Powers in the Lucas sequence when the index is divisible by three

Powers in the Lucas sequence when the index is divisible by three

Mª Dolores Lópeza,, Javier Rodrigob

a

Departamento de Matemática e Informática Aplicadas a la Ingeniería Civil de la E.T.S.I. Caminos, Canales y Puertos, Universidad Politécnica de Madrid marilo.lopez@ upm.es

b

Departamento de Matemática Aplicada, E.T.S. de Ingeniería, Universidad Pontificia Comillas de Madrid

*Corresponding author: Mª Dolores López, Departamento de Matemática e Informática Aplicadas a la Ingeniería Civil de la E.T.S.I. Caminos, Canales y Puertos, Universidad Politécnica de Madrid, c/ Profesor Aranguren s/n, Madrid 28040, Spain

Abstract:

In this paper the m-powers with m≥2 included in the Lucas sequences when the index satisfies some conditions are found.

1. Introduction

The Lucas and companion Lucas sequences have been widely studied in the scientific literature as a natural extension of the so called Fibonacci sequence. They are defined inductively by: 0, =0

Q P

U , 11 , =

Q P

U , nPQ

Q P n Q

P

n PU QU

U ,2

, 1 ,

− − −

= and 0 , =2

Q P

V , P

V₁P,Q = , V_nP ,Q =PV_nP₋,₁Q −QV_nP₋,₂Q respectively, where P , are integer numbers (the Q Fibonacci sequence is the particular case of U_nP,Q with P=1, Q=−1. The indices

Q

P , will be omitted when they are not necessary).

Concretely, the search of the powers included in these sequences has deserved a special attention by the number theorist. The main results at this respect for Un are:

- For odd relative primes P , , the only square in Q Un with n>2, n≠6 is 144

1 , 1 12 =

−

U

(see Ribenboim, Bremner&Tzanakis)

- For P divisible by 4, there is no squares in UnP,−1 for positive, even numbers n when

additional conditions on P are considered (Kagawa) - The only powers in 1 ,−1

n

U with n>2 are 1 ,1 8

6 =

−

U and 1 ,1 144 12 =

−

U (Bugeaud)

- The only power in U_n2,−1 with n>1 is U₇2 ,−1 =169 (Cohn)

In this paper the case P>0, Q=−1 is studied: it is proved that, for n divisible by three and satisfying some extra conditions, the powers in Un are always attained for

1

=

P and they are 16 ,1 =8

−

U and 121 ,1 =144

−

U . To do this, the diophantine equation

m P

n u

U ,−1 = with unknowns

(

n, P,u,m

)

, m≥2, is related with these diophantine equations according to the different cases:

- The Catalan equation xm −y2 =1 - The Thue equation 2 xm −ym =1 - The equation 2x2 −y4 =1

- _⎟⎟

⎠ ⎞ ⎜⎜ ⎝ ⎛

k n

for the binomial coefficient.

- e_p

( )

n for the exponent of the prime p in the factorization in prime numbers of the integer number n (if p is not a prime factor of n, it is assumed that ep

( )

n =0).

The following two identities for Un, Vn , Uk, Vk are used along the paper (see

Kagawa):

(

) (

)

k

n

k k n

n k n

P U V P

U

V 4 4

2 −1 + 2 + = + 2 + when k n and k, n are positive numbers (1)

( ) (

)( )

( )

k

k

k P U

V 2 − 2 +4 2 =4 −1 (2)

The paper is structured as follows: in section 2 the main result is stated and in section 3 some concluding remarks are given. Along section 2, k and n are positive integer numbers.

2. Development

To prove the result mentioned in the introduction, a lemma is needed: Lemma

If k n, then

k n U U U

GCD _k

k n

, _⎟⎟

⎠ ⎞ ⎜⎜

⎝ ⎛

Proof

First it is noted that if k n, then U_k U_n (see for example Ribenboim), so

k n

U U

is an integer number and the statement makes sense. Now, the following cases appear:

1) If V_k is an odd number, then by expanding (1) it is obtained that:

(

)

( )

_⎟⎟

⎠ ⎞ ⎜

⎜ ⎝ ⎛

+ =

+ + ⎟

⎟ ⎟

⎠ ⎞

⎜ ⎜ ⎜

⎝ ⎛ +

= − − −

−

k k

n

k k k

k n

k k

k n

k n

k n

U f V k n U P

U V k n U V k n

U 1 3 3 2 1

1

... 4 3

( )

x

f is a polynomial without independent term. Now, since _{k k}

k n U U U U

GCD , _⎟⎟

⎠ ⎞ ⎜⎜ ⎝ ⎛ ,

( )

1 1 2 , − − + = ⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ k n k k n k k n k k n U f V k n U U U U U

GCD , then _⎟⎟

⎠ ⎞ ⎜⎜ ⎝ ⎛ k k n U U U

GCD , divides to U_k,

( )

k k

n

k f U

V k

n −1₊

, so it divides to −1 +

( ) ( )

− = k−1 n k k k k n k V k n U f U f V k n .

Since V_k is an odd number, it is satisfied that GCD

(

V_k,U_k

)

=1 by identity (2) and then

k n U U U GCD _k k n , _⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ as desired.

2) If V_k and P are even numbers, then (1) and (2) yield respectively:

( )

⎟ ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎜ ⎝ ⎛ + ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ = + + ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ ⎟ ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎜ ⎝ ⎛ + ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ = − − − k k n k k k k n k k k n k

n g U

V k n U P U V k n U V k n U 1 2 3 3 1 2 ... 4 4 2 3 2

( )

k k k U P V 1 4 4 2 2 2 2 − = + − ⎟ ⎠ ⎞ ⎜ ⎝ ⎛

Where g

( )

x is a polynomial with integer coefficients and without independent term. The last identity implies that , 1

2 ⎟⎠=

⎞ ⎜ ⎝ ⎛ k k U V GCD .

Now, since _{k k}

k n U U U U

GCD , _⎟⎟

⎠ ⎞ ⎜⎜

⎝ ⎛

, k

( )

_k

n k k n k k n U g V k n U U U U U

GCD ⎟ +

⎠ ⎞ ⎜ ⎝ ⎛ = ⎟⎟ ⎠ ⎞ ⎜⎜ ⎝

⎛ −1

2

, , then

⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ k k n U U U

GCD , divides to

( ) ( )

1 1 2 2 − − ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ = − + ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ k n k k k k n k V k n U g U g V k n , so k n U U U GCD _k k n , _⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛

also in this case because , 1 2 ⎟⎠=

⎞ ⎜ ⎝ ⎛ k k U V GCD .

k n k k n n P U V P U V ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ _{+ +} = + + 4 2 2 4 2 2 2 2

. (3)

In an analogous way to case 1), it can be seen that

⎟ ⎟ ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎜ ⎜ ⎝ ⎛ ⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ k U U e k n U U U GCD k n , 2 2 divides to k n , so

in order to prove that

k n U U U GCD _k k n , _⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛

, it is sufficient to see that

⎟ ⎠ ⎞ ⎜ ⎝ ⎛ ≤ ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ ⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ k n e U U U GCD e _k k n 2

2 , . Three new cases appear:

- If k n

is an odd number, then expanding the right side of (3) and equaling the terms with P2 +4, it is obtained that:

(

)

(

)

2

1 1 2 1 2 2 3 1 4 2 ... 4 2 2 3 2 ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ − − − − + ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ + + + ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ ⎟ ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎜ ⎝ ⎛ + ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ = k n k n k k k n k k n k k n P U P U V k n V k n U U , so k n U U is

an odd number regardless 2

k

U

is an odd number or 2

k

V

is an odd number and then

⎟ ⎠ ⎞ ⎜ ⎝ ⎛ = = ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ ⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ k n e U U U GCD e k k n 2

2 , 0 .

-If k n

is an even number and 2

k

V

is an odd number, then 2

k

U

is an even number and it is obtained in an analogous way than in the previous case that:

(

)

(

)

2

Now, it is a known fact that ⎟ ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎜ ⎝ ⎛ ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ i k n i k n GCD k n ,

, so ⎟

⎠ ⎞ ⎜ ⎝ ⎛ ≥ ⎟ ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎜ ⎝ ⎛ ⎟ ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎜ ⎝ ⎛ k n e i k n

e_{2 2} when i is an odd

number. Since 2

k

U

is an even number, this implies that ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ = ⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ k n e U U e k n 2

2 , so

⎟ ⎠ ⎞ ⎜ ⎝ ⎛ ≤ ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ ⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ k n e U U U GCD e _k k n 2

2 , as desired.

-If k n and 2 k V

are even numbers, then 2

k

U

is an odd number, so

⎟ ⎠ ⎞ ⎜ ⎝ ⎛ ≤ ≤ ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ ⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ k n e U U U GCD e k k n 2

2 , 1 , as desired.

Now the theorem is stated: Theorem

If 1

2 , ⎟= ⎠ ⎞ ⎜ ⎝ ⎛ n P

GCD when n is an even number, n has not prime factors congruent with 1 modulus 4 and it has the factor three, then the only powers included in U_n are

8 1 , 1 6 = −

U and U₁₂1 ,−1 =144 Proof

It is satisfied that

(

)

3 2 3 3 1 U U P U U U

U n n

n = = + , with

3 1 , 2 3 n P U U GCD n ⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛

+ by the lemma.

Hence _⎟⎟

⎠ ⎞ ⎜⎜ ⎝ ⎛ +1 , 2 3 P U U

GCD n has not prime factors congruent with 1 modulus 4 since n has not prime factors congruent with 1 modulus 4 by hypothesis. It is also satisfied that

⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ +1 , 2 3 P U U GCD n

1) If , 2 1 1 3

= ⎟⎟ ⎠ ⎞ ⎜⎜

⎝ ⎛

+ P U U

GCD n

, then

(

)

n m

n u

U U P

U = + =

3 2

1 implies that P2 +1=vm with

N

v∈ , so the Catalan equation xm− y2 =1 has a solution in positive integers:

(

v,P

)

, a contradiction (see Mihailescu)

2) If , 2 1 2

3

= ⎟⎟ ⎠ ⎞ ⎜⎜

⎝ ⎛

+ P U U

GCD n , then n must be an even number because

3 2 1 , 2 3

n P

U U

GCD n _⎟⎟=

⎠ ⎞ ⎜⎜

⎝ ⎛

+ by the lemma, so

(

)

3 2 2

3 2 3

2 1

U U

U P

P U U

U U U

Un = n = + n .

Now,

(

)

n m

n u

U U

U P

P

U = + =

3 2 2

1 implies that P=wm with w∈N, P2 +1=2vm

with v∈N, because , 2 1 2 3

= ⎟⎟ ⎠ ⎞ ⎜⎜

⎝ ⎛

+ P U U

GCD n

and , 1

2 = ⎟⎟ ⎠ ⎞ ⎜⎜

⎝ ⎛

U U P

GCD n

by the lemma and

the hypothesis 1

2 , ⎟=

⎠ ⎞ ⎜ ⎝ ⎛ n

P

GCD . Then it is satisfied that 2vm−

( )

w2 m =1, so the diophantine equation 2 xm −ym =1 has a solution in positive integers:

(

v,w2

)

. But for

3 ≥

m the only solution in positive integers to this equation is x= y=1 (see Darmon). This implies that w2 =1 and then P=wm =1 (P is a positive number). But in this case the only m-power with m≥3 in Un for the required conditions on n is

3 1

, 1

6 =8=2

−

U ,

as stated in the introduction.

For m=2 it is satisfied that 2v2 −w4 =1, so the diophantine equation 2 x2 − y4 =1 has a solution in positive integers:

(

v,w

)

. But the only solution in positive integers to this equation is x= y=1 (see M. Le). This implies that P=wm =1. But in this case the only square in U_n for the required conditions on n is U₁₂1 ,−1 =144=122, as stated in the introduction and the proof is finished.

3. Concluding remarks

A result about the powers in the Lucas sequences Un has been stated for indices n

studies of Kagawa, Ribenboim about the squares in the Lucas sequences and of Bennett about the m-powers with m≥3 in the Lucas sequences.

A future line of work could be to relax the conditions on the index n in the theorem of section 2

References

1. M. A. Bennett, Powers in recurrence sequences: Pell equations, Transactions of the American Mathematical Society, 357 (2004), 1675–1691

2. A. Bremner and N. Tzanakis, Lucas sequences whose 12th or 9th term is a square, J. Number Theory, 107 (2004), 215-227.

3. Y. Bugeaud, M. Mignotte and S. Siksek, Classical and modular approaches to exponential Diophantine equations, I. Fibonacci and Lucas perfect powers. Annals of Math. 163 (2006), 969-1018.

4. Cohn, J. H. E. (1996). Perfect Pell powers. Glasgow Mathematical Journal 38 (1): 19–20.

5. H. Darmon and L. Merel, Winding quotients and some variants of Fermat’s Last Theorem, J. Reine Angew. Math. 490 (1997), 81–100.

6. Kagawa, T. and Terai, N.: Squares in Lucas sequences and some Diophantine equations. Manuscripta Math. 96, 195–202 (1998).

7. M. Le. On the diophantine equation

(

xm +1

)(

xn +1

)

= y2. Acta arithmetica, LXXXII.1, 17-26, 1997.