On a Diophantine equation involving powers of Fibonacci numbers

On a Diophantine equation involving powers of Fibonacci numbers

By Krisztia´n GUETH,^1ÞFlorian LUCA^2Þ,3Þ,4Þand La´szlo´ SZALAY^5Þ

(Communicated by Shigefumi MORI,M.J.A., March 12, 2020)

Abstract: This paper deals with the diophantine equationF₁^pþ2F₂^pþ þkF_k^p¼F_n^q, an equation on the weighted power terms of Fibonacci sequence. For the exponentsp; q2 f1;2gthe problem has already been solved in ad hoc ways using the properties of the summatory identities appear on the left-hand side of the equation. Here we suggest a uniform treatment for arbitrary positive integerspandqwhich works, in practice, for small values. We obtained all the solutions for p; q10by testing the new approach.

Key words: Fibonacci number; diophantine equation; weighted sum.

1. Introduction. As usual, fFmg_m0 de- notes the sequence of Fibonacci numbers F₀¼0, F1¼1 and F_mþ2¼F_mþ1þFm for all m0. Its companion sequencefLmg_m0is the Lucas sequence given byL₀¼2,L₁¼1and L_mþ2¼L_mþ1þL_m for all m0. We assume that the reader is familiar with their Binet formula.

In this paper, we determine the solutions to the Diophantine equation

F₁^pþ2F₂^pþ þkF_k^p¼F_n^q ð1Þ

in positive integers ðp; q; k; nÞ where p and q are small. This equation was first investigated by Ne´methet al.in [5] for the four possibilitiesfp; qg f1;2g, and all the solutions in these particular cases were obtained in elementary ad hoc ways. The purpose of this paper is to provide a uniform treatment independently from the values of p and q. As a particular case, we solve the above equation for all values ofp; qwhich do not exceed the upper bound 10.

We consider

F₁^p¼1¼F₁^q¼F₂^q; and F₁^pþ2F₂^p¼3¼F4

as trivial solutions to (1). The authors in [5] have made the following

Conjecture 1. Equation (1) has only the three non-trivial solutions

ðp; q; k; nÞ ¼ ð1;1;4;8Þ; ð1;2;3;4Þ; ð3;3;3;4Þ:

The above conjecture says, in particular, that there exist only finitely many solutions. Since the equation is not a standard equation, the finiteness of its number of solutions does not seem to follow in an easy way. Note that the first two quadruples were obtained in [5], while the last one is justified here in the sense that our present work confirms the conjecture by solving the equation for maxfp; qg 10. The result is recorded in

Theorem 2. Conjecture 1 is true whenever maxfp; qg 10.

Problems having similar flavour appear in the extensive literature of Fibonacci sequence. For instance, the sumF_n^sþF_nþ1^s (n0) gives Fibonacci numbers when s2 f1;2g. For larger exponents s, Marques and Togbe´ [4] proved that ifF_n^sþF_nþ1^s is a Fibonacci number for all sufficiently large n, then s¼1 or 2. Afterwards, Luca and Oyono [2] com- pleted the solution of the question by showing that apart from F₁^sþF₂^s¼F3 there is no solutions3 to the equationF_n^sþF_nþ1^s ¼F_m.

A naturally arising question is what would happen if we replace the Fibonacci numbers by other linear recurrence? In the case of non degen- erate binary recurrences with real roots it is likely our approach works. On the other hand, we do not think that the method extends to Tribonacci numbers or to other recurrences of order higher

doi: 10.3792/pjaa.96.007

#2020 The Japan Academy

2020 Mathematics Subject Classification. Primary 11B39;

Secondary 11D45.

1Þ University Eo¨tvo¨s Lora´nd, Savaria Centre, Ka´rolyi Ga´spa´r te´r 4, 9700 Szombathely, Hungary.

2Þ University of the Witwatersrand, School of Mathematics, 1 Jan Smuts Ave, Johannesburg, 2000, South Africa.

3Þ Research Group of Algebraic Structure & Applications, King Abdulaziz University, Jeddah, Saudi Arabia.

4Þ Department of Mathematics, Centro de Ciencias Mate- ma´ttics UNAM, Morelia, Mexico.

5Þ Jan Selye University, Institute of Mathematics and Informatics, Elekra´renska´ cesta 2, 94501 Koma´rno, Slovakia.

than 2 although we have made no efforts in this direction.

Now we collect some preliminary results we will use in the proof of Theorem 2. In what follows, log_b denotes the logarithm to baseb, whereb >1 is any real number, while :¼ ð1þ ffiffiffi

p5

Þ=2 is the dominant root of the Fibonacci sequence. Since the following three lemmata are widely known, we present them without proof.

Lemma 3. For n1, we have ⁿ² Fnⁿ¹, andⁿ¹ Ln.

Lemma 4. The inequality F_nþ1=F_n3=2 holds forn2.

Lemma 5. Assume that n is divisible by 4.

ThenF_nF₄¼F_ðn4Þ=2L_ðnþ4Þ=2.

At some stage of the proof of the theorem we will use the following estimates.

Lemma 6. Equation(1) implies . ðk2Þp <ðn1Þqif k2, and . ðn2Þq <ðk1Þpþlogð4kÞifk3.

Proof. Combining Lemma 3 andF_k^p< F_n^q(pro- vided by (1) and k2) leads immediately to the first statement.

For the second statement, Lemma 4 yields

Fk

F_ki¼Yⁱ¹

j¼0

Fkj

F_kj1

3 2

i1

for alli¼1;2;. . .; k1, wherek3. Note that the lower bound could be improved toð3=2Þⁱifik2 because we avoid the quotientF₂=F₁¼1. Put:¼ 2=3. Recalling Lemma 3, observe that

^ðn2Þq< F_n^q¼kF_k^pX^k1

j¼0

kj k

F_kj Fk

p

ð2Þ

< kF_k^pð1þ1þþ²þ þ^k2Þ

<4k^ðk1Þp:

Then the statement follows by taking logarithms.

Lemma 7. Suppose thatkandnare positive integers. Then5^kkFn if and only if 5^kkn.

Proof. See Lemma 1 in [1].

Lemma 8. Assume thatk,p andqare posi- tive integers,p is odd. If

Lpk²þ ðLp2Þk1¼ 5^pqL²_p holds, thenðp; q; kÞ ¼ ð1;1;1Þ;ð1;1;2Þ.

Proof. Since Lp is never a multiple of 5, and 5^pqL²_p is an integer, it follows that qp. Put r¼pq. If p¼1, then k²k1¼ 1 gives the

two solutions above. The condition p3 entails that the left-hand side is positive so the sign in the right-hand side must be +. Suppose now thatp¼3 or p¼5. The left-hand side is a quadratic poly- nomial in k with leading coefficient Lp6¼0. A verification with q2 f1;. . .; pg provides no more solutions.

In the sequel, we may assume p7. We will show that this assertion contradicts the equality in the lemma. Reducing

Lpk²þ ðLp2Þk1¼5^rL²_p ð3Þ

moduloLp, it leads to Lpj2kþ1. Thus, Lp is odd, sop is not a multiple of 3. Moreover 2kþ1¼aLp

holds for some odd integera. Thus,k¼ ðaLp1Þ=2.

Substituting this into (3), after some manipulations we obtain

a²L²_p ð4aþ1Þ ¼45^rL_p: ð4Þ

On one hand, this gives Lpj4aþ1, therefore a ðLp1Þ=47. On the other hand, since L_p29, we have4aþ1<5a < a²L²_p=2. Consequently

45^rLp¼a²L²_p ð4aþ1Þ>a²L²_p 2 ; therefore

a < 2³⁼²5^r=2 L¹⁼²p

:

This implies

ðLp1ÞL¹⁼²_p <2⁷⁼²5^r=2: ð5Þ

On the other hand, rewriting (4) as ðL²_pÞa²4a ð45^rLpþ1Þ ¼0;

and treating it as a quadratic ina, its discriminant is a perfect square. Subsequently,

4þL²_pð45^rLpþ1Þ ¼y²

holds for some positive integer y. So, 45^rL³_pþ ðL²_pþ4Þ ¼y². Since L²_pþ4¼5F_p² (because p is odd), we get

45^rL³_pþ5F_p²¼y²:

Letc; dbe such that5^ckFpand5^dky. ThenFp¼5^cu, y¼5^dv for some integers u; v with gcdðuv;5Þ ¼1 and

45^rL³_pþ5^2cþ1u²¼5^2dv²:

Since 5-Lp, 5^r is the exact power of 5 in the

factorisation of 45^rL³_p. From the above equation, we have that since2cþ16¼2d, eitherr¼2cþ1, or r¼2d, and in the last case r2cþ1. Clearly, in both cases r2cþ12 log₅pþ1, where in the second inequality we used Lemma 7. Thus, 5^r 5p². Returning to (5), we obtain

L¹⁼²_p ðLp1Þ 2⁷⁼²5^r=22⁷⁼²5¹⁼²p;

and sinceL_p^p1 (see Lemma 3), we conclude ^ðp1Þ=2ð^p11Þ 2⁷⁼²5¹⁼²p;

an inequality false for anyp6.

2. Proof of the theorem. Because we al- ready accounted for the trivial solutions, we may assume k3. First we handle the case k¼3 separately. In fact we will exploit k4 only in (12), but without this assumption, one has more difficulties in our argument after (11). With k¼3 we find F_n^q¼F₁^pþ2F₂^pþ3F₃^p¼3ð1þ2^pÞ, so 3jFn

and Fn is odd, so 4jn and 3-n. If q¼1, then 3 2^p¼F_n3¼F_nF₄¼F_ðn4Þ=2L_ðnþ4Þ=2 by Lemma 5. Thus F_ðn4Þ=2 has its largest prime factor at most 3. The Primitive Divisor Theorem implies ðn4Þ=212, so n28 and the remaining possi- bilities can be verified by hand. If q¼2, we have 3ð1þ2^pÞ ¼. Thus, 2^pþ1¼3. Distinguishing between p 0;1;2 (mod 3Þ, we get the equation 3y² ¼1þ2^rx³, where r2 f0;1;2g, which one can solve with MAGMA [3]. If q¼3, we handle similarly the equation 1þ2^p¼9y³. Distinguishing between p even and odd one has 9y³¼1þ2^rx², r2 f0;1g, and all integer solutions ðx; yÞ to these equations can again be computed with MAGMA [3].

Assume now that q4. Then 3⁴jF_n^q. Hence, 3³j1þ2^p, so p is odd and 9jp. In particular, 19j2⁹þ1j2^pþ1jF_n^q, so19jFn. Thus18jn, con- sequently3jn, a contradiction.

So, from now onk4. Let the integerp1be fixed (not necessarily inf1;2;. . .;10g), and consider the term F_j^p with j1. Since we have F_j¼ ð^{j j}Þ= ffiffiffi

p5

, where¼ ð1 ffiffiffi p5

Þ=2, it follows that F_j^p¼ð^jjÞ^p

5^p=2 ¼ ^jp 5^p=2þ_p;j; where

jp;jj<2^pðp1Þj

5^p=2 < ^jðp1Þ: Thus,

X^k

j¼1

jF_j^p ¼ 1 5^p=2

X^k

j¼1

j^jp

! þR₁; where jR1j<Pk

j¼1j^jðp1Þ< k^2kðp1Þ. The inner sum is

X^k

j¼1

jx^j¼x d dx

X^k

j¼1

x^j

!

¼x d

dx xx^k1 x1

¼kx^kþ2 ðkþ1Þx^kþ1þx ðx1Þ²

with x:¼^p. Thus, X^k

j¼1

jF_j^p ¼k^p ðkþ1Þ 5^p=2ð^p1Þ^{2 pðkþ1Þ} þ ^p

5^p=2ð^p1Þ²þR1: Letqbe a positive integer. Writing also

F_n^q¼ ^nq 5^q=2þR2; where

jR2j 2 5¹⁼² q

^nðq1Þ< ^nðq1Þ;

the above formulas lead to 5^ðqpÞ=2ðk^p ðkþ1ÞÞ

ð^p1Þ^{2 pðkþ1Þqn}

ð6Þ

¼5^q=2R25^q=2R15^ðqpÞ=2p ð^p1Þ²: Thus, in the right-hand side of (6) we see that

5^q=2jR1j 5^q=2k^2kpk; ð7Þ

and

5^ðqpÞ=2p

ð^p1Þ² 5^q=235^q=2kpkþ3: ð8Þ

In the latter case, we used the facts that ^p1 ^p=2 for all p2, and =ð1Þ² ¼³ (to include the casep¼1). Bounding5^q=2R2 takes a bit longer.

Clearly, in the exponent of the upper bound onjR2j we have nðq1Þ ¼ ðn2Þqþ2qn and further 5^q=22q¼ ð ffiffiffi

p5

²Þ^q<6^q. Combining these and (2), we obtain

5^q=2jR2j<5^q=2nðq1Þ ð9Þ

<5^q=24k^ðk1Þpþ2qn

<6^qþ1k^ðk1Þpn:

If kn, then the last term is not larger then 6^qþ1k^ðk1Þpk. Assume nowk > n. Obviously,pq leads to F_n^q< F_k^p, which contradicts (1). Hence, p < q. Putb¼maxfp; qg ¼q. The first statement of Lemma 6 provides

n >ðk2Þp

qþ1k2 b þ1;

which together with (9) implies

5^q=2jR2j<6^qþ1kðk1Þpðk2Þ=b1

<6^qþ1k^kpk=b:

Comparing it with the estimate obtained forkn, we get a general upper bound on5^q=2jR2j. Obvious- ly, in (7) and in (8) we can replace k in the exponents byk=b to unify the two upper bounds obtained for the casesknandk > n, respectively.

Let_p:¼ ð^p1Þ², and

zqðkÞ:¼5^q=23þ6^qþ1kþ5^q=2k²:

Putting all the above estimates together, we get that

5^ðqpÞ=2ðk^pk1Þ

_p ^qnpðkþ1Þ

< zqðkÞ ^k=bþp: ð10Þ

The next goal is to analyse the exponent :¼ qnpðkþ1Þ. We distinguish two situations. Sup- pose first that

5^ðqpÞ=2 3_p : ð11Þ

Here observe that fork3andp1 5^ðqpÞ=2

p

ðk^pk1Þ ð12Þ

5^ðqpÞ=2 p

ðkð^p1Þ 11=3Þ>5^ðqpÞ=2 p

:

It now follows from (10) and (12), that ^k=bþp<5^ðpqÞ=2_pz_qðkÞ;

ð13Þ

which gives us an upper bound on k for fixed p; q.

Later, we will see that the other branch provides also a bound which is larger.

In the sequel, we suppose that the opposite of (11) is true. First consider the case when the left- hand side in (10) is zero. After rearranging the corresponding relation, we take the norms inQ½ ffiffiffi

p5 and get to

k²ðÞ^p ðkþ1Þkð^pþ^pÞ þ ðkþ1Þ²

¼5^pqðÞ^qnpðkþ1ÞN_Q^pffiffi₅ðpÞ:

By¼ 1,^pþ^p¼L_p, and

N_Q½^pffiffi₅ð_pÞ ¼ ðð1Þ^pL_pþ1Þ²

the previous equality simplifies

ðð1Þ^pL_pþ1Þk²þ ð2L_pÞkþ1

¼ 5^pqðð1Þ^pLpþ1Þ²:

Ifp is even, the corresponding equation is

ðL_p2Þk²þ ðL_p2Þk1¼ 5^pqðL_p2Þ²: Hence,ðLp2Þ j1, leading toLp¼3, sop¼2. Now k²þk1¼ 5^2q;where the eligible values forqis 1 or 2. Each case leads to a trivial solution.

If p is odd, then Lemma 8 handles the equation

L_pk²þ ðLp2Þk1¼ 5^pqL²_p

and provides only ðp; q; kÞ ¼ ð1;1;1Þ;ð1;1;2Þ, deriv- ing trivial solutions to (1).

Assume now that the left-hand side in (10) is nonzero. We then have

jk^p ðkþ1Þ 5^ðpqÞ=2pj ð14Þ

<5^ðpqÞ=2_pz_qðkÞ

^k=bþp :

Changingtoin the left above we get an amount jk^p ðkþ1Þ 5^ðpqÞ=2ðpÞj

ð15Þ

<ð2kþ1Þ þ95^pqp;

whereðpÞ ¼ ð^p1Þ²<3. We also used the fact that the opposite of (11) is true, therefore

jj¼<35^ðpqÞ=2p:

The product of the left-hand sides of (14) and (15) is the norm of a nonzero algebraic integer inKso it is 1. We thus get that

^k=bþp<5^ðpqÞ=2_pz_qðkÞðð2kþ1Þ þ95^pq_pÞ:

ð16Þ

Note that (13) is weaker in (16), which therefore gives a general bound for kirregardless of whether (11) holds or not.

Taking maxfp; qg 10, (16) gives k1104.

Now one can easily check when X^k

j¼1

jF_j^p

!1=q

is a Fibonacci number for positive integer variables p; q2 f1;. . .;10g and k1104 getting only the

solutions from the statement of the theorem. The proof is complete.

Acknowledgements. The work on this pa- per started when the last author visited School of Mathematics of the Wits University. He thanks this Institution for support, and also thanks Kruger Park for excellent working conditions. F. L. was supported in part by the Number Theory Focus Area Grant of CoEMaSS at Wits (South Africa).

Part of this work was done when this author visited the Max Planck Institute for Mathematics in Bonn, Germany from September 2019 to February 2020.

He thanks this Institution for hospitality and support. For L. Sz. the research was supported by Hungarian National Foundation for Scientific Re- search Grant No. 128088. This presentation has been made also in the frame of the ‘‘EFOP-3.6.1-16- 2016-00018 – Improving the role of the research + development + innovation in the higher education

through institutional developments assisting intel- ligent specialization in Sopron and Szombathely’’.

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