Diophantine triples in linear recurrences of Pisot type
Abstract
The study of Diophantine triples taking values in linear recurrence sequences is a variant of a problem going back to Diophantus of Alexandria which has been studied quite a lot in the past. The main questions are, as usual, about existence or finiteness of Diophantine triples in such sequences. Whilst the case of binary recurrence sequences is almost completely solved, not much was known about recurrence sequences of larger order, except for very specialised generalisations of the Fibonacci sequence. Now, we will prove that any linear recurrence sequence with the Pisot property contains only finitely many Diophantine triples, whenever the order is large and a few more not very restrictive conditions are met.
Keywords
Diophantine triples Linear recurrence sequences Diophantine equations Application of the Subspace theoremMathematics Subject Classification
Primary 11D72 11B39 Secondary 11J871 Introduction
Now it is an interesting variation of the original problem of Diophantus to consider a linear recurrence sequence instead of the sequence of squares. So we ask for bounds m on the size of tuples of integers \(\{a_1, a_2, a_3, \dots , a_m\}\) with \(a_i a_j + 1\) being members of a given linear recurrence for \(1\le i<j\le m\). We shall call this set a Diophantine mtuple with values in the linear recurrence (or a Diophantine mtuple in the recurrences, for short). Here, the first result was due to Fuchs, Luca and Szalay, who proved in [12] that for a binary linear recurrence sequence \((u_n)_{n \ge 0}\), there are only finitely many Diophantine triples, if certain conditions are met. The Fibonacci sequence and the Lucas sequence both satisfy these conditions and all Diophantine triples with values in these sequences were computed in [22] and [23]. Further results in this direction can be found in [2, 20] and [21]. Moreover, in [1] it is shown that there are no balancing Diophantine triples; see also [3] for a related result. In [4] it is shown that there are no Diophantine triples taking values in Pellans sequence.
The result in this paper deals with a significantly larger class of linear recurrence sequences:
Let \((F_n)_{n\ge 0}\) be a sequence of integers satisfying a linear recurring relation. Assume that the recurrence is of Pisot type, i.e., that its characteristic polynomial is the minimal polynomial (over \(\mathbb {Q}\)) of a Pisot number. We denote the power sum representation (Binet formula) by \(F_n = f_1 \alpha _1^n + \cdots + f_k \alpha _k^n\). Assume w.l.o.g. that \(\alpha =\alpha _1\) is the Pisot number; i.e., \(\alpha \) is a real algebraic integer of degree k satisfying \(\alpha >1\) and if \(\alpha _2,\ldots ,\alpha _k\) denote the conjugates of \(\alpha \) over \(\mathbb {Q}\) then \(\max \{\alpha _2, \dots , \alpha _k\} < 1\). We remark that by a result of Mignotte (cf. [24]) it immediately follows that the sequence is nondegenerate, and that the characteristic roots are all simple and irrational.

Neither the leading coefficient \(f_1\) nor \(f_1\alpha \) is a square in \(K=\mathbb {Q}(\alpha _1, \dots , \alpha _k)\).

\(k \ge 2\) and \(\alpha \) is not a unit.

\(k\ge 4\).
For example, let us consider the irreducible polynomial \(X^3  X  1\), which has the Pisot property. Its Pisot root \(\theta := 1.3247179572 \dots \) is the smallest existing Pisot number by [6]. This number is also known as the plastic constant. Its corresponding linear recurrence sequence \((F_n)_{n \ge 0}\), given by \(F_{n+3} = F_{n+1} + F_n\), is of Pisot type. If the initial values are not \(F_0 = 6, F_1 = 9, F_2 = 2\), then neither the leading coefficient nor the leading coefficient times \(\theta \) are squares in the splitting field of \(X^3X1\) over \(\mathbb {Q}\). So the theorem can be applied and we obtain, that there are only finitely many Diophantine triples with values in this sequence. However it is yet not clear, what happens in the case \(F_0 = 6, F_1 = 9, F_2 = 2\).
We quickly discuss the main shape of the recurrences we study in this paper. Let \((F_n)_{n\ge 0}\) be a recurrence of Pisot type as described above. Let us denote \(K=\mathbb {Q}(\alpha _1,\ldots ,\alpha _k)\). Since \(F_n\in \mathbb {Z}\) it follows that each element of the Galois group of K over \(\mathbb {Q}\) permutes the summands in the power sum representation of \(F_n\). Moreover, each summand is a conjugate of the leading term \(f_1\alpha _1^n\) over \(\mathbb {Q}\) and each conjugate of it appears exactly once in the Binet formula. Therefore \(F_n\) is just the trace \({\text {Tr}}_{K/\mathbb {Q}}(f_1 \alpha _1^n)\). Since \(f_1\) might not be integral, we write \(f_1 = f/d\) with \(d \in \mathbb {Z}\) and f being an integral element in K. Thus, conversely starting with a Pisot number \(\alpha \), an integer \(d \in \mathbb {Z}\) and an integral element f in the Galois closure K of \(\alpha \) over \(\mathbb {Q}\) such that \(dF_n=\text{ Tr }_{K/\mathbb {Q}}(f\alpha ^n)\) for every \(n\in \mathbb {N}\), we can easily construct further examples for which our result applies.
The proof will be given in several steps: First, a more abstract theorem is going to be proved, which guarantees the existence of an algebraic equality, that needs to be satisfied, if there were infinitely many Diophantine triples. This works on utilizing the Subspace theorem (cf. [10]) and a parametrization strategy in a similar manner to that of [14]. If the leading coefficient is not a square in \(\mathbb {Q}(\alpha _1, \dots , \alpha _k)\), we obtain the contradiction quite immediately from this equality. In a second step, we will use divisibility arguments and algebraic parity considerations in order to show that this equality can also not be satisfied if the order k is large enough. Let us now state the results.
2 The results
We start with a general and more abstract statement which gives necessary conditions in case infinitely many Diophantine triples exist. It is derived by using the Subspace theorem (cf. [10]).
Theorem 1
The proof is given in Sect. 4.
This theorem looks quite abstract. However, it can be applied to a huge family of linear recurrences. Firstly, it can be applied to all linear recurrences, in which the leading coefficient is not a square:
Theorem 2
The proof of this theorem is given in Sect. 5.
Another consequence of Theorem 1 applies to linear recurrences of sufficiently large order. Namely if \(k \ge 4\), the existence of such a \(c(\ell )\) leads to a contradiction. The same holds already for \(k = 2,3\), if we assume that the Pisot element \(\alpha _1\) is not a unit in the ring of integers of \(\mathbb {Q}(\alpha _1,\ldots ,\alpha _k)\). Thus, we obtain the following result.
Theorem 3
 (i)
\(k\ge 2\) and \(\alpha _1\) is not a unit.
 (ii)
\(k\ge 4\).
This theorem is proved in Sect. 6.
Before we give the proofs we first start with several useful lemmas that will be used in the sections afterwards.
3 Some useful lemmas
Assume that we have infinitely many solutions \((x,y,z) \in \mathbb {N}^{3}\) to (1) with \(1<a<b<c\). Obviously, we have \(x< y < z\). First, one notices that not only for z, but for all three components, we necessarily have arbitrarily “large” solutions.
Lemma 1
Let us assume, we have infinitely many solutions \((x,y,z) \in \mathbb {N}^{3}\) to (1). Then for each N, there are still infinitely many solutions \((x,y,z) \in \mathbb {N}^{3}\) with \(x > N\).
Proof
It is obvious that we must have arbitrarily large solutions for y and for z, since otherwise, a, b, c would all be bounded as well, which is an immediate contradiction to our assumption.
If we had infinitely many solutions (x, y, z) with \(x < N\), then there is at least one fixed x which forms a solution with infinitely many pairs (y, z). Since \(F_x = ab + 1\), we have a bound on these two variables as well and can use the same pigeon hole argument again to find fixed a and b, forming a Diophantine triple with infinitely many \(c \in \mathbb {N}\).
Next, we prove the following result, which generalizes Proposition 1 in [13]. Observe that the upper bound depends now on k.
Lemma 2
Proof
The next lemma states the irreducibility (over \(\mathbb {C}\)) of a certain polynomial. This lemma will be used in the proof of Theorem 3.
Lemma 3
Proof
Corollary 1
Assume that \(k\ge 1\). If \(n\ge 2\), the polynomial \(c_1X_1^k+\cdots +c_nX_n^k1\) is irreducible.
Proof
Lemma 4
Proof
 (i)
f(X) is irreducible over \({\mathbb {C}}[X_2,\ldots ,X_n]\);
 (ii)
If \(\alpha \) is a root of f(X), then \(\alpha \) is not of the form \(\beta ^q\) for some element \(\beta \in {\mathbb {C}}(X_2,\ldots ,X_n)(\alpha )\) and any \(q\mid k\).
4 Proof of Theorem 1
The aim of this section is to prove Theorem 1.
Proof
We first show that if there are infinitely many solutions to (1), then all of them can be parametrized by finitely many expressions as given in (14) for c below. The arguments in this section follow the arguments from [13] and [14].
From now on, we assume w.l.o.g. that \(\alpha _1=\alpha _1 > \alpha _2 \ge \cdots \ge \alpha _k\).
We work with the field \(K={\mathbb {Q}}(\alpha _1,\ldots ,\alpha _k)\) and let S be the finite set of places (which are normalized so that the Product Formula holds, cf. [10]), that are either infinite or in the set \(\{ v \in M_K: \alpha _1_v \ne 1 \vee \cdots \vee \alpha _k_v \ne 1\}\). Observe that we may choose \(\alpha _1,\ldots ,\alpha _k\) in \(\mathbb {C}\) and therefore view K as a subfield of \(\mathbb {C}\). We denote by \(\vert \cdot \vert _\infty \) the unique place such that \(\vert \beta \vert _\infty =\vert \beta \vert =\sqrt{\mathfrak {R}(\beta )^2+\mathfrak {I}(\beta )^2}\) for all \(\beta \in \mathbb {C}\). According to whether \(x+y+z\) is even or odd, we set \(\epsilon = 0\) or \(\epsilon = 1\) respectively, such that \(\alpha _1^{(x+y+z\epsilon )/2} \in K\). By going to a still infinite subset of the solutions, we may assume that \(\epsilon \) is always either 0 or 1.
Likewise, we can find finite expressions of this form for a and b.
Next we use the following parametrization lemma:
Lemma 5
Proof
We may assume that infinitely many of the solutions \(\mathbf{x }\) are nondegenerate solutions of (16) by replacing the equation by a new equation given by a suitable vanishing subsum if necessary.
We may assume, that \((L_{1,i}, \dots , L_{\ell ,i}) \ne (L_{1,j}, \dots , L_{\ell ,j})\) for any \(i \ne j\), because otherwise we could just merge these two terms.
5 Linear recurrences with nonsquare leading coefficient
The aim of this section is to prove Theorem 2.
Proof
6 Linear recurrences of large order
We now prove Theorem 3.
Proof
(P) Parity: \(r_1+r_2+r_3\equiv 0\pmod 2\). This is clear from degree considerations since \(2{{\text {deg}}_{X_1}}(Q_h)={{\text {deg}}_{X_1}}(P_i)+{{\text {deg}}_{X_1}} (P_j){{\text {deg}}_{X_1}} (P_h)=r_i+r_jr_h\).
Observe now that by the corollary to Lemma 3 (proved in Sect. 3) we know that the polynomials \(P_i(\mathbf{X })\) are irreducible (as a polynomial in \(\mathbb {C}[\mathbf{X }]\)). We have that \(r_1\le r_2\le r_3\). From (21) for \((i,j,h)=(1,2,3)\) it follows that \(P_h(\mathbf{X })\) divides \(P_i(\mathbf{X })\) or \(P_j(\mathbf{X })\). By degree considerations, \(P_h(\mathbf{X })\) divides \(P_j(\mathbf{X })\) (otherwise \(r_1=r_2=r_3\) and we can take them all to be 1, which is impossible by (P)). So, again by degree considerations, \(P_h(\mathbf{X })\) and \(P_j(\mathbf{X })\) are associated and \(P_i(\mathbf{X })\) is a square which contradicts Lemma 3.
Notes
Acknowlegements
Open access funding provided by Austrian Science Fund. The authors are most grateful to an anonymous referee for very careful reading of the text and for pointing out several inaccuracies which are now corrected in the final version. Clemens Fuchs and Christoph Hutle were supported by FWF (Austrian Science Fund) Grant No. P24574 and by the Sparkling Science Project EMMA Grant No. SPA 05/172. Florian Luca was supported in part by Grant CPRR160325161141 and an Arated scientist award both from the NRF of South Africa and by Grant No. 1702804S of the Czech Granting Agency. Part of this work was done when this author was in residence at the Max Planck Institute for Mathematics in Bonn in 2017.
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