , then from (3.2) we obtain . Completing the square gives
Working on modulo 8 shows that (3.3) is impossible.
If , then from (3.2) we immediately have . Completing the square gives
Working on modulo 8 shows that (3.4) is impossible.
Lastly, we consider (1.3) for the case when . If , then from (3.2) we get and . Substituting into and completing the square give
This equation is of the form
Let be a solution of the equation . Since the fundamental solution of this equation is , we get , and therefore and , where . It can be easily seen that and . Equation (3.6) has exactly two solution classes and the fundamental solutions are and . So, the general solution of (3.6) is given by
From (2.2), if we write instead of and rearrange the above equation, then we get . This means that by (3.5). Dividing both sides of the equation by 3 gives . However, this is impossible for and . Another possibility is that , implying that
It can be shown by the induction method that
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 . Since and , it follows that . In view of (2.2), we readily obtain . By (3.5), we get , 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
which is impossible. So, a is even, and therefore for some positive integer m. Substituting 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 . We now return to (3.12). Since n is even, for some . Then (3.12) becomes
By (2.3) and (2.1), it can be seen that and therefore
By using (2.2), we get . In view of the recurrence relation of the sequence , we immediately have
Dividing both sides of the above equation by 13 and rearranging the equation gives
Since , it follows that
By Theorem 2.2, since , clearly, . This implies that either
for some positive integer c, where . By (2.4) and (2.5), it can be seen that is always odd. Therefore (3.14) is impossible. By Theorem 2.1, (3.15) is impossible for the case when . Hence, we have . On the other hand, since is even, from (2.4), it follows that r is even. Since r is even and , we get . Substituting this value of n into (3.12), we obtain
Since and , a simple computation shows that . Moreover, since and , we get and therefore . Substituting into gives . Hence, the theorem is proved, the elliptic curve has only the integral points and , which is the largest integral point on it. This completes the proof of the main theorem. □
Kalman D, Mena R: The Fibonacci numbers-exposed. Math. Mag. 2003, 76: 167–181.
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.
Muskat JB: Generalized Fibonacci and Lucas sequences and rootfinding methods. Math. Comput. 1993, 61: 365–372.
Rabinowitz S: Algorithmic manipulation of Fibonacci identities. 6. In Application of Fibonacci Numbers. Kluwer Academic, Dordrecht; 1996:389–408.
Ribenboim P: My Numbers, My Friends. Springer, New York; 2000.
Baker A:The Diophantine equation . J. Lond. Math. Soc. 1968, 43: 1–9.
Stroeker RJ, Tzanakis N: On the elliptic logarithm method for elliptic Diophantine equations: reflections and an improvement. Exp. Math. 1999, 8: 135–149.
Stroeker RJ, Tzanakis N: Computing all integer solutions of a genus 1 equation. Math. Comput. 2003, 72: 1917–1933.
Zagier D: Large integral points on elliptic curves. Math. Comput. 1987, 48: 425–436.
Zhu H, Chen J:Integral points on . J. Math. Study 2009, 42(2):117–125.
Wu H:Points on the elliptic curve . Acta Math. Sin., Chin. Ser. 2010, 53(1):205–208.
He Y, Zhang W: An elliptic curve having large integral points. Czechoslov. Math. J. 2010, 60(135):1101–1107.
Mignotte M, Pethő A: Sur les carrés dans certanies suites de Lucas. J. Théor. Nr. Bordx. 1993, 5(2):333–341.
Ribenboim P: An algorithm to determine the points with integral coordinates in certain elliptic curves. J. Number Theory 1999, 74: 19–38.
Nagell T: Introduction to Number Theory. Wiley, New York; 1981.
Dedicated to Professor Hari M Srivastava.
The authors declare that they have no competing interests.
Both authors contributed equally in the preparation of this article. Both authors read and approved the final manuscript.
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Karaatlı, O., Keskin, R. Integral points on the elliptic curve . J Inequal Appl 2013, 221 (2013). https://doi.org/10.1186/1029-242X-2013-221
- elliptic curves
- integral point
- generalized Fibonacci and Lucas sequences