January 01, 1970
Jacobsthal numbers are an example of Lucas sequence defined by a recurrence relation analogous to that of the Fibonacci numbers but with different parameters. In this paper, we show that the only perfect powers among Jacobsthal numbers are the trivial ones, namely \(0\) and \(1\). Using Binet formulas, our problem reduces to an exponential Diophantine equation in three unknowns. It is resolved by the modular approach, following the framework developed by Bennett and Skinner for ternary Diophantine equations.
Note added in Version 2. After the first version of this manuscript was posted, it was brought to the author’s attention that the main result follows as a special case of Theorem 3 of Laishram, Ngairangbam and Maibam [1]. Moreover, the Diophantine equation upon which this manuscript is centered follows from Lemma 2.6 of [1], and also from Theorem 2 of Bugeaud and Mignotte [2].
A non-negative integer \(w\) is called a perfect power if there exist non-negative integers \(x\) and \(m\geq2\) such that \(w=x^m\). The cases \(w=0\) and \(w=1\) are referred to as trivial perfect powers. Determining perfect powers in linear recurrence sequences is a classical problem in number theory. The most prominent example is the Fibonacci numbers, defined by the recurrence \[F_0=0,~F_1=1,\quad F_n=F_{n-1}+F_{n-2}\] for all positive integers \(n\geq2\). In 2006, a landmark paper by Bugeaud, Mignotte, and Siksek [3] finally resolved a long-standing conjecture by proving that the only perfect powers among the Fibonacci numbers are \[F_0=0,\quad F_1=F_2=1,\quad F_6=8, \quad F_{12}=144.\] This breakthrough subsequently inspired extensive research on related Diophantine equations involving Fibonacci numbers, such as \(F_n+1=x^m\) or \(F_n-F_k=2^m\) (see, e.g., [4]–[14]).
The Fibonacci numbers are a particular case of the family of Lucas sequences. Given two integer parameters \(P\) and \(Q\), the Lucas sequence \(U_n(P,Q)\) of the first kind is defined by \[U_0(P,Q)=0,\quad U_1(P,Q)=1,\quad U_n(P,Q)=P\,U_{n-1}(P,Q)-Q\,U_{n-2}(P,Q)\] for \(n\geq2\). Note that the Fibonacci numbers correspond to \(U_n(1,-1)\). Closely related to \(U_n(P,Q)\) is its companion sequence \(V_n(P,Q)\), called Lucas sequence of the second kind. It is defined by \[V_0(P,Q)=2,\quad V_1(P,Q)=P,\quad V_n(P,Q)=P\,V_{n-1}(P,Q)-Q\,V_{n-2}(P,Q)\] for \(n\geq2\). The first terms of \(V_n(1,-1)\) are \[2,1,3,4,7,11,18,29,47,76, ...\] The terms of \(V_n(1,-1)\) are known as Lucas numbers. In the same paper of [3], it was shown that the only perfect powers in \(V_n(1,-1)\) are \(1\) and \(4\). Other well-known examples of Lucas sequences are Pell numbers, corresponding to \(U_n(2,-1)\), and the Pell-Lucas numbers, corresponding to the companion sequence \(V_n(2,-1)\). In 1991, Pethő [15] proved that the only perfect powers among Pell numbers are \(0,1\) and \(169\) (a substantially simpler proof was later found by Cohn [16]), while in 2015, Bravo, Das, Guzmán, and Laishram [17] showed that there are no perfect powers among Pell-Lucas numbers. Other results on perfect powers in linear recurrence sequences can be found in [18]–[26].
The characteristic polynomial of the sequences \(U(P,Q)\) and \(V(P,Q)\) is \[\lambda(x)=x^2-Px+Q\] with the roots \[\alpha=\frac{P+\sqrt{D}}{2} \quad\text{and}\quad \beta=\frac{P-\sqrt{D}}{2},\] where \(D\) is the discriminant of \(\lambda(x)\). The Binet formulas show closed-form expressions for the \(n\)-th term of Lucas sequences. In particular \[U_n(P,Q) = \frac{\alpha^n - \beta^n}{\alpha - \beta} \quad\text{and}\quad V_n(P,Q) = \alpha^n + \beta^n.\] Thus, the problem of determining perfect powers in Lucas sequences, via their Binet formulas, reduces to the study of Diophantine equations of the form \[x^m=U_n(P,Q) \quad \text{or} \quad x^m=V_n(P,Q),\quad x\geq0,~m\geq2.\] Depending on the choice of parameters \(P\) and \(Q\), the complexity of such equations varies considerably. For example, in the Fibonacci case, the problem reduces to the Diophantine equation \[x^m=\frac{1}{\sqrt{5}}\left(\left(\frac{1+\sqrt{5}}{2}\right)^n - \left(\frac{1-\sqrt{5}}{2}\right)^n\right).\] The proof that this equation admits integer solutions with \(m \geq 2\) only for \(n=0,1,2,6,12\) required an innovative combination of lower bounds for linear forms in three logarithms with the modular approach, extending techniques that were originally developed in the proof of Fermat’s Last Theorem. In contrast, the \(n\)-th term of the sequence \(U_n(3,2)\) is given by \(M_n=2^n - 1\), which is the sequence of Mersenne numbers. The corresponding Diophantine equation \[x^m=2^n - 1\] has no solutions with \(m \geq 2\). This follows directly from Mihăilescu’s proof of Catalan’s conjecture [27], which states that the only non-trivial consecutive perfect powers are \(3^2\) and \(2^3\).
In this paper, we consider Lucas sequences with \(P=1\) and \(Q=-2\). The terms of \(U_n(1,-2)\) are called Jacobsthal numbers (after German mathematician Ernst Jacobsthal) and are denoted by \(J_n\), while the terms of \(V_n(1,-2)\) are called Jacobsthal-Lucas numbers and are denoted by \(j_n\). The Jacobsthal numbers satisfy the recurrence relation \[J_0=0,~J_1=1,\quad J_n=J_{n-1}+2J_{n-2},\quad n\geq 2.\] The sequence \(J_n\) begins with \[\label{Jl1} 0,1,1,3,5,11,21,43,85,171,\ldots,\tag{1}\] and admits the Binet formula \[\label{nth} J_n=\frac{2^n-(-1)^n}{3}.\tag{2}\] The Jacobsthal-Lucas numbers satisfy the recurrence relation \[j_0=2,~j_1=1,\quad j_n=j_{n-1}+2j_{n-2},\quad n\geq 2.\] The sequence \(j_n\) begins with \[\label{jl1} 2,1,5,7,17,31,65,127,257,511,\ldots,\tag{3}\] and admits the Binet formula \[\label{jl2} j_n=2^n+(-1)^n.\tag{4}\] It is known that Jacobsthal and Jacobsthal-Lucas numbers have applications in coding theory and graph theory [28]–[30]. Numerical properties of these sequences can be found in [31], [32], while various generalizations are discussed in [33]–[35]. Several Diophantine problems involving Jacobsthal numbers have been studied in [36], [37].
In the present work, we contribute to the research on perfect powers in linear recurrence sequences by investigating separately the Diophantine equations \[x^m = J_n\quad\text{and}\quad x^m = j_n.\] In fact, due to Mihăilescu’s [27] proof of Catalan’s conjecture, we know that the equation \(x^m=2^n + (-1)^n\) has no solutions in positive integers for \(m\geq 2\). In view of 3 and 4 , this means that the only perfect power among Jacobsthal-Lucas numbers is \(j_1=1\), corresponding to \(1^m=2^1-1\). We prove that the analogous conclusion holds for the Jacobsthal numbers, namely that the only perfect powers are the trivial ones. In view of 2 , this is a consequence of the following result.
Theorem 1. The Diophantine equation \[3x^m=2^n-(-1)^n\] has no integer solution \((x,m,n) \in \mathbb{Z}^3\) with \(m\geq2\) and \(n\geq3\).
By inspecting the values of \(J_n\) for \(n=0,1,2\) in 1 , we obtain the following corollary.
Corollary 2. The only perfect powers among the Jacobsthal numbers are \(0\) and \(1\).
The main case of Theorem 1 is to prove that for \(m\geq2\) the Diophantine equation \[3x^m=2^n+1\] has no integer solutions when \(n\geq3\) is odd. By transforming this equation into \[\label{me} 3x^m-2^r\cdot (2^t)^m=1^2,\tag{5}\] we treat it in Lemma [wiles] by applying the results of Bennett and Skinner [38] on ternary Diophantine equations of the form \[\label{put} Ax^m+By^m=Cz^2.\tag{6}\] In their pioneering work, Bennett and Skinner showed that to any primitive solution \((x,y,z)=(a,b,c)\) of 6 with \(m\geq7\) an odd prime, one can associate an elliptic curve \(E\), called a Frey curve, whose discriminant is an \(m\)-th power up to a constant factor. Then, via Ribet’s level-lowering techniques, they further showed that the corresponding Galois representation arises from a cuspidal newform of weight \(2\) and some relatively small level \(N\). For the equation 5 related to Jacobsthal numbers, we obtain \(N=6\). Since it is known that there are no newforms of weight \(2\) and level \(6\), it follows that there is no non-trivial solution to 5 . The cases with small exponents, namely \(m<7\), are handled by reducing them to Thue equations and solving them with a computer algebra system, similarly as in [3].
The paper is structured as follows. In Section 2, we prove two key lemmas, which allow for a more concise and transparent proof of Theorem 1 in Section 3.
If \(n\) is odd, the \(n\)-th Jacobsthal number is given by \[J_n=\frac{2^n+1}{3}.\] Thus, if \(x^m=J_n\), then the problem converts to the study of Diophantine equation \[3x^m-2^n=1.\] The following lemma treats the case \(n=m\).
Let \(m\ge 5\) be a prime. Then the Diophantine equation \[\label{smkp} 3x^m-y^m=1\tag{7}\] has no integer solutions \((x,y)\in\mathbb{Z}^2\) with \(xy\neq0\).
Proof. Following Cohen [39], we refer to Diophantine equations of the form \[\label{smke} x^m+L^r y^m+z^m=0,\tag{8}\] where \(r<m\) and \(m\ge5\) is prime, as Serre-Mazur-Kraus equations. According to [39], the solution \((x,y,z)=(a,b,c) \in \mathbb{Z}^3\) to 8 is called non-trivial primitive if \(abc\neq0\) and \(a,b,c\) are pairwise coprime. From [39] follows that equation 8 has no non-trivial primitive solutions for \(L=3\). Now suppose on the contrary, that \((x,y)=(a,b)\) is a solution to 7 with \(ab\neq0\). Thus \[3a^m-b^m=1,\] which can be rewritten as \[(-b)^m+3^1\cdot a^m+(-1)^m=0.\] However, that implies that \((x,y,z)=(-b,a,-1)\) is a non-trivial primitive solution to 8 with \(L=3\) and \(r=1\), a contradiction. Hence the proof is complete. ◻
The following lemma is necessary to tackle the case \(n\ne m\). Its proof proceeds via the modular approach following the algorithm described by Bennett and Skinner [38] to systematically investigate Diophantine equations of the form \[\label{ter} Ax^m+By^m=Cz^2,\tag{9}\] where \(m\ge 7\) is prime, and \(A, B, C\) are fixed non-zero pairwise coprime integers. The ideas underlying the modular approach rely on deep results of Wiles [40], Ribet [41], Mazur [42], and others (see, e.g., [43]–[45]). Accordingly, the introduction to the methods used requires a substantial amount of background. Since our argument is essentially a direct application of the algorithm, we do not reproduce the definitions or technical details here and instead refer the reader to [38].
According to [38], a solution \((x,y,z)=(a,b,c)\in \mathbb{Z}^3\) to 9 is called primitive if \(Aa\), \(Bb\), and \(Cc\) are all non-zero and pairwise coprime.
Let \(r\) be a fixed positive integer, and let \(m\ge 7\) be prime. Then the Diophantine equation \[\label{mn} 3x^m-2^r y^m=z^2\tag{10}\] has no primitive integer solutions \((x,y,z)\in\mathbb{Z}^3\) such that \(xy\neq1\) and \(y\) even.
Proof. Observe that 10 matches the form of 9 with \[A=3,\quad B=-2^r,\quad C=1.\] Suppose on the contrary, that there exist a primitive integer solution \((x,y,z)=(a,b,c)\) to 10 such that \(ab\neq1\) and \(b\) is even . Thus \[\label{sol} 3a^m-2^r b^m=c^2.\tag{11}\] According to the classification given in [38], equation 11 falls into class (v), characterized by the properties \[\operatorname{ord}_2(Bb^m)\geq6\quad\text{and}\quad c\equiv C \pmod 4.\] The condition \(\operatorname{ord}_2(Bb^m)\geq6\) is satisfied due to the assumption that \(b\) is even, and \(m\geq7\). With respect to \(c\), note that \(c^2\) is necessarily odd (due to \(3a^m\) and \(2^rb^m\) being coprime) so that if \(c\) is not equal to \(1\) modulo \(4\), then \(c\) can be substituted for \(-c\) to satisfy the condition \(c\equiv C \pmod 4\). The algorithm proceeds further by associating to equations from class (v) a Frey elliptic curve \[E~:~Y^2+XY=X^3+\frac{cC-1}{4}X^2+\frac{BCb^m}{64}X\] which in our case becomes \[\label{e3} E~:~Y^2+XY=X^3+\frac{c-1}{4}X^2-\frac{2^r b^m}{64}X.\tag{12}\] To the elliptic curve \(E\), according to [38] we associate a number \(N_m(E)\) (related to the conductor \(N(E)\) of \(E\)), which in our case is obtained as follows \[\label{lvl} N_m(E)=\prod_{p|C,~ p\neq m} p^2 \cdot \prod_{q|AB, ~q \neq m} q=3\cdot 2 =6,\tag{13}\] since in our case \(C=1\) and \(AB=-3\cdot 2^r\) with \(r\geq1\). Let \(\rho_m^E\) be a Galois representation \[\rho_m^E:\operatorname{Gal}(\overline{\mathbb{Q}}/{\mathbb{Q}}) \to \operatorname{GL}_2(\mathbb{F}_m)\] arising from the action on the \(m\)-torsion points \(E[m]\) of \(E\). Since \(m\geq7\) is prime, and \(ab\neq1\), [38] implies that the representation \(\rho_m^E\) is irreducible. Consequently, by [38], the representation \(\rho_m^E\) arises from a cuspidal newform of weight \(2\), level \(N_m^E\), and trivial Nebentypus character. From 13 we have that \(N_m^E=6\). However, this leads to a contradiction as there are no newforms of weight \(2\) and level \(6\) [38]. Hence the proof is complete. ◻
Proof of Theorem 1. Suppose on the contrary, that Diophantine equation \[\label{de} 3x^m=2^n-(-1)^n\tag{14}\] has at least one integer solution \((x,m,n) \in \mathbb{Z}^3\) with \(m\geq2\) and \(n\geq3\).
We begin with the case when \(n\) is even. Thus, 14 becomes \(3x^m=2^n-1\). Since \(n\geq3\), we have that \(n=2k\) for some positive integer \(k>1\). Consequently \[3x^m=(2^{k}-1)(2^{k}+1).\] It is clear that \(x\) must be odd. Hence \(2^{k}-1\) and \(2^{k}+1\) are coprime. Therefore \[2^{k}-1=3^sx_1^m\quad\text{and}\quad 2^{k}+1=3^{1-s} x_2^m,\] where \(s\in \{ 0,1\}\). Consequently, either \(2^k-1=x_1^m\) or \(2^k+1=x_2^m\). From Mihăilescu’s proof of Catalan’s conjecture [27] it follows that the only possible solution to either of these equations is \(2^3+1=3^2\). However, if \(k=3\), then \(x^m=21\), which is impossible due to \(m\geq2\). This settles the case when \(n\) is even.
We continue with the case when \(n\) is odd. Thus, 14 becomes \[\label{refq} 3x^m=2^n+1.\tag{15}\] Since \(2^n+1 \equiv 1 \pmod{4}\) it follows that \(3x^m\equiv 1 \pmod{4}\). This, and the fact that \(x\) is clearly odd, imply that \(m\) is odd too. If \(m\) is not prime, then \(x^m=(x^l)^p\) for some prime odd divisor of \(m\), which means that from any solution \((x,m,n)\) one can obtain a solution \((x^l,p,n)\). Therefore, it is sufficient to prove that 15 has no solutions of the required type, when \(m\) is an odd prime. Also, due to \(3x^m=2^n+1\geq 2^2+1=5\), and the fact that \(x\) is odd, we can assume that \(x>2\). Thus \[2^n+1=3x^m > 3 \cdot 2^m.\] Therefore, \(n>m\). Write \(n=tm+r\), where \(t\geq1\) is some positive integer and \(r \in \{0, \ldots, m-1\}\). Then 15 can be rewritten as \[\label{rew} 3x^m-2^r\cdot (2^t)^{m}=1.\tag{16}\] First we deal with the case, when \(m\geq7\) (also recall that \(m\) can be assumed to be prime). If 16 holds for \(r=0\) with some \(x=a\), then \((x,y)=(a,2^t)\) is a solution to \[\label{sup1} 3x^m-y^m=1\tag{17}\] with \(xy\neq0\), which was shown to be impossible in Lemma [smk]. If 16 holds with \(r \in \{1,\ldots, m-1\}\) with some \(x=a\), then \((x,y,z)=(a,2^t,1) \in \mathbb{Z}^3\) is a solution \[\label{sup2} 3x^m-2^r y^m=z^2\tag{18}\] with \(xy\neq1\) and \(y\) even. Moreover, this solution is primitive, as as \(3a, 2^r\cdot 2^t, 1\) and are clearly non-zero and pairwise coprime (recall that \(a\) is odd and \(t\geq1\)). However, this is again a contradiction, as in Lemma [wiles] it was demonstrated, that such solutions do not exist.
To conclude the proof, it remains to consider equation 16 when \(m<7\) is an odd prime, i.e, \(m \in \{3,5\}\). In order to treat these cases it suffices to solve
eight Thue equations with any suitable computer algebra system. Indeed, if \(m=5\), then we have five equations to consider, namely \[3x^5-2^r\cdot y^5=1,~~\text{where}~~r=0,1,2,3,4.\] If
\(m=3\), then we have three equations \[3x^3-2^r\cdot y^3=1,~~\text{where}~~r=0,1,2.\] By solving these equations in the computer algebra system PARI/GP (for example, \(3x^5-4y^5=1\) is solved using the command thue(3*x^5-4,1)) we find that the only possible solutions \((x,y)\in\mathbb{Z}^2\) that satisfy at least one of these equations are \((1,1)\), \((-1,-1)\) or \((0,-1)\). However, since \(y=2^t\) with \(t\geq1\), we always have
\(2^t\neq \pm 1\). This completes the proof of the theorem. ◻
Proof of Corollary 2. The \(n\)-th Jacobsthal number is given by the formula \[J_n=\frac{2^n-(-1)^n}{3},\quad n=0,1,2\ldots\] If \(x^m=J_n\) for some \(m\geq2\), then \[3x^m=2^n-(-1)^n.\] By Theorem 1, this equation has no integer solutions with \(m\geq2\) and \(n\geq3\). Thus, no \(J_n\) is a perfect power when \(n\geq3\). On the other hand \[J_0=0,~J_1=1,~J_2=1.\] Therefore, \(0\) and \(1\) are the only perfect powers among the Jacobsthal numbers. ◻