Open Access

An upper bound for solutions of the Lebesgue-Nagell equation \(x^{2}+a^{2}=y^{n}\)

Journal of Inequalities and Applications20162016:209

https://doi.org/10.1186/s13660-016-1154-5

Received: 15 July 2016

Accepted: 26 August 2016

Published: 2 September 2016

Abstract

Let a be a positive integer with \(a>1\), and let \((x, y, n)\) be a positive integer solution of the equation \(x^{2}+a^{2}=y^{n}\), \(\gcd(x, y)=1\), \(n>2\). Using Baker’s method, we prove that, for any positive number ϵ, if n is an odd integer with \(n>C(\epsilon)\), where \(C(\epsilon)\) is an effectively computable constant depending only on ϵ, then \(n<(2+\epsilon)(\log a)/\log y\). Owing to the obvious fact that every solution \((x, y, n)\) of the equation satisfies \(n>2(\log a)/\log y\), the above upper bound is optimal.

Keywords

exponential diophantine equation Lebesgue-Nagell equation upper bound for solutions Baker’s method

MSC

11D61

1 Introduction

Let \(\mathbb{Z}\), \(\mathbb{N}\) be the sets of all integers and positive integers, respectively. Let D be a positive integer. In 1850, Lebesgue [1] proved that if \(D=1\), then the equation
$$ x^{2}+D=y^{n},\quad x, y, n\in\mathbb{N}, \gcd(x,y)=1, n>2 $$
(1.1)
has no solutions \((x,y,n)\), which solved a type important case of the famous Catalan’s conjecture. From then on, Nagell [24] dealt with the solution of (1.1) more systematically for the case of \(D>1\). Therefore, equation (1.1) is called the Lebesgue-Nagell equation (see [5]).
In this paper, we shall discuss an upper bound for solutions of (1.1) when \(D>1\), that is, \(D=a^{2}\), where a is a positive integer with \(a>1\). So equation (1.1) can be expressed as
$$ x^{2}+a^{2}=y^{n},\quad x, y, n\in\mathbb{N}, \gcd(x,y)=1, n>2. $$
(1.2)
This is a type of Lebesgue-Nagell equation leading to more discussions (see [6]). Let \((x,y,n)\) be a solution of (1.2). In 2004, Tengely [7] proved that if \(y>50{,}000\) and n is an odd prime with \(n>9{,}511\), then
$$ n< \frac{4\log a}{\log50{,}000}. $$
(1.3)
Using Baker’s method, the following result is proved.

Theorem

For any positive number ϵ, if n is an odd number with \(n>C(\epsilon)\), then
$$ n< \frac{(2+\epsilon)\log a}{\log y}, $$
(1.4)
where \(C(\epsilon)\) is an effectively computable constant depending only on ϵ.
Owing to (1.2) every solution \((x, y, n)\) of the equation satisfies \(a^{2}< y^{n}\), then we have
$$ n>\frac{2(\log a)}{\log y}. $$
(1.5)
Hence comparing (1.4) and (1.5), we see that the upper bound we get in this paper is optimal.

2 Preliminaries

Lemma 2.1

For a positive odd integer n, every solution \((X, Y, Z)\) of the equation
$$ X^{2}+Y^{2}=Z^{n},\quad X, Y, Z\in\mathbb{N}, \gcd(X,Y)=1 $$
(2.1)
can be expressed as
$$\begin{aligned} &Z=f^{2}+g^{2}, \qquad X+Y\sqrt{-1}= \lambda_{1}(f+\lambda_{2}g\sqrt{-1})^{n},\quad f, g \in\mathbb{N}, \\ &\gcd(f, g)=1,\quad \lambda_{1}, \lambda_{2}\in\{1,-1\}. \end{aligned}$$
(2.2)

Proof

See Section 15.2 of [8]. □

Let α be an algebraic number of degree d, c be a leading coefficient of the defined polynomial of α, \(\alpha^{(j)}\) (\(j=1,\ldots,d\)) be the whole conjugate numbers of α. Then
$$ h (\alpha )=\frac{1}{d} \Biggl(\log c+\sum_{j=1}^{d} \log\max \bigl\{ 1,\bigl\vert \alpha^{(j)}\bigr\vert \bigr\} \Biggr) $$
(2.3)
is called the Weil height of α.

Lemma 2.2

For the positive integers \(b_{1}\) and \(b_{2}\), assume
$$ \Lambda=b_{1}\log\alpha-b_{2}\pi\sqrt{-1}, $$
(2.4)
where logα is principal value of the logarithm of α. If \(|\alpha|=1\) and α is not a unit root, then
$$ \log|\Lambda|\geq-8.87AB^{2}, $$
(2.5)
where
$$\begin{aligned}& A=\max \biggl\{ 20, 10.98|\log\alpha|+ \frac{1}{2}dh(\alpha) \biggr\} , \\& B=\max \biggl\{ 17, \frac{\sqrt{d}}{40}, 5.03+2.35 \biggl(\frac{d}{2} \biggr)+\frac{d}{2} \biggl(\frac {b_{1}}{68.9}+\frac{b_{2}}{2A} \biggr) \biggr\} . \end{aligned}$$

Proof

See Theorem 3 of [9]. □

3 Proof of theorem

Let \((x, y, n)\) be a solution of equation (1.2) with n being odd and satisfying
$$ n>\frac{(2+\epsilon)\log a}{\log y}. $$
(3.1)
By (1.2), we see that equation (2.1) has the solution \((X,Y,Z)=(x,a,y)\). So from Lemma 2.1, we get
$$\begin{aligned}& y=f^{2}+g^{2},\quad f,g\in\mathbb{N}, \gcd(f,g)=1, \end{aligned}$$
(3.2)
$$\begin{aligned}& x+a\sqrt{-1}=\lambda_{1}(f+\lambda_{2}g \sqrt{-1})^{n},\quad \lambda_{1}, \lambda_{2}\in \{1,-1\}. \end{aligned}$$
(3.3)
Assume
$$ \theta=f+g\sqrt{-1},\qquad \bar{\theta}=f-g\sqrt{-1}. $$
(3.4)
From (3.2) and (3.4), we have
$$ \theta\bar{\theta}=y,\qquad |\theta|=|\bar{\theta}|=\sqrt{y}. $$
(3.5)
Let \(\alpha=\theta/\bar{\theta}\). From (3.4) and (3.5), we see that α satisfies \(|\alpha|=1\) and
$$ y\alpha^{2}-2\bigl(f^{2}-g^{2}\bigr)\alpha+y=0. $$
(3.6)
Since \(\gcd(x,y)=1\) by (1.1) and \(n> 2\), we have \(\gcd(x,a)=1\) and y is odd. And since \(\gcd(f,g)=1\) from (3.2), we see f is odd, g is even, so \(\gcd(f^{2}+g^{2},f^{2}-g^{2})=\gcd(f^{2}+g^{2}, 2(f^{2}-g^{2}))=1\). Hence \(y>1\) and we see that α is not a unit root. And since the discriminant of the polynomial \(yz^{2}-2(f^{2}-g^{2})z+y\in\mathbb{Z}[z]\) is equal to \(-16f^{2}g^{2}\), we see that α is a quadratic algebraic number, α and \(\alpha^{-1}\) are its whole conjugate numbers. Thus by (2.3), we deduce that the Weil height of α is
$$ h(\alpha)=\frac{1}{2}\log y. $$
(3.7)
Since by (3.3) we have
$$ x-a\sqrt{-1}=\lambda_{1}(f-\lambda_{2}g \sqrt{-1})^{n}, $$
(3.8)
from (3.3), (3.4), (3.5), and (3.8), we obtain
$$ a=\biggl\vert \frac{\theta^{n}-\bar{\theta}^{n}}{2\sqrt{-1}}\biggr\vert =\frac {1}{2}\bigl\vert \theta^{n}-\bar{\theta}^{n}\bigr\vert =\frac{1}{2} \bigl\vert {\bar{\theta}}^{n}\bigr\vert \biggl\vert \biggl( \frac{\theta}{\bar {\theta}} \biggr)^{n}-1\biggr\vert =\frac{y^{n/2}}{2}\bigl\vert \alpha^{n}-1\bigr\vert . $$
(3.9)
According to the maximum modulus principle, for any complex number z, we are sure that
$$ \bigl\vert e^{z}-1\bigr\vert \geq\frac{1}{2} $$
(3.10)
or
$$ \bigl\vert e^{z}-1\bigr\vert \geq\frac{2}{\pi} \vert z-k\pi \sqrt{-1}\vert ,\quad k\in\mathbb{Z}. $$
(3.11)
Assume \(\alpha=e^{z}\). If (3.10) holds, then from (3.9), we can deduce that
$$ a\geq\frac{y^{n/2}}{4}. $$
(3.12)
Combining (3.1) and (3.12), we get
$$ 4>y^{\epsilon n/2(2+\epsilon)}. $$
(3.13)
However, since \(y\geq5\) by (3.2), we see that (3.13) does not hold when \(n>2(2+\epsilon)/\epsilon\). Hence, we only need to discuss the case when (3.11) holds.
Owing to \(a^{2+\epsilon}< y^{n}\) by (3.1), if (3.11) holds, then from (3.9) and (3.11) we have
$$ y^{n/(2+\epsilon)}>a\geq\frac{y^{n/2}}{\pi} \vert n\log \alpha-k\pi\sqrt{-1} \vert ,\quad k\in\mathbb{N}, k\leq n. $$
(3.14)
Let
$$ \Lambda=n\log\alpha-k\pi\sqrt{-1}. $$
(3.15)
By (3.14) and (3.15), we see
$$ \log\pi-\log|\Lambda|\geq\frac{\epsilon n}{2(2+\epsilon)}\log y. $$
(3.16)
Since we have proved that α is not only a quadratic algebraic number but also a non-unit root with \(|\alpha|=1\), and the degree of α is 2, from Lemma 2.2, by (3.7), we see that Λ satisfies (2.5), where
$$\begin{aligned}& A=\max \biggl\{ 20, 10.98|\log\alpha|+ \frac{1}{2}\log y \biggr\} , \end{aligned}$$
(3.17)
$$\begin{aligned}& B=\max \biggl\{ 17, 7.38+\log \biggl(\frac{n}{2A}+\frac{k}{68.9} \biggr) \biggr\} . \end{aligned}$$
(3.18)
Since \(y\geq5\) and the principal value of the logarithm of α satisfies \(|\log\alpha|\leq\pi\), we deduce by (3.17) that
$$ A\leq10.98\pi+\frac{1}{2}\log y. $$
(3.19)
By (3.14) and (3.17), we have \(k\leq n\) and \(1/(2A)\leq0.025\), respectively, therefore if \(n>68.9\times10^{8}\), then by (3.18) we get
$$ B< 7.38+\log(0.04n)< 4.17+\log n. $$
(3.20)
Hence from (2.5), (3.16), (3.19), and (3.20), we have
$$ \log\pi+8.87 \biggl(10.98\pi+\frac{1}{2}\log y \biggr) (4.17+\log n )^{2}>\frac{\epsilon n}{2(2+\epsilon)}\log y. $$
(3.21)
Since \(y\geq5\), we see by (3.21) that
$$ \frac{2(2+\epsilon)}{\epsilon} \bigl(1+194.56 (4.17+\log n )^{2} \bigr)>n. $$
(3.22)
From (3.22), we get \(n< C'(\epsilon)\), where \(C'(\epsilon)\) is an effectively computable constant depending only on ϵ. Let
$$ C(\epsilon)=\max \biggl\{ 68.9\times10^{8}, \frac{2(2+\epsilon)}{\epsilon}, C'(\epsilon) \biggr\} . $$
(3.23)
We see by (3.23) that \(C(\epsilon)\) is also an effectively computable constant depending only on ϵ, and to sum up, we can deduce when \(n>C(\epsilon)\), the solution \((x, y, n)\) of equation (1.2) does not satisfy (3.1), so (1.4) holds definitely. Therefore, we completed the proof of the theorem.

Declarations

Acknowledgements

This work is supported by the National Natural Science Foundation of China (No. 11526162) and the Natural Science Foundation of Shaanxi Province (No. 2016JQ1040).

Open Access This article is distributed under the terms of the Creative Commons Attribution 4.0 International License (http://creativecommons.org/licenses/by/4.0/), which permits unrestricted use, distribution, and reproduction in any medium, provided you give appropriate credit to the original author(s) and the source, provide a link to the Creative Commons license, and indicate if changes were made.

Authors’ Affiliations

(1)
Xi’an Medical University

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© Pan 2016