Abstract
Let \(a,\ b,\ c,\ m\) be positive integers such that \(a+b=c^2, 2\mid a, 2\not \mid c\) and \(m>1\). In this paper we prove that if \(c\mid m \) and \(m>36c^3 \log c\), then the equation \((am^2+1)^x+(bm^2-1)^y=(cm)^z\) has only the positive integer solution \((x,\ y,\ z)\)=\((1,\ 1,\ 2)\).
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1 Introduction
In recent years, many papers investigated pure ternary exponential diophantine equations (see [6,7,8,9,10,11,12,13,14,15]).
Let \(a,\ b,\ c,\ m\) be positive integers such that
In this paper we discuss the equation
In 2012, Terai [13] proved that if \((a,\ b,\ c)=(4,\ 5,\ 3)\), then (1.2) has only the solution \((x,\ y,\ z)=(1,\ 1,\ 2)\) under some conditions. Recently, Wang et al. [17] improved that for \((a,\ b,\ c)=(4,\ 5,\ 3)\) and \(3\not \mid m\). Then (1.2) has only the solution \((x,\ y,\ z)=(1,\ 1,\ 2)\). In this paper we prove a general result as follows:
Theorem 1.1
If \(c\mid m \) and \(m>36c^3 \log c\), then (1.2) has only the solution \((x,\ y,\ z)=(1,\ 1,\ 2)\).
As a direct consequence we get:
Corollary 1.2
If \((a,\ b,\ c)=(4,\ 5,\ 3)\) and \(3\mid m\) and \(m>1068\), then (1.2) has only the solution \((x,\ y,\ z)=(1,\ 1,\ 2)\).
Thus, combining the result of [17] and our corollary we get that (1.2) is basically solved for \((a,\ b,\ c)=(4,\ 5,\ 3)\).
2 Preliminaries
For any nonnegative integer n, let \(F_n\) and \(L_n\) be the n-th Fibonacci and Lucas number, respectively.
Lemma 2.1
([3]). The equation
has only the solutions \((n,\ X)=(1,\ 1),(2,\ 1)\) and \((12,\ 12)\).
For any positive integer D, let \(h(-4D)\) denote the class number of positive binary quadratic forms of discriminant \(-4D\).
Lemma 2.2
([4], Theorems 11.4.3, 12.10.1 and 12.14.3]).
Let \(D,\ D_1,\ D_2,\ k\) be positive integers such that \(\min \left\{ D,\ D_1,\ D_2\right\} >1, \gcd \left( D_1,\ D_2\right) =1,\ 2\not \mid k\) and \(\gcd (D,\ k)=\gcd \left( D_1,\ D_2,\ k\right) =1.\)
Lemma 2.3
([5], Theorems 1 and 2]). If the equation
has solutions \((X,\ Y,\ Z)\), then every solution \((X,\ Y,\ Z)\) of (2.1) can be expressed as
where \(X_1, Y_1, Z_1\) are positive integers satisfying \(X_1^2+DY_1^2=k^{z_1},\ \gcd (X_1,\ Y_1)=1\) and \(h(-4D)\equiv 0(\bmod Z_1)\).
Lemma 2.4
([5], Lemma 1]). For a fixed solution \((X,\ Y,\ Z)\) of the equation
there exists a unique positive integer l such that
where \(\alpha ,\ \beta \) are integers with \(\beta X-\alpha Y=1\).
The positive integer l defined as in Lemma 2.4 is called the characteristic number of the solution \((X,\ Y,\ Z)\) and is denoted by \(<X,\ Y,\ Z>\).
Lemma 2.5
([5], Lemma 6]). If \(<X,\ Y,\ Z>=l\), then \(D_1X\equiv -lY(\bmod k)\).
For a fixed positive integer \(l_0\), if (2.2) has a solution \((X_0,\ Y_0,\ Z_0)\) with \(<X_0,\ Y_0,\ Z_0>=l_0\), then the set of all solutions \((X,\ Y,\ Z)\) of (2.2) with \(<X,\ Y,\ Z>\equiv \pm l_0(\bmod k)\) is called a solution class of (2.2) and is denote by \(S(l_0)\).
Lemma 2.6
([5], Theorems 1 and 2]). For any fixed solution class \(S(l_0)\) of (2.2), there exists a unique solution \((X_1,\ Y_1,\ Z_1)\in S(l_0)\) such that \(X_1>0, Y_1>0\) and \(Z_1\le Z\), where Z runs through all solutions \((X,\ Y,\ Z)\in S(l_0)\). The solution \((X_1,\ Y_1,\ Z_1)\) is called the least solution of \(S(l_0)\). Every solution \((X,\ Y,\ Z)\in S(l_0)\) can be expressed as
Lemma 2.7
([2], Theorem 2]). Let \((X_1,\ Y_1,\ Z_1)\) be the least solution of \(S(l_0)\). If (2.2) has a solution \((X,\ Y,\ Z)\in S(l_0)\) satisfying \(X>0\) and \(Y=1\), then \(Y_1=1\). Further, if \((X,\ Z)\ne (X_1,\ Z_1)\), then one of the following conditions is satisfied:
-
(i)
\(D_1X^2_1=\frac{1}{4}(k^{Z_1}\pm 1),\ D_2=\frac{1}{4}(3k^{Z_1}\mp 1),\ (X,\ Z)=(X_1|D_1X^2_1-3D_2|,\ 3Z_1)\).
-
(ii)
\(D_1X^2_1=\frac{1}{4}F_{3r+3\varepsilon },\ D_2=\frac{1}{4}L_{3r},\ k^{Z_1}=F_{3r+\varepsilon },\ (X,\ Z)=(X_1|D^2_1X^4_1-10D_1D_2X^2_1+5D^2_2|,\ 5Z_1)\), where r is a positive integer, \(\varepsilon \in \{1, -1\}\).
Let \(\alpha ,\ \beta \) be algebraic integers. If \(\alpha +\beta \) and \(\alpha \beta \) are nonzero coprime integers and \(\frac{\alpha }{\beta }\) is not a root of unity, then \(\left( \alpha ,\ \beta \right) \) is called a Lucas pair. Further, let \(A=\alpha +\beta \) and \(C=\alpha \beta \). Then we have
where \(B=A^2-4C\). We call \((A,\ B)\) the parameters of the Lucas pair \((\alpha ,\ \beta )\). Two Lucas pairs \((\alpha _1,\ \beta _1)\) and \((\alpha _2,\ \beta _2)\) are equivalent if \(\frac{\alpha _1}{\alpha _2}=\frac{\beta _1}{\beta _2}=\pm 1\). Given a lucas pair \((\alpha ,\ \beta )\), one defines the corresponding sequence of Lucas numbers by
For equivalent Lucas pairs \((\alpha _1,\ \beta _1)\) and \((\alpha _2,\ \beta _2)\), we have \(L_n(\alpha _1,\ \beta _1)=\pm L_n(\alpha _2,\ \beta _2) (n=0,\ 1,\ldots ).\) A prime p is called a primitive divisor of \(L_n(\alpha ,\ \beta )(n>1)\) if
A Lucas pair \((\alpha ,\ \beta )\) such that \(L_n(\alpha ,\ \beta )\) has no primitive divisors will be called an n-defective Lucas pair. Further, a positive integer n is called totally non-defective if no Lucas pair is n-defective.
Lemma 2.8
([1], Theorem 1.4]). If \(n>30\), Then n is totally non-defective.
Lemma 2.9
([16]). Let n satisfy \(4<n\le 30\) and \(n\ne 6\). Then, up to equivalence, all parameters of n-defective Lucas pairs are given as follows:
-
(i)
\(n=5,\ (A,\ B)=(1,\ 5),\ (1,\ -7),\ (2,\ -40),\ (1,\ -11),\ (1,\ -15),\ (12,\ -76), (12,\ -1364)\).
-
(ii)
\(n=7,\ (A,\ B)=(1,\ -7),\ (1,\ -19)\).
-
(iii)
\(n=8,\ (A,\ B)=(2,\ -24),\ (1,\ -7)\).
-
(iv)
\(n=10,\ (A,\ B)=(2,\ -8),\ (5,\ -3),\ (5,\ -47)\).
-
(v)
\(n=12,\ (A,\ B)=(1,\ 5),\ (1,\ -7),\ (1,\ -11),\ (2,\ -56),\ (1,\ -15),\ (1,\ -19)\).
-
(vi)
\(n\in \{13,\ 18, \ 30 \},\ (A,\ B)=(1,\ -7)\).
3 Proof of theorem
We now assume that \((x,\ y,\ z)\) is a solution of (1.2) with \((x,\ y,\ z)\ne (1,\ 1,\ 2)\). Since \(c\mid m\), we have \(cm\mid m^2\), and by (1.2), we get
Since \(m>1\) and \(z>2\), by (1.2), we have \(0\equiv (cm)^z\equiv (am^2+1)^x+(bm^2-1)^y \equiv 1+(-1)^y(\bmod m^2)\) and
Further, since \(z\ge 3\), by (1.2) and (3.2), we get \(0\equiv (cm)^z\equiv (am^2+1)^x+(bm^2-1)^y \equiv (ax+by)m^2(\bmod m^3)\) and
Notice that \(2\mid a, \ 2\not \mid c\) and \(2\not \mid b\) by (1.1). We see from (3.2) and (3.3) that \(2\not \mid ax+by\) and
So we have
We first consider the case of \(2\mid x\). Then, by (3.2), the equation
has the solution
By (3.1) and (3.5), applying Lemma 2.3 to (3.6) and (3.7), we have
where \(X_1,\ Y_1,\ Z_1\) are positive integers satisfying
If \(2\mid t\), let
By Lemma 2.3, \(X_2\) and \(Y_2\) are integers satisfying
Substitute (3.12) into (3.9), we have \((am^2+1)^{\frac{x}{2}}+(bm^2-1)^{\frac{y-1}{2}}\sqrt{1-bm^2}=\lambda _1(X_2+Y_2\sqrt{1-bm^2})^2\) and
By (3.1) and (3.13), we get \(\gcd \left( X_2,\ bm^2-1\right) =1\). Therefore, we see from (3.14) that
Substitute (3.15) into (3.13), we get
Since \(z>2\), we have \(\frac{{z}}{2}\ge 2\). By (3.2) and (3.16), we get \(0\equiv (cm)^\frac{{z}}{2}\equiv 1+\frac{1}{4}(bm^2-1)^y\equiv 1-\frac{1}{4}\equiv \frac{3}{4}(\bmod m^2)\) and \(m^2\mid 3\), a contradiction. So we have
Let
By (3.10) and (3.18), we have \(\alpha +\beta =2X_1,\ \alpha -\beta =2Y_1\sqrt{1-bm^2},\ \alpha \beta =(cm)^{Z_1}\) and \(\frac{\alpha }{\beta }\) satisfies \((cm)^{Z_1}(\frac{\alpha }{\beta })^2-2\left( X^2_1-(bm^2-1)Y^2_1\right) (\frac{\alpha }{\beta })+(cm)^{Z_1}=0\). It implies that \((\alpha ,\ \beta )\) is a Lucas pair with parameters
Further, let \(L_n(\alpha ,\ \beta )\ (n=0,\ 1,\ldots )\) be the corresponding Lucas numbers. By (2.3), (3.9) and (3.18), we have
We see from (3.19) and (3.20) that the Lucas number \(L_t(\alpha ,\ \beta )\) has no primitive divisors. Therefore, by Lemmas 2.8 and 2.9, we get from (3.17) and (3.19) that
By (3.8), (3.11) and (3.21), we have
Applying Lemma 2.2 to (3.22), we get
Further, since \(b<a+b=c^2\), by (3.23), we have
On the other hand, since \(2\mid x\), if \(z=3\), then \((cm)^3>(am^2+1)^x\ge (am^2+1)^2>a^2m^4\), whence we get \(c^3>a^2m>m>36c^3\log c\), a contradiction. It implies that \(z\ge 4\) and \(0\equiv (cm)^z\equiv (am^2+1)^x+(bm^2-1)^y\equiv (ax+by)m^2(\bmod m^4)\), whence we obtain \(ax+by\equiv 0(\bmod m^2)\) and
Since \(m>36c^3\log c\), by (1.2), we have
Hence, by (3.25) and (3.26), we get
The combination of (3.24) and (3.27), we have
But, since \(m>36c^3\log c\), (3.28) is false. Thus, (1.2) has no solutions \((x,\ y,\ z)\) with \(2\mid x\).
Finally, we consider the case of \(2\not \mid x\). Then, by (1.2) and (3.2), the equation
has the solution
Let \(l=\langle (am^2+1)^{\frac{x-1}{2}},\ (bm^2-1)^{\frac{y-1}{2}},\ z\rangle \). Since \(cm\mid m^2\), by Lemma 2.5, l satisfies
On the other hand, since \((x,\ y,\ z)\ne (1,\ 1,\ 2)\), (3.29) has an other solution
Let \(l_0=\langle 1,\ 1,\ 2\rangle \). Then we have
Obviously, since \(z\ge 2\) for any solution \((x,\ y,\ z)\) of (3.29), the least solution of \(S(l_0)\) is
Compare (3.31) and (3.33), we have \(l\equiv \pm l_0 (\bmod cm)\). It implies that the solution (3.30) belongs to \(S(l_0)\). Therefore, using Lemma 2.6, we get from (3.30) and (3.32) that
By (3.35), we have
Further, since \(2\mid bm^2-1\) and \(2\not \mid (am^2+1)t\), we see from (3.36) that \(y=1\) and \((bm^2-1)^{\frac{y-1}{2}}=1\). It implies that (3.30) is a solution of \(S(l_0)\) satisfying \(X>0, \ Y=1\) and \((X,\ Z)\ne (X_1,\ Z_1)=(1,\ 2)\). Therefore, by Lemma 2.7, we get either
or
When (3.37) holds, since \(c\mid m\), we have \(1\equiv am^2+1\equiv \frac{1}{4}\left( (cm)^2\pm 1\right) \equiv \pm \frac{1}{4}(\bmod c^2)\). But, since \(c^2\ge 9\), it is impossible. On the other hand, since \(cm>1\) and \(2\not \mid cm\), by Lemma 2.1, (3.38) is false. Thus, (1.2) has only the solution \((x,\ y,\ z)=(1,\ 1,\ 2)\) with \(2\not \mid x\). the theorem is proved.
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Acknowledgements
The authors would like to thank the referees for their valuable suggestions. This work is supported by N.S.F.(11226038, 11371012) of P.R. China , the Education Department Foundation of Shaanxi Province(14JK1311) and Scientific Research Foundation for Doctor of Xi’an Shiyou University (2015BS06)
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Fu, R., Yang, H. On the exponential diophantine equation \(\left( am^{2}+1\right) ^{x}+\left( bm^{2}-1\right) ^{y}=(cm)^{z}\) with \( c\mid m \) . Period Math Hung 75, 143–149 (2017). https://doi.org/10.1007/s10998-016-0170-z
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DOI: https://doi.org/10.1007/s10998-016-0170-z
Keywords
- Exponential diophantine equation
- Existence of primitive divisor of Lucas and Lehmer numbers
- Application of BHV theorem