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# 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:

## 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

• 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  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  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 ).
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 ). Let $$(x,y,n)$$ be a solution of (1.2). In 2004, Tengely  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 . □

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 . □

## 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). 