On the equation ∑j=1kjFjp=Fnq

doi:10.1016/j.jnt.2020.05.005

Journal of Number Theory

Volume 217, December 2020, Pages 256-277

https://doi.org/10.1016/j.jnt.2020.05.005 Get rights and content

Abstract

In this paper, we show that the title equation, where $F_{m}$ is the mth Fibonacci number, in positive integers $(k, n, p, q)$ with $k > 1$ entails $\max {k, n, p, q} \leq 10^{2500}$ .

Introduction

Let ${F_{n}}_{n \geq 0}$ be the Fibonacci sequence given by $F_{0} = 0, F_{1} = 1$ and the recurrence $F_{n + 2} = F_{n + 1} + F_{n}$ for all $n \geq 0$ . A number of recent papers have studied exponential diophantine equations involving Fibonacci numbers. For example, the equation $F_{n}^{x} + F_{n + 1}^{x} = F_{m}^{y}$ in positive integers $n, m, x, y$ has been studied in [5], [6] and [8]. The equation $F_{n}^{x} + \dots + F_{n + k - 1}^{x} = F_{m}$ in positive integers $n, k, m, x$ has been studied in [7] while the equation $F_{1}^{k} + \dots + F_{n - 1}^{k} = F_{n + 1}^{ℓ} + \dots + F_{n + r}^{ℓ}$ in positive integers $n, r, k, ℓ$ has been studied in [2] and [3]. For all these equations, all their positive integer solutions are now known. The title equation has been first studied, to our knowledge, in the paper [10]. Since $F_{1} = F_{2} = 1$ , the title equation has the solutions $(k, n, p, q) = (1, 1, p, q), (1, 2, p, q)$ (for any p and q), as well as $(2, 4, p, 1)$ for any p. We call such solutions trivial. From now on, we assume that $k \geq 3$ . In that paper the authors suggested the following conjecture:

Conjecture 1.1

The only nontrivial solutions of the title equation are given by $(k, n, p, q) = (3, 4, 1, 2), (3, 4, 3, 3), (4, 8, 1, 1)$ .

In [10], the authors found all solutions when ${p, q} \subseteq {1, 2}$ by ad-hoc methods. A general method to find all solutions of the title equation when p and q are given was proposed in [4]. As an application, the authors of [4] confirmed Conjecture 1.1 for the range of the exponents ${p, q} \subset [1, 10]$ . None of the above papers addressed the question of whether the title equation has only finitely many positive integers solutions and whether the current methods allow us to bound them or even compute them. The aim of our paper is to prove that there are only finitely many effectively computable solutions. Our concrete result is the following.

Theorem 1.1

The title equation has only finitely many positive integer solutions $(k, n, p, q)$ with $k \geq 3$ . They all satisfy $\max {k, n, p, q} \leq 10^{2500}$ .

We made no efforts to try to reduce the above bound. In the last section of the paper we explain why we believe that the computations are not feasible and that new ideas will be required to confirm computationally Conjecture 1.1.

Section snippets

The tools

Most arguments on exponential diophantine equations go via some linear forms in logarithms of algebraic numbers. We are no exception to this. So, let us recall the terminology and the results we need.

Let η be an algebraic number of degree d with minimal primitive polynomial over the integers $a_{0} x^{d} + a_{1} x^{d - 1} + \dots + a_{d} = a_{0} \prod_{i = 1}^{d} (x - η^{(i)}),$ where the leading coefficient $a_{0}$ is positive and the $η^{(i)}$ s are the conjugates of η. Then the logarithmic height of η is given by $h (η) : = \frac{1}{d} (\log a_{0} + \sum_{i = 1}^{d} \log (\max {| η^{(i)} |, 1})) .$

The set up

Recall that the equation is $\sum_{j = 1}^{k} j F_{j}^{p} = F_{n}^{q} .$ As we already mentioned, we take $k \geq 3$ . Thus, $n \geq 3$ also holds. Put $M : = \sum_{j = 1}^{k} j F_{j}^{p} = F_{n}^{q} and X : = \log M .$ We want to bound X. We write $(α, β) : = (\frac{1 + \sqrt{5}}{2}, \frac{1 - \sqrt{5}}{2})$ for the roots of the characteristic equation $x^{2} - x - 1$ of the Fibonacci sequence. Then it is well-known that $F_{ℓ} = \frac{α^{ℓ} - β^{ℓ}}{α - β} holds for all ℓ \geq 0 .$ This is sometimes referred to as the Binet formula. We also use the fact that the inequalities $α^{ℓ - 2} \leq F_{ℓ} \leq α^{ℓ - 1}$ hold for all $ℓ \geq 1$ . At one point we will need the companion Lucas sequence ${L_{ℓ}}_{ℓ}$

The sizes of kp versus nq

Lemma 4.1

We have $10^{- 1} X < k p < 10 X and 10^{- 1} X < n q < 10 X .$

Proof

We have ${(α^{k / 3})}^{p} \leq {(α^{k - 2})}^{p} < F_{k}^{p} < M; M < k {(\sum_{j = 1}^{k} F_{j})}^{p} < k {(F_{k + 2})}^{p} < k {(α^{k + 1})}^{p} < α^{(k + 1) p + 3 \log k} < α^{5 k p} .$ In the above, we used the identity $F_{1} + F_{2} + \dots + F_{k} = F_{k + 2} - 1 < F_{k + 2} .$ Since $1 / \log α \in (2, 3)$ , it follows that $k p / 3 < 3 M and 2 M < 5 k p,$ which gives the desired bounds for kp versus M. Similarly, ${(α^{n / 3})}^{q} \leq {(α^{n - 2})}^{q} < F_{n}^{q} = M < {(α^{n - 1})}^{q} < α^{n q},$ so $n q / 3 < 3 M$ and $2 M < n q$ , which gives the desired bounds for nq versus M. □

The linear forms

Our proof exploits five linear forms in logarithms together with their lower bounds (by linear forms in logarithms) provided that they are nonzero: $\begin{matrix} Γ_{1} & : = & k F_{k}^{p} F_{n}^{- q} - 1; & \log | Γ_{1} | & > & - c_{1} k n {(\log 10 X)}^{2}; \\ Γ_{2} & : = & k F_{k}^{p} α^{- n q} {\sqrt{5}}^{q} - 1; & \log Γ_{2} | & > & - c_{1} k {(\log 10 X)}^{2}; \\ Γ_{3} & : = & α^{p (k + 1)} (\frac{k α^{p} - (k + 1)}{5^{p / 2} {(α^{p} - 1)}^{2}}) F_{n}^{- q} - 1; & \log | Γ_{3} | & > & - c_{1} p n {(\log 10 X)}^{2}; \\ Γ_{4} & : = & (\frac{k α^{p} - (k + 1)}{{(α^{p} - 1)}^{2}}) 5^{(q - p) / 2} α^{p (k + 1) - q n} - 1; & \log | Γ_{4} | & > & - c_{1} p {(\log 10 X)}^{2}; \\ Γ_{5} & : = & k α^{k p - q n} 5^{(q - p) / 2} - 1; & \log | Γ_{5} | & > & - c_{1} {(\log 10 X)}^{2} . \end{matrix}$ We will justify the fact that they are non-zero later. For now, let us assume that they are.

The nonvanishing of the first 4 forms

It is easy to see that $Γ_{1}, Γ_{2}$ are nonzero. Indeed, since $\sum_{j = 1}^{k - 1} j F_{j}^{k} = - (k F_{k}^{p} - F_{n}^{q})$ and $k > 1$ , it follows that $Γ_{1} < 0$ . If $Γ_{2} = 0$ , then $α^{2 n q} = {(k F_{k}^{p})}^{2} 5^{q} \in Q$ , a contradiction.

For $Γ_{3}$ , assuming that it is zero, we get $k α^{p} - (k + 1) = F_{n}^{q} Δ_{p}^{2} 5^{p / 2} α^{- p (k + 1)},$ where $Δ_{p} : = α^{p} - 1$ . Taking norms in $Q [\sqrt{5}]$ , we get $| {(- 1)}^{p} - L_{p} + 1) k^{2} + (2 - L_{p}) k + 1 | = F_{n}^{2 q} 5^{p} | {(- 1)}^{p} - L_{p} + 1 |^{2} .$ For p even, we get $| (L_{p} - 2) k^{2} + (L_{p} - 2) k - 1 | = F_{n}^{2 q} 5^{p} | L_{p} - 2 |^{2} .$ The case $L_{p} - 2 = 0$ is impossible since it leads to $1 = 0$ . Thus, both sides above are nonzero. If p is odd, we get $| L_{p} k^{2} + (L_{p} - 2) k - 1 | = F_{n}$

The terminology

Definition

We say that $F_{m}^{j} = {(\frac{α^{m}}{\sqrt{5}})}^{j} {(1 + \frac{{(- 1)}^{m}}{α^{2 m}})}^{j}$ is “expandable”, if the right parenthesis above satisfies ${(1 + \frac{{(- 1)}^{m}}{α^{2 m}})}^{j} = 1 + ζ_{j, m}, with | ζ_{j, m} | < \frac{1}{α^{1.5 m}} .$

Lemma 7.1

If $κ > 0$ is any constant and $j < m^{κ}$ , then estimate (11) above holds provided $m > c_{2} : = c_{2} (κ)$ , where $c_{2} (κ)$ is such that $2 m^{κ} < α^{0.5 m} holds for all m > c_{2} (κ) .$

Proof

Suppose that m is even. Then ${(1 + \frac{{(- 1)}^{m}}{α^{2 m}})}^{j} < {(1 + \frac{1}{α^{2 m}})}^{j} < \exp (\frac{j}{α^{2 m}}) < 1 + \frac{2 j}{α^{2 m}},$ provided $j / α^{2 m} < 1 / 2$ . Thus, we need $2 j / α^{2 m} < 1 / α^{1.5 m}$ , or $2 j < α^{0.5 m}$ , so $2 m^{κ} < α^{0.5 m}$ , which clearly holds for $m > c_{2} (κ)$ . A similar argument works for m odd. □

The strategy

We will start with the linear form $Γ_{1}$ which is small. We look for positive constants $κ_{1}, κ_{2}, κ_{3}$ all smaller than 1 such that $k > X^{κ_{1}}, p > X^{κ_{2}}, n > X^{κ_{3}},$ for $X > X_{0}$ . For this subsection we shall ignore the implied constants (but we will make them explicit when time comes) and we use the Vinogradov symbols ≫ and ≪. If we succeed say first that $k > X^{κ_{1}}$ , it then follows, by Lemma 4.1, that $p ≪ X / k ≪ X^{1 - κ_{1}} ≪ k^{(1 - κ_{1}) / κ_{1}}$ so, by Lemma 7.1, $F_{k}^{p}$ is expandable. This allows us to transform first $Γ_{1}$ into $Γ_{3}$ . If this further

The value of $κ_{1}$

Lemma 9.1

We can take $κ_{1} = 1 / 4$ for $X > 10^{200}$ .

Proof

We write $X_{0}$ for some number increasing from one iteration to the next which is a lower bound on X resulting from some inequality. At the end we collect the largest $X_{0}$ that we have encountered along the way. We assume that $k < X^{1 / 4}$ and we search for an acceptable $X_{0}$ . From Lemma 4 in [4], we have that $F_{j} / F_{j + 1} \leq 2 / 3$ for all $j \geq 2$ . In particular, $F_{k - j} / F_{k} \leq {(2 / 3)}^{j}$ for all $j = 1, 2, \dots, k - 2$ , while for $j = k - 1$ this must be replaced by $F_{1} / F_{k} \leq {(2 / 3)}^{k - 2}$ . Thus, in the left-hand side of (3),

The value of $κ_{2}$

Lemma 10.1

We can take $κ_{2} = 1 / 9$ for $X > 10^{2000}$ .

Proof

We follow the same convention about $X_{0}$ as in Lemma 9.1. We assume that $p < X^{1 / 9}$ . We use Lemma 9.1 and Lemma 7.1 to conclude that $F_{j}^{p}$ is expandable for all $j \geq ℓ : = ⌊ k / 2 ⌋ + 1 (\geq 2)$ . Indeed, for us, $p < 10 X / k < 10 k^{3} = 100 {(k / 2)}^{3} < {(k / 2)}^{4},$ since $k / 2 > (1 / 2) X^{1 / 4} > 100$ for $X > 10^{200}$ . Thus, it suffices, by Lemma 7.1, that $κ / 2 > c_{2} (4)$ , where $c_{2} (4)$ is such that $2 n^{4} < α^{0.5 n} holds for all n > c_{2} (4),$ and we can take $c_{2} (4) : = 100$ . Thus, $F_{j}^{p} = \frac{α^{p j}}{5^{p / 2}} (1 + O (\frac{1}{α^{k / 2}})) for j \in [ℓ, k] .$ The constant inside O can be taken to be

The value of $κ_{3}$

Lemma 11.1

We have $κ_{3} = 1 / 10$ for $X > 10^{2499}$ .

Proof

We return to estimates (13), use Lemma 7.1 and the fact that $F_{k}^{p}$ is expandable. Indeed, for this note that $p < 10 X / k < 10 k^{3} < k^{4},$ and one checks that $k > c_{2} (4)$ in our range for $k > X_{0}^{1 / 4}$ . Thus, we get that $M = k F_{k}^{p} (1 + O (\frac{1}{{1.5}^{p}})) = \frac{k α^{p k}}{5^{p / 2}} (1 + O (\frac{1}{α^{k}})) (1 + O (\frac{1}{{1.5}^{p}})) = \frac{k α^{p k}}{5^{p / 2}} (1 + O (\frac{1}{{1.5}^{\min {p, k}}})) .$ The constant inside the O can be taken to be 3. Thus, equation (3) is $\frac{k α^{p k}}{5^{p / 2}} (1 + O (\frac{1}{{1.5}^{\min {p, k}}})) = F_{n}^{q},$ and can be rewritten as $| k α^{p k} 5^{- p / 2} F_{n}^{- q} - 1 | = O (\frac{1}{{1.5}^{\min {p, k}}}),$ where the constant inside the last

The case when $Γ_{5} \neq 0$

Lemma 12.1

We have $X \leq 10^{2499}$ provided $Γ_{5} \neq 0$ .

Proof

Assume $X > 10^{2499}$ . By Lemma 11.1, $F_{n}^{q}$ is also expandable. Indeed we can check easily that $q < 10 X / n < 10 n^{9} < n^{10}$ and $n > c_{2} (10)$ in our range. So, equation (26) is $\frac{k α^{k p}}{5^{p / 2}} (1 + O (\frac{1}{{1.5}^{\min {p, k}}})) = F_{n}^{q} = \frac{α^{q n}}{5^{q / 2}} (1 + O (\frac{1}{α^{n}})) .$ The constants inside both O can be taken to be 10. This can be rearranged as $| k α^{p k - q n} 5^{(q - p) / 2} - 1 | = O (\frac{1}{{1.5}^{\min {k, n, p}}}),$ where the constant inside the last O can be taken to be 100. In the left–hand side, we recognise $Γ_{5}$ which we assume it is nonzero. Thus, $\log | Γ_{5} | ≪ - \min$

The case when $Γ_{5} = 0$

Lemma 13.1

If $Γ_{5} = 0$ , then $X \leq 10^{2499}$ .

Before giving the proof, note that this finishes the proof of our theorem since by Lemma 4.1, we have $\max {k, n, p, q} < 10 X < 10^{2500}$ .

Proof

This is messy. We write $X_{0} : = 10^{2499}$ . The condition that $Γ_{5} = 0$ implies the equality $α^{k p - n q} = (5^{(p - q) / 2} / k)$ . Squaring this we get $α^{2 (k p - n q)} = 5^{p - q} / k^{2} \in Q$ . The only possibility is that both sides of the above equality are 1. Thus, $k p = n q$ and $k = 5^{(p - q) / 2}$ . In particular, $p > q$ , $p - q$ is even and $p - q = O (\log (10 X))$ , where the constant in the O can be taken to be 10. We

Comments on computations

We did not make any attempts to reduce the bounds. However, we believe new ideas will be needed to lower the bounds to the range where one can just enumerate the solutions. Indeed, in order to reduce the variables, in the most fortunate case where $Γ_{5} \neq 0$ , one is lead to a final inequality of the type $| x \log α - y \log 5^{1 / 2} + \log k | ≪ \exp {- δ z},$ where $(x, y, z) : = (k p - q n, q - p, \min {k, p, q}),$ and $δ > 0$ in (30) is some small number. At this step, one applies a reduction method due to Baker and Davenport which only

Acknowledgments

We thank the referee for comments which helped us improve the quality of our paper.

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General SectionOn the equation ∑j=1kjFjp=Fnq

Abstract

Introduction

Section snippets

The tools

The set up

The sizes of kp versus nq

The linear forms

The nonvanishing of the first 4 forms

The terminology

The strategy

The value of κ1

The value of κ2

The value of κ3

The case when Γ5≠0

The case when Γ5=0

Comments on computations

Acknowledgments

Logarithmic forms and group varieties

J. Reine Angew. Math.

Balancing with Fibonacci powers

Fibonacci Q.

On a conjecture regarding balancing with powers of Fibonacci numbers

Integers

On a diophantine equation involving powers of Fibonacci numbers

Proc. Jpn. Acad., Ser. A, Math. Sci.

On the Diophantine equations Fn+Fn+1x=Fmy

Rocky Mt. J. Math.

General Section
On the equation $\sum_{j = 1}^{k} j F_{j}^{p} = F_{n}^{q}$

The value of $κ_{1}$

The value of $κ_{2}$

The value of $κ_{3}$

The case when $Γ_{5} \neq 0$

The case when $Γ_{5} = 0$

On the Diophantine equations $F_{n}^{+} F_{n + 1}^{x} = F_{m}^{y}$