On sums of two Fibonacci numbers that are powers of numbers with limited Hamming weight

In 2018, Luca and Patel conjectured that the largest perfect power representable as the sum of two Fibonacci numbers is $3864^2 = F_{36} + F_{12}$. In other words, they conjectured that the equation \begin{equation}\tag{$\ast$}\label{eq:abstract} y^a = F_n + F_m \end{equation} has no solutions with $a\geq 2$ and $y^a>3864^2$. While this is still an open problem, there exist several partial results. For example, recently Kebli, Kihel, Larone and Luca proved an explicit upper bound for $y^a$, which depends on the size of $y$. In this paper, we find an explicit upper bound for $y^a$, which only depends on the Hamming weight of $y$ with respect to the Zeckendorf representation. More specifically, we prove the following: If $y = F_{n_1}+ \dots + F_{n_k}$ and equation \eqref{eq:abstract} is satisfied by $y$ and some non-negative integers $n,m$ and $a\geq 2$, then \[ y^a \leq \exp\left(C{(\varepsilon)} \cdot k^{(3+\varepsilon)k^2} \right). \] Here, $\varepsilon>0$ can be chosen arbitrarily and $C(\varepsilon)$ is an effectively computable constant.


Introduction
The Fibonacci numbers, defined by F 0 = 0, F 1 = 1 and F k+2 = F k+1 +F k for k ≥ 0, might be the most popular linear recurrence sequence of all.They have a great many beautiful properties and a vast amount of research has been done on problems involving Fibonacci numbers.For instance, it was a long-standing conjecture that 0, 1, 8 and 144 are the only Fibonacci Numbers that are perfect powers.This conjecture was proven in 2003 by Bugeaud, Mignotte and Siksek [3].In view of this result, it was a natural next step to search for all perfect powers that are sums of two Fibonacci numbers, i.e. to try and solve the equation (1) F n + F m = y a , where n, m, y, a are non-negative integers with a ≥ 2. There are 18 solutions known with n ≥ m ≥ 0, the largest being F 36 + F 12 = 3864 2 .In 2018, Luca and Patel [9] conjectured that these are the only solutions to equation (1).They proved their conjecture in the case that n ≡ m (mod 2).The general conjecture, however, remains open.Let us summarize further existing partial results on this conjecture.If m = 0, then we have F n = y a , which, as mentioned above, was solved in [3].For m = 1, 2 we have the equation F n + 1 = y a , which was solved by Bugeaud, Luca, Mignotte, Siksek in 2006 [2].For any fixed y it is in principal possible to solve equation (1) completely.For example, Bravo and Luca [1] solved the equation F n + F m = 2 a .In the general case with fixed y, explicit upper bounds for n, m and a in terms of y were established recently in [5] and in [6].Moreover, Kebli, Kihel, Larone and Luca [5] proved that the abc-conjecture implies that (1) has only finitely many solutions.Most recently, Ziegler [11] proved that for any fixed y equation (1) has at most one solution with a ≥ 1, unless y = 2, 3, 4, 6, 10.In particular, his result implies that if y can be represented as y = F n1 + F n2 , then equation (1) has no solutions with a ≥ 2 (with the exceptions y = 2, 3,4,6,10).
In this paper, we want to make another step towards solving equation (1), and generalize the above mentioned results in the following way: Instead of fixing y or requiring that it have the form y = F n1 + F n2 , we allow arbitrary y with bounded Hamming weight with respect to the Zeckendorf representation (i.e.y = F n1 + • • • + F n k with bounded k).More specifically, we give an explicit upper bound for any perfect power y a that is a sum of two Fibonacci numbers y a = F n + F m , and the upper bound does not depend on the size of y, but only on the Hamming weight of the Zeckendorf representation of y.We now state our main result.
Theorem 1.Let ε > 0. Then there exists an effectively computable constant C(ε) such that the following holds.If the equations are satisfied by some non-negative integers y, a, n 1 , . . ., n k , n, m with a ≥ 2, then (2) Remark 1.The constant C(ε) can indeed be computed from our proof.However, it will be extremely large and not useful in practice.This is because we chose to write the upper bound in a way that is both simple and asymptotically good.So if one chooses an ε > 0 and computes the constant C(ε) such that (2) holds for all k, the bound will be extremely bad for small k.
For computing an actual upper bound for a given k, we recommend going to equation ( 16) in Section 6 and computing the maximum of the expressions T k+1 over all 1 ≤ ℓ 0 ≤ k.Then, one can proceed as described in Remark 2 in Section 2.1 and solve the inequality n < 6 • 10 29 • T 4 k+1 .Let us outline the rest of the paper and the strategy of our proof.In Section 2, we state some preliminary results related to Fibonacci numbers and our problem, as well as lower bounds for linear forms in logarithms and an inequality.In Section 3, we construct a total of 2k "basic" linear forms in logarithms from equations (A) and (B).Each of these linear forms will contain the unknown logarithm log y.The main idea of the proof is the following: In several steps (k or k + 1 steps), we take two of the "basic" linear forms at a time and eliminate log y.Then we apply lower bounds for linear forms in logarithms to the new linear form and obtain an upper bound for one of the expressions n − m, n 1 − n 2 , . . ., n 1 − n k , n 1 .Depending on how large n − m is compared to n 1 − n 2 , . . ., n 1 − n k , n 1 , we need to do a slightly different succession of steps.An overview of these steps can be found in Figure 1 in Section 4. The exact bounds, that are obtained in each step, are computed in Section 5. We "walk the steps" in Section 6: Depending on which path of steps we walk, we end up with a different bound for n 1 in the last step, see Figure 2. In Section 6, we compute these bounds and find a common upper bound for n 1 of the shape n 1 < c • (log n) x .In particular, this implies log y < c • (log n) x .Finally, in Section 7 we combine this bound with the bound from [5], which is of the shape n < c(log y) 4 .Thus, we end up with an inequality of the shape n < c • (log n) x , which implies an upper bound for n (see Remark 2).Let us moreover point out that we use the result from [9] to exclude the case n ≡ m (mod 2).This is very helpful for checking that our linear forms in logarithms don't vanish.
Of course, the strategy of applying lower bounds for linear forms in logarithms to exponential Diophantine equations has been well known and extensively used for a long time.In this paper, we use two particular tricks: the elimination of unknown logarithms and a kind of finite induction.Both of these tricks have been used in several papers before (see e.g.[8] for the elimination of unknown logarithms and [7] for the finite induction method), however, to the authors' best knowledge, this is the first time that they are used in a combined way.
Moreover, the induction is not just a straightforward induction over k steps, but different cases lead to quite different bounds.It is interesting to see how the bounds depend on the cases, and to then determine an overall asymptotically good bound.

Preliminary results
In this section we, start by recalling some basic properties of Fibonacci numbers.Moreover, we argue why we may assume n k ≥ 2 and n i+1 ≥ n i + 2 for i = 1, . . ., k − 1, as well as n − 2 ≥ m ≥ 2 in the rest of the paper.Then we state some known results and some elementary results related to equations (A) and (B).
In the second subsection, we state one of Matveev's lower bounds for linear forms in logarithms.Furthermore, we prove an elementary lemma that will allow us to deduce an absolute bound for n from a bound of the shape n ≤ c • (log n) x .

Results related to Fibonacci numbers
For the Fibonacci numbers we have the well known Binet formula In Theorem 1 the representations F n1 + • • • + F n k and F n + F m are not necessarily Zeckendorf representations, i.e. we might have consecutive or identical indices, or indices equal to 0 or 1.However, in the rest of this paper we will assume that n k ≥ 2 and n i+1 ≥ n i + 2 for i = 1, . . ., k − 1, as well as n − 2 ≥ m ≥ 2. Let us justify now why we can do this.
Since the Zeckendorf representation is minimal (see e.g.[4, Theorem 1.1]) we have k ′ ≤ k and we will get the same or an even stronger result.If F n + F m is not a Zeckendorf representation, then either n = m, or the Zeckendorf representation in fact only consists of one Fibonacci number and we have y a = F n ′ .As mentioned in the Introduction, the latter is by [3] only possible for y a ≤ 144, in which case we are done.If n = m, then y a = 2F n implies F n = 2 s (y ′ ) a , for suitable y ′ and s.From [2, Theorem 4] it follows that this is only possible for n = 1, 2, 3, 6, 12 (cf.[9,Theorem 2]).In fact, y a = 2F n only works for n = 3, 6 and thus y a ≤ 2F 6 = 16, and we are done as well.
Finally, note that we may assume y ≥ 2, since Theorem 1 is trivial for y = 0, 1. Next, we state the result due to Luca and Patel [9], that we mentioned in the Introduction.We will use it to exclude the case n ≡ m (mod 2).
Given integers y, a, n > m that satisfy equation (B), one can use lower bounds for linear forms in logarithms to obtain a bound for n in terms of y.This was done explicitly by Kebli et al. [5, Theorem 1] and we will use their bound in our proof.

Proof. By the properties of the Zeckendorf expansion, we have y
Remark 2. In view of Theorem B and Lemma 1, we are done if we manage to prove a bound for n 1 of the shape n 1 ≪ (log n) x : Then we have n < 6 • 10 29 (log y) 4 ≤ 6 • 10 29 n 4 1 ≪ (log n) 4x , which immediately gives us an effective upper bound for n.This will indeed be our strategy.
We will use the following not very sharp but simple estimates.Note that there is no benefit from making them sharper, as the constants coming from these estimates will be "swallowed" by a much larger constant in Section 5.

Lemma 2. For any
Proof.Inequality (3) follows from a simple estimation: Then (4) follows easily as well: The next lemma will be used to bound the coefficients in the linear forms in Section 5.For simplicity, the estimates are very rough.Sharper estimates would improve the bound (17) at the end of Section 6 only marginally.Proof.Since y a > y, it follows immediately from the properties of the Zeckendorf representation, that n 1 < n.
For the second inequality note that we have 2 a ≤ y

Result related to the application of lower bounds for linear forms in logarithms
Our proof will heavily rely on lower bounds for linear forms in logarithms.In order to switch from expressions of the shape In order to compute lower bounds for expressions of the shape we will use Matveev's popular result [10,Corollary 2.3] because it is very good and easy to apply.
Let us first recall the definition of the height and some basic properties.Let η be an algebraic number of degree d over the rationals, with minimal polynomial Then the absolute logarithmic height of η is given by For any algebraic numbers η 1 , . . ., η t ∈ Q and z ∈ Z, the following well known properties hold: Theorem C (Matveev, 2000).Let η 1 , . . ., η t be positive real algebraic numbers in a number field K of degree D, let b 1 , . . .b t be rational integers and assume that In order to be able to apply Theorem C, one has to check that the linear form Λ does not vanish.This often requires some tricks.In Section 5 we will make use of the following lemma.
Lemma 5. Let x be an odd positive integer.Then 2 ).Proof.Let x be an odd positive integer.If α x + 1 were divisible by √ 5, then the norm N Q( √ 5)/Q (α x + 1) would be divisible by 5. Let us compute the norm: where L x is the x-th Lucas number (the Lucas numbers are defined by L 0 = 2, L 1 = 1 and L n = L n−1 + L n−2 for n ≥ 2, and they indeed have the Binet representation L n = α n + β n ).Modulo 5 the Lucas sequence looks like this: 2, 1, 3, 4, 2, 1, . . .and in particular no Lucas number is divisible by 5. Thus N Q( √ 5)/Q (α x + 1) is not divisible by 5 for odd x, and α x + 1 cannot be divisible by √ 5.
Finally, after the repeated application of lower bounds for linear forms in logarithms, we will end up with an inequality of the shape n ≤ c • (log n) x by the end of Section 6.In order to obtain an absolute upper bound for n, we will use the following lemma in Section 7. Lemma 6.Let δ > 0 and let n, c, x ∈ R ≥1 satisfy the inequality Proof.Let n, c, x be numbers ≥ 1 that satisfy (5).First, note that if n ≤ e e , then we have n < exp(exp((1 + δ −1 ) 2 )) and we are done immediately.Therefore, we may assume that n > e e , which implies (6) log log log n > 0.
Now we consider inequality (5) and take logarithms, obtaining (8) log n ≤ log c + x log log n.
Assume for the moment that x log log n ≤ (log n)/2.Then the above inequality implies log n ≤ 2 log c, which plugging back into (5) immediately yields n ≤ 2 x • c • (log c) x and we are done as well.
Finally, we consider the case that x log log n > (log n)/2, i.e. log n < 2x log log n.Taking logarithms, we obtain log log n < log(2x) + log log log n.Together with (7) this yields and then log log n < (1 + δ) log(2x).Plugging this into (8), we obtain

Constructing the basic linear forms in logarithms
In this section, we construct k linear forms in logarithms from equation (A) and k linear forms from equation (B).In the third subsection we give an overview of all the linear forms in logarithms and their upper bounds.

Linear forms coming from equation (A)
Using the Binet formula, we can rewrite equation (A) as Multiplication by √ 5 yields To obtain the first linear form in logarithms, we shift the largest power α n1 to the left hand side, take absolute values and use Lemma 2: Now we divide by α n1 and obtain If n 1 − n 2 ≥ 6, then the above expressions are ≤ 0.5, so with Lemma 4 we get This is our first linear form in logarithms.Next, we construct more of them by shifting more powers of α to the left hand side.
Let us go back to (9) and shift the largest ℓ powers α n1 , . . ., α n ℓ to the left (2 ≤ ℓ ≤ k − 1).Then, as above, we obtain Dividing by Then, if n 1 − n ℓ+1 ≥ 6, by Lemma 4 we have Finally, let us construct the last linear form in logarithms by shifting all powers of α to the left hand side.Then from equation ( 9) we obtain Dividing by Then, as above, if n 1 ≥ 6, with Lemma 4 we obtain

Linear forms coming from Equation (B)
We rewrite equation (B) using the Binet formula: Then, by the same procedure as above, if n − m ≥ 6 and n ≥ 6, we obtain the two linear forms

Overview of basic linear forms
Let us sum up all the linear forms in logarithms that we have constructed and name them in an appropriate way: Let us denote the k linear forms coming from equation (A) by Λ A1 , . . ., Λ Ak , and the two linear forms coming from equation (B) by Λ B1 , Λ B2 .Specifically, for 2 ≤ l ≤ k − 1, we have The upper bounds are all of the shape 12α −X .Mind that each of the inequalities only holds if X ≥ 6.However, the goal will always be to bound X from above, so if X ≤ 6, we will just skip the corresponding step.

Eliminating y from the linear forms and overview of the steps
In each of the linear forms Λ A1 , . . ., Λ Ak , Λ B1 , Λ B2 we have one logarithm that is unknown, namely log y.Therefore, we will always take two distinct linear forms and eliminate log y.For example, if we take Λ A1 and Λ B1 and eliminate log y, we get a new linear form Λ * A1 in fixed logarithms with a bound of the shape . After computing a lower bound for |Λ * A1 | with Matveev's theorem, we then obtain an upper bound either for n 1 − n 2 or for n − m, depending on which one is smaller.This leads to different cases and in each case we have to continue with slightly different steps.Figure 1 shows an overview of the steps.Each rectangular box stands for a step, where we take the two linear forms written in the box and eliminate y.The obtained upper bound is written along the arrow that points to the next step.In the steps on the left (Steps A1 to Ak), there are always two cases, depending on whether n 1 − n ℓ (or n 1 ) or n − m is smaller.If we ever cross over to the steps on the right (Steps B1 to Bk), then we just follow the arrows pointing downwards, always obtaining bounds for n 1 − n ℓ (or n 1 ).The purpose of Figure 1 is to give a rough idea of the proof.There will be a more detailed figure in Section 6.We now construct and estimate all the linear forms in logarithms that are used in the steps.
Step Bk: where we used a ≥ 2. In the same way we also obtain the linear forms for Steps A2 to Ak: Analogously, we construct and estimate the linear forms for Steps B1 to Bk.Note that n 1 − n ℓ < n 1 < n, so clearly α −n < α −(n1−n ℓ ) for any ℓ and we don't need to write any minima.
Mind that each of these estimates only holds if we have X ≥ 6 for the exponent in the upper bound 18aα −X .

Application of Matveev's theorem
In this section we apply Matveev's theorem to all the linear forms Λ * Aℓ , Λ * Bℓ , that we obtained in the previous section.Moreover, we compare the lower bounds to the upper bounds and compute general bounds for the exponents n 1 − n ℓ , n 1 or n − m.
First, we check that the linear forms Λ * A1 , . . ., Λ * Ak , Λ * B1 , . . ., Λ * Bk are non-zero.Note that in each linear form we have the expression (a − 1) log √ 5 and a − 1 = 0. Thus, if a linear form were zero, log √ 5 would have to be canceled out by the other logarithms.In particular, since √ 5 is prime in Z[α], it would have to divide at least one argument of a logarithm.We only have three other types of logarithms: log α, log(α n−m +1) and log(1+α , so log α does not contribute to the cancellation of log √ 5. Second, by Lemma 5, the expression α n−m + 1 is not divisible by √ 5 unless n − m is even.If n − m is even, we are immediately done with Theorem A, so let us assume that n − m is odd.Thus log(α n−m + 1) does not contribute to the cancellation of log √ 5 either.And third, the expression ( ) associated with the prime ideal ( √ 5), normalized by v √ 5 ( √ 5) = 1.Therefore, all the linear forms Λ * A1 , . . ., Λ * Ak , Λ * B1 , . . ., Λ * Bk are non-zero and we can apply Matveev's theorem to each of them.Each linear form has an upper bound of the shape 18aα −X and in each step we compare the lower and the upper bound to obtain a bound for the expression X.
We start by describing the last step, because the lower bound for the Λ * Bk will be the weakest, so for simplicity we will be able to reuse it in the other steps.Of course, one can obtain sharper bounds by considering each linear form separately, in particular for the linear form Λ * A1 in only two logarithms.

Step Bk
Assume that we already have an upper bound T k for n 1 − n k and an upper bound S for n − m.We want to apply Matveev's theorem (Theorem C) to Λ * Bk with t = 4 and The four numbers η 1 , η 2 , η 3 , η 4 are real, positive and belong to K = Q( √ 5), so we can take D = 2.As stated in Lemma 3, we have a ≤ n and and we can set B = n 2 .Since h( √ 5) = (log 5)/2 and h(α) = (log α)/2, we can set A 1 = log 5 and A 2 = log α.Finally, we estimate (rather roughly) the heights of η 3 and η 4 .Recall that we are assuming Thus we can set We can estimate 1 + log n 2 = 1 + 2 log n ≤ 3 log n (this estimate holds for n ≥ 3; if n < 3, we are immediately done).Then we simplify the above lower bound to (11) log Together with (10) this yields Note that we are allowed to omit the expressions log 18 and log a because log a ≤ log n and the constant C is extremely large and was estimated very roughly.Moreover, note that the upper bound coming from (10) only holds if n ≥ 6.However, if n < 6, then (12) is trivially fulfilled.An analogous argument will implicitly be used in all other steps as well.

Now let us have a look at
Step B1.The linear form Λ * B1 has only three logarithms (instead of four), so it is possible to get a better lower bound.However, for simplicity, we can set T 1 = 1 and see that a bound analogous to (11)

Thus, in
Step B1 we can also obtain the bound (13) for ℓ = 1 and T 1 = 1.

Steps A1 to Ak
Let ℓ ∈ {1, . . ., k}.In Step Aℓ we assume that we already have a bound R ℓ ≥ n 1 − n ℓ (if ℓ = 1, we have no bound yet).We consider the linear form Λ * Aℓ , which is almost exactly equal to the corresponding linear form Λ * Bℓ , except that the logarithm log(α n−m + 1) is missing.Thus, we only have three logarithms (or even two logarithms if ℓ = 1) and we can obtain even better lower bounds than in (11).However, for simplicity we set S = 1 and see that the lower bound log

Walking the steps
Depending on how large n − m is compared to n 1 − n 2 , . . ., n 1 − n k , n 1 , we do a specific series of steps, starting with Step A1.
Say we are in Step Aℓ.If min{n 1 − n ℓ+1 , n − m} = n 1 − n ℓ+1 , then we get a bound R ℓ+1 for n 1 − n ℓ+1 and we continue with Step Aℓ + 1.In the other case that n − m is the minimum, we get a bound n − m ≤ S ℓ and we continue with Step Bℓ, and after that, we continue with Steps Bℓ + 1 all the way until Step Bk.This is illustrated in Figure 2. Note that the question marks stand for numbers and letters that depend on the case, i.e. which path we are coming from.For example, if n − m ≤ n 1 − n 2 , then we do Step A1 -Step B1 -Step B2 -• • • .In this case, in Step B2 we get If n 1 < n − m, we finish with Step Ak, obtaining the bound n 1 ≤ R k+1 .In all other cases, we finish with Step Bk, obtaining a bound n 1 ≤ T k+1 .
In order to compute the maximal such bound n 1 ≤ R k+1 or n 1 ≤ T k+1 , we need to see what exactly happens when we "walk the steps".Case 1: n 1 < n − m ("Walking down the left side").In this case we start with R 1 = 1 and at each Step Aℓ we compute the next bound as described in (14), namely by From this recursion, we immediately see that the last bound is Case 2: n 1 ≥ n − m ("Crossing over").
In this case, we cross from left to right at some point in Figure 2, i.e. we go from Step Aℓ 0 to Step Bℓ 0 for some ℓ 0 ∈ {1, . . ., k}.This means that we get bounds in the following order: R 1 = 1, R 2 , . . ., R ℓ0 , S ℓ0 , T ℓ0+1 , . . ., T k+1 .By the same reasoning as in Case 1, the last bound on the left will be R Then, in Step Aℓ 0 , the bound S ℓ0 is computed in the same way, and we obtain Step A1: Step A2: Step Ak: Step B1: Step B2: Step Bk: After that, for 1 ≤ i ≤ k − ℓ 0 , we have the recursion From this recursion, we obtain the last bound This bound depends on ℓ 0 .In order to obtain an overall upper bound for all 1 ≤ ℓ 0 ≤ k, we first compute the maximum of the exponent (ℓ and the maximum is (k 2 + 6k + 1)/4.Thus we have Finding the maximum of the expression (ℓ 0 !) k−ℓ0+1 is much harder.We bound the expression in a rough way.Note that for 1 ≤ ℓ ≤ k, we have Again, the exponent ℓ(k − ℓ + 1) = ℓ(k + 1 − ℓ) is maximal in ℓ = (k + 1)/2, and the maximum is (k 2 + 2k + 1)/4.Thus we have Finally, for simplicity, we estimate k! ≤ k k .Then we obtain from (16) that Thus, we have proven that no matter at which point we cross from the left to the right (Step Aℓ 0 -Step Bℓ 0 ), we always end up with the bound above.Since this bound is of course larger than the bound (15) from Case 1, we overall obtain Let ε > 0 be given.We want to apply Lemma 6 to inequality (18), setting c = C k 2 +6k+3 • k k 2 +2k+5 and x = k 2 + 6k + 1.Moreover, we fix a 0 < δ < 1, which we will specify in a moment.

Figure 2 .
Figure 2. Overview of steps