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An upper bound for solutions of the Lebesgue-Nagell equation \(x^{2}+a^{2}=y^{n}\)
Journal of Inequalities and Applications volume 2016, Article number: 209 (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.
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
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
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
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
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
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
can be expressed as
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
is called the Weil height of α.
Lemma 2.2
For the positive integers \(b_{1}\) and \(b_{2}\), assume
where logα is principal value of the logarithm of α. If \(|\alpha|=1\) and α is not a unit root, then
where
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
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
Assume
Let \(\alpha=\theta/\bar{\theta}\). From (3.4) and (3.5), we see that α satisfies \(|\alpha|=1\) and
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
Since by (3.3) we have
from (3.3), (3.4), (3.5), and (3.8), we obtain
According to the maximum modulus principle, for any complex number z, we are sure that
or
Assume \(\alpha=e^{z}\). If (3.10) holds, then from (3.9), we can deduce that
Combining (3.1) and (3.12), we get
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
Let
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
Since \(y\geq5\) and the principal value of the logarithm of α satisfies \(|\log\alpha|\leq\pi\), we deduce by (3.17) that
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
Hence from (2.5), (3.16), (3.19), and (3.20), we have
Since \(y\geq5\), we see by (3.21) that
From (3.22), we get \(n< C'(\epsilon)\), where \(C'(\epsilon)\) is an effectively computable constant depending only on ϵ. Let
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.
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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).
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Pan, X. An upper bound for solutions of the Lebesgue-Nagell equation \(x^{2}+a^{2}=y^{n}\) . J Inequal Appl 2016, 209 (2016). https://doi.org/10.1186/s13660-016-1154-5
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DOI: https://doi.org/10.1186/s13660-016-1154-5