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 [2–4] 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.