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

Abstract

We give a new proof that the elliptic curve ${y}^{2}={x}^{3}+27x-62$ has only the integral points $\left(x,y\right)=\left(2,0\right)$ and $\left(x,y\right)=\left(28\text{,}844\text{,}402,±15\text{,}491\text{,}585\text{,}540\right)$ 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 $\left({U}_{n}\left(P,Q\right)\right)$ and the Lucas sequence $\left({V}_{n}\left(P,Q\right)\right)$ are defined by the following recurrence relations:

and

${U}_{n}\left(P,Q\right)$ is called the n th generalized Fibonacci number and ${V}_{n}\left(P,Q\right)$ is called the n th generalized Lucas number. Also, generalized Fibonacci and Lucas numbers for negative subscripts are defined as

(1.1)

respectively. Taking $\alpha =\left(P+\sqrt{{P}^{2}+4Q}\right)/2$ and $\beta =\left(P-\sqrt{{P}^{2}+4Q}\right)/2$ to be the roots of the characteristic equation ${x}^{2}-Px-Q=0$, we have the well-known expressions named Binet’s formulas

${U}_{n}\left(P,Q\right)=\left({\alpha }^{n}-{\beta }^{n}\right)/\left(\alpha -\beta \right)\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}{V}_{n}\left(P,Q\right)={\alpha }^{n}+{\beta }^{n}$
(1.2)

for all $n\in \mathbb{Z}$. Instead of ${U}_{n}\left(P,Q\right)$ and ${V}_{n}\left(P,Q\right)$, we use ${U}_{n}$ and ${V}_{n}$, respectively. For $P=Q=1$, the sequence $\left({U}_{n}\right)$ is the familiar Fibonacci sequence $\left({F}_{n}\right)$ and the sequence $\left({V}_{n}\right)$ is the familiar Lucas sequence $\left({L}_{n}\right)$. If $P=2$ and $Q=1$, then we have the well-known Pell sequence $\left({P}_{n}\right)$ and Pell-Lucas sequence $\left({Q}_{n}\right)$. For $Q=-1$, we represent $\left({U}_{n}\right)$ and $\left({V}_{n}\right)$ by $\left({u}_{n}\right)$ and $\left({v}_{n}\right)$, 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

${u}_{-n}=-{u}_{n}\left(P,-1\right)\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}{v}_{-n}={v}_{n}\left(P,-1\right)$

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

${y}^{2}={x}^{3}+27x-62$
(1.3)

is $\left(x,y\right)=\left(28\text{,}844\text{,}402,±154\text{,}914\text{,}585\text{,}540\right)$. 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 $\left(x,y\right)=\left(2,0\right)$ and $\left(28\text{,}844\text{,}402,±154\text{,}914\text{,}585\text{,}540\right)$ 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 $\left(x,y\right)=\left(28\text{,}844\text{,}402,±154\text{,}914\text{,}585\text{,}540\right)$ 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 $\left({u}_{n}\right)$ and $\left({v}_{n}\right)$, 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 $c\in \left\{1,2,3,6\right\}$ and $n>3$, then $\left(n,P,c\right)=\left(4,338,1\right)$ or $\left(6,3,1\right)$.

The following theorem is a well-known theorem (see [14]).

Theorem 2.2 Let $m\ge 1$ and $n\ge 1$. Then $\left({u}_{m},{u}_{n}\right)={u}_{\left(m,n\right)}$.

The well-known identities for $\left({u}_{n}\right)$ and $\left({v}_{n}\right)$ are as follows:

${u}_{2n}={u}_{n}{v}_{n},$
(2.1)
${v}_{n}={u}_{n+1}-{u}_{n-1},$
(2.2)
${u}_{2k+1}-1={u}_{k}{v}_{k+1}.$
(2.3)

Moreover, if P is even, then

(2.4)
(2.5)

3 Proof of the main theorem

The main theorem we deal with here is as follows.

Theorem 3.1 The elliptic curve ${y}^{2}={x}^{3}+27x-62$ has only the integral points $\left(x,y\right)=\left(2,0\right)$ and $\left(28\text{,}844\text{,}402,±154\text{,}914\text{,}585\text{,}540\right)$.

Proof Assume that $\left(x,y\right)$ 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 $\left(x,y\right)=\left(2,0\right)$ 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

${y}^{2}=k\left({k}^{2}+6k+39\right).$
(3.1)

Since $y\ne 0$, it is obvious that ${y}^{2}>0$. On the other hand, since ${k}^{2}+6k+39={\left(k+3\right)}^{2}+30>0$, we conclude that $k>0$. Clearly, . So, we get from (3.1) that

$k=d{a}^{2},\phantom{\rule{2em}{0ex}}{k}^{2}+6k+39=d{b}^{2},\phantom{\rule{2em}{0ex}}y=±dab$
(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 ${\left({a}^{2}+3\right)}^{2}+30={b}^{2}$. This implies that $\left[b-\left({a}^{2}+3\right)\right]\left[b+\left({a}^{2}+3\right)\right]=30$. 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

${b}^{2}-3{\left({a}^{2}+1\right)}^{2}=10.$
(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

${\left(13{a}^{2}+3\right)}^{2}-13{b}^{2}=-30.$
(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

${\left(39{a}^{2}+3\right)}^{2}+30=39{b}^{2}.$
(3.5)

This equation is of the form

${u}^{2}-39{v}^{2}=-30.$
(3.6)

Let ${x}_{n}+{y}_{n}\sqrt{39}$ 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 ${x}_{n}=\left({\alpha }^{n}+{\beta }^{n}\right)/2$ and ${y}_{n}=\left({\alpha }^{n}-{\beta }^{n}\right)/2\sqrt{39}$, where $\beta =25-4\sqrt{39}$. It can be easily seen that ${x}_{n}={v}_{n}\left(50,-1\right)/2$ and ${y}_{n}=4{u}_{n}\left(50,-1\right)$. 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

${a}_{n}+{b}_{n}\sqrt{39}=\left(3-\sqrt{39}\right)\left({x}_{n}+{y}_{n}\sqrt{39}\right),$
(3.7)
${a}_{n}+{b}_{n}\sqrt{39}=\left(3+\sqrt{39}\right)\left({x}_{n}+{y}_{n}\sqrt{39}\right),$
(3.8)

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

${a}_{n}=\left(3{v}_{n}-312{u}_{n}\right)/2.$

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 $n\ge 1$. 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

(3.10)

and

${u}_{n}\equiv n\left(mod8\right).$
(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 ${a}_{n}=\left(3{v}_{n}+312{u}_{n}\right)/2$. 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

${u}_{n+1}+27{u}_{n}\equiv -n-1+27n\equiv -1\left(mod13\right),$

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

${u}_{n+1}+27{u}_{n}\equiv n+1+3n\equiv 4n+1\equiv 6\left(mod8\right),$

i.e.,

$4n\equiv 5\left(mod8\right),$

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

${u}_{n+1}+27{u}_{n}\equiv n+1+3n\equiv 4n+1\equiv 5\left(mod8\right),$

which implies that

$n\equiv 1\left(mod2\right).$

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

$13{a}^{2}={u}_{2r+1}-1+27{u}_{2r}.$

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

$13{a}^{2}={u}_{r}\left({v}_{r+1}+27{v}_{r}\right).$

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

$13{a}^{2}={u}_{r}\left(3\text{,}848{u}_{r}-104{u}_{r-1}\right).$

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

${a}^{2}=8{u}_{r}\left(37{u}_{r}-{u}_{r-1}\right).$

Since $4|a$, it follows that

$2{\left(a/4\right)}^{2}={u}_{r}\left(37{u}_{r}-{u}_{r-1}\right).$

By Theorem 2.2, since $\left({u}_{r},{u}_{r-1}\right)=1$, clearly, $\left({u}_{r},37{u}_{r}-{u}_{r-1}\right)=1$. 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 ${u}_{r}={u}_{r}\left(50,-1\right)$. 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

$13{a}^{2}+1={u}_{5}+27{u}_{4}.$

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 $y=±15\text{,}491\text{,}585\text{,}540$. Hence, the theorem is proved, the elliptic curve ${y}^{2}={x}^{3}+27x-62$ has only the integral points $\left(x,y\right)=\left(2,0\right)$ and $\left(x,y\right)=\left(28\text{,}844\text{,}402,±15\text{,}491\text{,}585\text{,}540\right)$, 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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Correspondence to Olcay Karaatlı.

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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 ${y}^{2}={x}^{3}+27x-62$. J Inequal Appl 2013, 221 (2013). https://doi.org/10.1186/1029-242X-2013-221