ON THE DIOPHANTINE EQUATION Ln =x
5SZ. TENGELY
Dedicated to Professors K. Győry and A. Sárközy on their 70th birthdays
and Professors A. Pethő and J. Pintz on their 60th birthdays

Abstract. In this paper we` ´determine all integral solutions (n, x) of the Diophantine equation Ln = x5 , where Ln is the n-th Lucas number which isdefined as follows, L0 = 2, L1 = 1 and Ln = Ln−1 + Ln−2 for n > 1. We follow ideas described in [10], that is we combine Baker’s method and the so-called
Mordell-Weil sieve to show that the only positive solution is (n, x) = (1, 5).

1. introduction
There are many articles concerning the Diophantine equation
Rn = P (x),where Rn is a linear recursive sequence and P ∈ Z[X] is a polynomial. Several
papers have been published identifying perfect powers, products of consecutive
integers, binomial coefficients, figurate numbers and power sums in the Fibonacci,
Lucas, Pell and associated Pell sequences. These binary recurrence sequences aredefined by
F0 = 0,F1 = 1,

Fn = Fn−1 + Fn−2 for n ≥ 2,

L0 = 2,L1 = 1,

Ln = Ln−1 + Ln−2 for n ≥ 2,

P0 = 0,P1 = 1,

Pn = 2Pn−1 + Pn−2 for n ≥ 2,

Q0 = 1,Q1 = 1,

Qn = 2Qn−1 + Qn−2 for n ≥ 2.

It follows from a result by Ljunggren [24] that the only squares in the Fibonaccisequence are F0 = 0, F1 = F2 = 1, F12 = 144. Later it was rediscovered by Cohn
[16, 17] and Wyler [46]. Alfred [1] and Cohn [18] determined the perfect squares
in the Lucas sequence. In case of the Pell sequence Pethő [32] and independently
Cohn [19] obtained the complete list of perfect squares. London and Finkelstein[25] and Pethő [30] proved that the only cubes in the Fibonacci sequence are F0 =0, F1 = F2 = 1 and F6 = 8. London and Finkelstein [25] also showed the theonly cube in the Lucas sequence is 1. Higher powers were determined by Pethő
[31]. Bugeaud, Mignotte and Siksek [13] applied a combination of Baker’s method,
modular approach and some classical techniques to show that the perfect powers
in the Fibonacci sequence are 0,1,8 and 144, and the perfect powers in the Lucassequence are 1 and 4.
Another interesting problem is to determine triangular numbers, numbers of thein binary recurrence sequences. Ming [26] proved that the only
form Tx = x(x+1)
2
triangular numbers in the Fibonacci sequence are F0 = 0, F1 = F2 = 1, F4 = 3, F8 =21 and F10 = 55. It was shown by Ming [27] that L1 = 1, L2 = 3 and L18 = 5778
are the triangular numbers in the Lucas sequence. In case of the Pell sequence
2000 Mathematics Subject Classification. Primary 11D61; Secondary 11Y50.
Key words and phrases. Diophantine equations, recurrence sequences.
Research supported in part by OTKA PD75264 and János Bolyai Research Scholarship of the
Hungarian Academy of Sciences.
1

2

SZ. TENGELY
McDaniel [29] proved that the only triangular number is 1. Since Tx = x2 , it wasa natural question to ask for all solutions of the Diophantine equations
 
 
x
x
Fn =
, Ln =
,
k
k
 
 
x
x
Pn =
, Qn =.
kk
It was Szalay [42] who solved the equations
Fn , Ln , Pn = x3 . Later Szalay [41] alsoPxtreated the equations Fn , Ln = x3 and Fn , Ln , Pn = i=1 i3 . Kovács [22] solved
completely some related combinatorial Diophantine equations, e.g.
 
x
Pn =
4
and
Fn = Π4 (x) = x(x + 1)(x + 2)(x + 3).
Tengely [43] determined the g-gonal numbers in the Fibonacci, Lucas, Pell andassociated Pell sequences for g ≤ 20.
In this paper we consider the Diophantine equation
 
x
(1)
Ln =
.
5
We prove the following result.
Theorem 1. The only positive solution of equation (1) is (n, x) = (1, 5).
2. auxiliary results
Consider the hyperelliptic curve(2)C:

y 2 = F (x) := x5 + b4 x4 + b3 x3 + b2 x2 + b1 x + b0 ,where bi ∈ Z. Let α be a root of F and J(Q) be the Jacobian of the curve C. We
have thatx − α = κξ 2
where κ, ξ ∈ K = Q(α) and κ comes from a finite set. By knowing the Mordell-Weilgroup of the curve C it is possible
Pm to provide a method to compute such a finite set.
To each coset representative i=1 (Pi − ∞) of J(Q)/2J(Q) we associate
κ=

m
Y


γi − αd2i ,i=1where the set {P1 , . . . , Pm } is stable under the action of Galois, all y(Pi ) are nonzero and x(Pi ) = γi /d2i where γi is an algebraic integer and di ∈ Z≥1 . If Pi , Pj areconjugate then we may suppose that di = dj and so γi , γj are conjugate. We havethe following lemma (Lemma 3.1 in [10]).
Lemma 1. Let K be a set of κ values associated as above to a complete set of coset
representatives of J(Q)/2J(Q). Then K is a finite subset of OK and if (x, y) is an
integral point on the curve (2) then x − α = κξ 2 for some κ ∈ K and ξ ∈ K.
As an application of his theory of lower bounds for linear forms in logarithms,
Baker [2] gave an explicit upper bound for the size of integral solutions of hyperelliptic curves. This result has been improved by many authors (see e.g. [3], [4], [6],[11], [34], [36], [37] and [45]).
In [10] an improved completely explicit upper bound were proved combining
ideas from [11], [12], [13], [23], [28], [33], [45], [44]. Now we will state the theorem
which gives the improved bound. We introduce some notation. Let K be a number

ON THE DIOPHANTINE EQUATION Ln =

`x´

3

5

field of degree d and let r be its unit rank and R its regulator. For α ∈ K we denoteby h(α) the logarithmic height of the element α. Let log 2
if d = 1, 2,
d∂K = 1  log log d 3

if d ≥ 3
4
log d
and
0
∂K
=



π2
1+ 2
∂K

1/2
.

Define the constants(r !)2,2r−1 drdrc3 (K) = c1 (K)
,
∂K
rr+1
c5 (K) = r−1 .
2∂K
c1 (K) =
c2 (K) = c1 (K)

d
∂K

r−1
,

c4 (K) = rdc3 (K),Let
∂L/K = max

0[L : Q] , [K : Q]∂K,

0.16[K : Q]∂K

,

where K ⊆ L are number fields. DefineC(K, n) := 3 · 30n+4 · (n + 1)5.5 d2 (1 + log d).
The following theorem will be used to get an upper bound for the size of the integral
solutions of our equation. It is Theorem 3 in [10].
Theorem 2. Let α be an algebraic integer of degree at least 3 and κ be an integer
belonging to K. Denote by α1 , α2 , α3 distinct conjugates of α and by κ1 , κ2 , κ3
the corresponding conjugates of κ. Let√√√
K1 = Q(α1 , α2 , κ1 κ2 ), K2 = Q(α1 , α3 , κ1 κ3 ), K3 = Q(α2 , α3 , κ2 κ3 ),and√√
L = Q(α1 , α2 , α3 , κ1 κ2 , κ1 κ3 ).

In what follows R stands for an upper bound for the regulators of K1 , K2 and K3
and r denotes the maximum of the unit ranks of K1 , K2 , K3 . Letc∗j = max cj (Ki )
1≤i≤3

and
N = max1≤i,j≤3

NormQ(αi ,αj )/Q (κi (αi − αj ))2andH ∗ = c∗5 R +log N+ h(κ).
min1≤i≤3 [Ki : Q]

DefineA∗1 = 2H ∗ · C(L, 2r + 1) · (c∗1 )2 ∂L/L ·2rmax ∂L/Ki

1≤i≤3· R2 ,

andA∗2 = 2H ∗ + A∗1 + A∗1 log{(2r + 1) · max{c∗4 , 1}}.
If x ∈ Z\{0} satisfies x − α = κξ 2 for some ξ ∈ K thenlog|x| ≤ 8A∗1 log(4A∗1 ) + 8A∗2 + H ∗ + 20 log 2 + 13 h(κ) + 19 h(α).

4SZ. TENGELY

To obtain a lower bound for the possible unknown integer solutions we are goingto use the so-called Mordell-Weil sieve. The Mordell-Weil sieve has been successfully
applied to prove the non-existence of rational points on curves (see e.g. [7], [9], [21]and [35]).
Let C/Q be a smooth projective curve (in our case a hyperelliptic curve) of
genus g ≥ 2. Let J be its Jacobian. We assume the knowledge of some rational
point on C, so let D be a fixed rational point on C and let  be the corresponding
Abel-Jacobi map:
 : C → J,
P 7→ [P − D].
Let W be the image in J of the known rational points on C and D1 , . . . , Dr generators for the free part of J(Q). By using the Mordell–Weil sieve we are going toobtain a very large and smooth integer B such that
(C(Q)) ⊆ W + BJ(Q).
Let
φ : Zr → J(Q),φ(a1 , . . . , ar ) =X

ai Di ,

so that the image of φ is the free part of J(Q). The variant of the Mordell-Weilsieve explained in [10] provides a method to obtain a very long decreasing sequenceof lattices in Zr
BZr = L0 ) L1 ) L2 ) · · · ) Lk
such that
(C(Q)) ⊂ W + φ(Lj )for j = 1, . . . , k.
The next lemma [10, Lemma 12.1] gives a lower bound for the size of rational
points whose image are not in the set W.
Lemma 2. Let W be a finite subset of J(Q) and L be a sublattice of Zr . Supposethat (C(Q)) ⊂ W + φ(L). Let µ1 be a lower bound for h − ĥ andqĥ(w) : w ∈ W .
µ2 = max
Denote by M the height-pairing matrix for the Mordell–Weil basis D1 , . . . , Dr andlet λ1 , . . . , λr be its eigenvalues. Letnpoµ3 = minλj : j = 1, . . . , rand m(L) the Euclidean norm of the shortest non-zero vector of L. Then, for any
P ∈ C(Q), either (P ) ∈ W or2h((P )) ≥ (µ3 m(L) − µ2 ) + µ1 .
3. proof of Theorem 1
In this section first we prove a lemma and than we use it to prove Theorem 1.
Lemma 3. (a) The integral solutions of the equation(3)

C + : Y 2 = X 2 (X + 15)2 (X + 20) + 180000000

are(X, Y ) ∈ {(25, −15000), (25, 15000)}.
(b) There are no integral solution of the equation(4)

C − : Y 2 = X 2 (X + 15)2 (X + 20) − 180000000.

ON THE DIOPHANTINE EQUATION Ln =

`x´
5

5

Proof of Lemma 3. We start with the proof of part (a). Let J(Q)+ be the Jacobianof the genus two curve (3). Using MAGMA [5] we obtain that J(Q)+ is free of rank
1 with Mordell-Weil basis given by
D = (25, 15000) − ∞.
The MAGMA programs used to compute these data are based on Stoll’s papers[38], [39], [40]. The rank of the Jacobian of C + is 1, so classical Chabauty’s method(see e.g. [14], [15], [20]) can be applied. The Chabauty procedure of MAGMA
provides an upper bound for the number of rational points on the curve and in thiscase it is equal to the number of known points. Therefore
C + (Q) = {∞, (25, ±15000)}.
Now we deal with part (b). Let J(Q)− be the Jacobian of the genus two curve(4). Using MAGMA we determine a Mordell-Weil basis which is given by
D1 = (ω1 , −200ω1 ) + (ω 1 , −200ω 1 ) − 2∞,
D2 = (ω2 , 120000) + (ω 2 , 120000) − 2∞,
where ω1 is a root of the polynomial x2 − 5x + 1500 and ω2 is a root of x2 + 195x +13500.
Let f = x2 (x + 15)2 (x + 20) − 180000000 and α be a root of f . We will choose forP2
coset representatives of J(Q)− /2J(Q)− the linear combinations i=1 ni Di , whereni ∈ {0, 1}. Thenx − α = κξ 2 ,where κ ∈ K and K is constructed as described in Lemma 1. We have that K ={1, α2 − 5α + 1500, α2 + 195α + 13500, α4 + 190α3 + 14025α2 + 225000α + 20250000}.
By local arguments it is possible to restrict the set K further (see e.g. [7], [8]). In
our case one can eliminate
α2 − 5α + 1500,

α2 + 195α + 13500

by local computations in Q2 andα4 + 190α3 + 14025α2 + 225000α + 20250000
by local computations in Q3 . It remains to deal with the case κ = 1. We apply
Theorem 2 to get a large upper bound for log |x|. A MAGMA code were written to
obtain the bounds appeared in [10], it can be found at
http://www.warwick.ac.uk/∼maseap/progs/intpoint/bounds.m. We used theabove Magma functions to compute an upper bound corresponding to the case
κ = 1. It turned out to be
1.58037 × 10285 .
The set of known rational points on the curve (4) is {∞}. Let W be the image ofthis set in J(Q)− . Applying the Mordell-Weil implemented by Bruin and Stoll andexplained in [10] we obtain that (C(Q)) ⊆ W + BJ(Q)− , whereB = 26 · 32 · 52 · 72 · 112 · 132 · 19 · 23 · 31 · 41 · ·43 · 47 · 61 · 67 · 79 · 83 · 109 · 113 · 127,
that isB = 678957252681082328769065398948800.
Now we use an extension of the Mordell-Weil sieve due to Samir Siksek to obtaina very long decreasing sequence of lattices in Z2 . After that we apply Lemma 2 toobtain a lower bound for possible unknown rational points. We get that if (x, y) is
an unknown integral point, thenlog |x| ≥ 7.38833 × 101076 .
This contradicts the bound for log |x| we obtained by Baker’s method.

6

SZ. TENGELY

Finally we prove Theorem 1.
Proof of Theorem 1. We will use the following well known property of the sequences
Fn and Ln :L2n − 5Fn2 = 4(−1)n .
We have that 2x
± 4 = 5Fn2 .
5
The above equation can be reduced to two genus two curves given by (3) and (4),where Y = 53 5!Fn and X = 5x2 − 20x. By Lemma 3 we have that X = 25 and we
also have that X = 5x2 − 20x. That is it remains to solve a quadratic equation in
x. We obtain that x ∈ {−1, 5}. Hence the only positive solution of equation (1) is(n, x) = (1, 5), that is 
5
1 = L1 =.
5
Remark. A similar method could be used to treat the equation Fn = x5 , where
Fn is the n-th Fibonacci number. We obtain two genus two curves, one of them canbe solved by using the ideas described in [10], but in case of the remaining curve
the rank of the Jacobian is 4 and we could not obtain a good enough lower boundfor the possible unknown integral solutions.
Acknowledgement. The work is supported by the TÁMOP 4.2.1./B-09/1/KONV2010-0007 project. The project is implemented through the New Hungary Development Plan, co-financed by the European Social Fund and the European Regional
Development Fund.
The author is very greatful to Professor Michael Stoll for useful conversation
relating to the arithmetic of genus 2 curves and the Mordell-Weil sieve during theworkshop ”Rational Points 3” organized in Schloss Thurnau in 2010.
Special thanks are due to the anonymous referees for their careful reading of themanuscript and numerous pertinent remarks.
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Mathematical Institute
University of Debrecen
P.O.Box 124010 Debrecen
Hungary
E-mail address: tengely@math.klte.hu