TRIBONACCI NUMBERS THAT ARE PRODUCTS OF TWO FIBONACCI

NUMBERS

FLORIAN LUCA, JAPHET ODJOUMANI, AND ALAIN TOGBÉ

Abstract. Let Tm be the mth Tribonacci number and Fn be the nth Fibonacci number. In
this paper, we solve the Diophantine equation

Tm = FnFk

in positive integer unknowns m, n, and k.

1. Introduction

The Tribonacci sequence {Tm}m≥0 is given by T0 = 0, T1 = T2 = 1, and the recurrence

Tm+3 = Tm+2 + Tm+1 + Tm for all m ≥ 0.

Its characteristic polynomial is X3 −X2 −X − 1 = (X − α)(X − β)(X − β̄), where

α =
1 + r1 + r2

3
, β =

2− (r1 + r2) + i
√
3(r1 − r2)

6
,

with

r1 =
3

√
19 + 3

√
33 and r2 =

3

√
19− 3

√
33.

The Fibonacci sequence {Fn}n≥0 is given by F0 = 0, F1 = F2 = 1, and

Fn+2 = Fn+1 + Fn, for n ≥ 0.

Its characteristic polynomial is X2 −X − 1 = (X − γ)(X − δ), where

γ =
1 +

√
5

2
and δ =

1−
√
5

2
.

In this paper, we study the Diophantine equation

Tm = FnFk (1)

in positive unknowns m, n, and k.

Theorem 1. The only nonzero Tribonacci numbers that are products of two Fibonacci numbers
are

1, 2, 4, 13, and 24.

Our method of proof involves the application of Baker’s lower bounds for nonzero linear
forms in logarithms of algebraic numbers, and the Baker-Davenport reduction procedure.
Computations are done with the help of a computer program in Mathematica.

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TRIBONACCI NUMBERS THAT ARE PRODUCTS OF TWO FIBONACCI NUMBERS

2. Preliminary Results

2.1. Results on Tribonacci and Fibonacci Numbers. Here, we recall some well-known
formulas and inequalities on Tribonacci and Fibonacci sequences. Binet’s formula for the
Tribonacci sequence is

Tm = aαm + bβm + b̄β̄m for all m ≥ 0,

where

a =
5α2 − 3α− 4

22
, and b =

5β2 − 3β − 4

22
.

The minimal polynomial of a over integers is 44X3 − 2X − 1, with zeros a, b, and b̄ with
max{|a|, |b|, |b̄|} < 1. We have the numerical estimates

1.83 < α < 1.84;

0.73 < |β| = α−1/2 < 0.74; (2)

0.33 < |a| < 0.34;

0.25 < |b| < 0.27.

For m ≥ 1, denoting e(m) := Tm − aαm, we have

|e(m)| < α−m
2 . (3)

Furthermore,

αm−2 ≤ Tm ≤ αm−1 for all m ≥ 1. (4)

Binet’s formula for the Fibonacci sequence is

Fn =
γn − δn√

5
for all n ≥ 0.

One has γδ = −1. Furthermore, for n ≥ 1, one has

γn−2 ≤ Fn ≤ γn−1. (5)

For a number field K, let dK be its degree over Q, also usually denoted by [K : Q]. Because
dQ(α) = 3 and dQ(γ) = 2, it follows that Q(α) ̸= Q(γ). Further, Q(α) = Q(a). The numbers
α, γ, and a are positive and belong to the real field K = Q(α, γ) of degree dK = 6.

2.2. Auxiliary Results on Linear Forms in Logarithms of Algebraic Numbers. In
this subsection, we point out some useful results from the theory of lower bounds for nonzero
linear forms in logarithms of algebraic numbers. Let η ̸= 0 be an algebraic number of degree
d and let

a0(X − η(1)) · · · (X − η(d)) ∈ Z[X]

be the minimal polynomial over Z of η = η(1) with positive leading coefficient a0 ≥ 1. Then,
the absolute logarithmic Weil height is defined by

h(η) :=
1

d

(
log a0 +

d∑
i=1

max{0, log |η(i)|}

)
.

The height has the following basic properties. For algebraic numbers η, γ, and s ∈ Z, we have

h(η ± γ) ≤ h(η) + h(γ) + log 2, (6)

h(ηγ±1) ≤ h(η) + h(γ), (7)

h(ηs) = |s|h(η) (s ∈ Z). (8)

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THE FIBONACCI QUARTERLY

In the case that η is a rational number, say η = p/q ∈ Q with coprime integers p, q ≥ 1, we
have h(p/q) = max{log |p|, log q}.

Let K be a real number field, η1, . . . , ηt ∈ K and b1, . . . , bt ∈ Z \ {0}. Let B ≥
max{|b1|, . . . , |bt|} and put

Λ := ηb11 · · · ηbtt − 1.

Let A1, . . ., At be real numbers with

Ai ≥ max{dKh(ηi), | log ηi|, 0.16}, i = 1, 2, . . . , t.

With these basic notations, we have the following version of a lower bound for a nonzero linear
form in logarithms from Matveev [6].

Theorem 2. [3, Theorem 9.4] Assume that Λ ̸= 0. Then,

log |Λ| > −1.4 · 30t+3 · t4.5 · d2K · (1 + log dK) · (1 + logB) ·A1 · · ·At.

We also need the following lemma.

Lemma 1. [5, Lemma 7] If l ≥ 1, H >
(
4l2
)l
, and H > L/(logL)l, then

L < 2lH(logH)l.

Applying these results to some appropriate linear forms in logarithms resulting from equa-
tion (1), we end up with a large upper bound on max{k,m, n}, which we need to reduce. For
that, we use the following result of Dujella and Pethő [4], which is a variant of the reduction
method from Baker and Davenport [1].

Lemma 2. Let M be a positive integer, p/q be a convergent of the continued fraction expansion
of the irrational number τ such that q > 6M , and A, B, and µ be some real numbers with
A > 0 and B > 1. If

ε := ∥µq∥ −M · ∥τq∥ > 0,

then there is no solution to the inequality

0 < |uτ − v + µ| < AB−w

in positive integers u, v, and w with

u ≤ M and w ≥ log(Aq/ε)

logB
.

3. The Proof of the Main Result

Let (m,n, k) be a solution of Diophantine equation (1). We suppose 2 ≤ n ≤ k and m ≥ 2
because F1 = F2 = T1 = T2 = 1. We assume that n ≥ 3, because for n = 2, we get Tm = Fk

and the only common terms of the Fibonacci and Tribonacci sequence are 1, 2 = F3 = T3,
and 13 = F7 = T6 (see, for example, Theorem 1 in [2] for a more general result). From now
on, we assume k ≥ n ≥ 3 and m ≥ 4.

From (4) and (5), we have

αm−2 < Tm = FnFk < γn+k−2 and γn+k−4 < αm−1.

This implies that
log γ

logα
(n+ k)− 2.2 < m <

log γ

logα
(n+ k) + 0.5. (9)

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TRIBONACCI NUMBERS THAT ARE PRODUCTS OF TWO FIBONACCI NUMBERS

Furthermore, from equation (1), we have∣∣∣∣5aαm

γn+k
− 1

∣∣∣∣ = ∣∣∣−5e(m)γ−(n+k) − (−1)nδ2n − (−1)kδ2k + (−1)n+kδ2(n+k)
∣∣∣

< 5|e(m)|γ−(n+k) + |δ|2n + |δ|2k + |δ|2(n+k)

< 5α
−m
2 γ−(n+k) + γ−2n + γ−2k + γ−2(n+k)

< (5α−1 + 3)γ−2n,

because m ≥ 4 and k ≥ n. We thus obtain∣∣∣∣5aαm

γn+k
− 1

∣∣∣∣ < 5.74γ−2n. (10)

Let Λ1 := 5aαmγ−n−k − 1. We have Λ1 ̸= 0. To see this, if Λ1 = 0, then

5a · αm · 2n+k = (1 +
√
5)n+k ∈ Q(

√
5),

which is false (aαm is a real number larger than 1 that has two complex conjugates of absolute
values smaller than 1, which cannot be in the normal real field Q(

√
5)).

We can now apply Theorem 2 to Λ1 with t := 3 and

(η1, b1) := (α, m), (η2, b2) := (γ, −n− k), and (η3, b3) := (5a, 1).

We compute upper bounds on the absolute logarithmic Weil height of each of the above
algebraic numbers and we get

h(η1) = h(α) < 0.204,

h(η2) = h(γ) < 0.241,

h(η3) = h(5a) ≤ log 5 + h(a) < 2.871.

(11)

We have dK = 6 and we can choose

A1 = 1.23, A2 = 1.45, A3 = 17.23, and B = n+ k.

We then get

log |Λ1| > −8.85 · 1014 log(n+ k).

In the above, we used that 1 + log(n + k) < 2 log(n + k) because n + k ≥ 4. Combining the
above bound with inequality (10), we obtain

2n log γ < 8.85 · 1014 log(n+ k) + log 5.74 < 8.85 · 1014 log(n+ k) + 1.748,

so

n log γ < 4.43 · 1014 log(n+ k). (12)

We rewrite equation (1) as

aαm + e(m) = Fn

(
γk − δk√

5

)
.

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THE FIBONACCI QUARTERLY

Because n ≥ 3, we have ∣∣∣∣aαm

Fn
− γk√

5

∣∣∣∣ < |e(m)|
Fn

+
|δ|k√
5∣∣∣∣∣

√
5aαm

Fnγk
− 1

∣∣∣∣∣ <
√
5

γk

(
1

Fn
+

1√
5γk

)

<

(√
5

2
+

1

γ3

)
γ−k,

because k ≥ 3 and Fn ≥ 2. Thus, putting

Λ2 := (
√
5a/Fn)α

mγ−k − 1,

we obtain

|Λ2| < 1.4γ−k. (13)

Of course, we have Λ2 ̸= 0 again because aαm ̸∈ Q(
√
5). So, we can apply Theorem 2 again

to Λ2 with t := 3 and

(η1, b1) := (α, m), (η2, b2) := (γ, −k), and (η3, b3) :=

(√
5a

Fn
, 1

)
.

We have

h

(√
5a

Fn

)
≤ h(

√
5) + h(a) + log(Fn) < h(a) + log(

√
5) + log

(
γn−1

)
< 2.5n log γ,

where the last inequality above holds because n ≥ 3. We again have dK = 6 and we can choose

A1 = 1.23, A2 = 1.45, and A3 = 15n log γ, B = n+ k

and get

log |Λ2| > −7.7 · 1014n log γ log(n+ k).

Combining this with inequality (13), we obtain

k log γ < 7.7 · 1014 · n log γ log(n+ k) + log(1.35)

leading to

k log γ < 7.71 · 1014(n log γ) log(n+ k). (14)

Therefore, considering the upper bound of n log γ from inequality (12), we get

k <

(
(7.71 · 1014) · (4.43 · 1014)

log γ

)
log2(n+ k) < 7.1 · 1029 log2(2k).

or

2k < 14.2 · 1030 log(2k)2.
Applying Lemma 1 with l = 2, L = 2k, and H = 14.2 · 1030, we obtain

k < 1.37 · 1034. (15)

We next reduce the above bound by applying Lemma 2. Recall that for a positive real x, if
|x− 1| < 0.5, then |log x| < 1.5 |x− 1| (see [7, Lemma 4]). Hence, we have from inequality
(10) (note that because n ≥ 3 the right side of inequality (10) is indeed smaller than 0.5),

0 < |m logα− (n+ k) log γ + log(5a)| < 8.61γ−2n,

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TRIBONACCI NUMBERS THAT ARE PRODUCTS OF TWO FIBONACCI NUMBERS

which implies that

0 <

∣∣∣∣m logα

log γ
− (n+ k) +

log(5a)

log γ

∣∣∣∣ < 17.893γ−2n. (16)

Note that α and γ are multiplicatively independent. Indeed, if αx = γy for integers x and
y, then this common value lives in Q(α)∩Q(γ) = Q because Q(α) has degree 3 and Q(γ) has
degree 2. Thus, αx = γy is a rational unit so it is ±1, and this is possible only when x = y = 0.
Thus, logα/ log γ is an irrational. From inequalities (9) and (15), we have

m <
log γ

logα
· (2 · k) + 0.5 <

log(0.5 + 0.5
√
5)

log 1.83
· (2 · 13700 + 0.5) · 1030;

i.e.,

m < 2.22 · 1034. (17)

We apply Lemma 2 with w := 2n,

τ :=
logα

log γ
, µ :=

log(5a)

log γ
, A := 17.893, B := γ, and M := 2.22× 1034.

With the help of Mathematica, we find that the 71st convergent of τ is

p71
q71

=
452544523220541439982411039079661113

357364106913532334879636629737733870
.

Its denominator satisfies q71 > 6M and ε > 0.247647 > 0. Thus, inequality (16) has no
solution for

2n ≥ log(17.893q71/ε)

log γ
> 179.01,

which implies that

n ≤ 89.

Substituting this upper bound for n into inequality (14), we obtain

k < 6.9 · 1016 log(n+ k) ≤ 6.9 · 1016 log(2k).

We again apply Lemma 1 with l = 1, H = 13.8 · 1016, and L = 2k and get

k < 5.49 · 1018.

From here, because n ≤ k and from inequality (9), we have

m < 8.75 · 1018.

We consider Λ2 and we get from inequality (13)

0 <

∣∣∣∣∣m logα− k log γ + log

(√
5a

Fn

)∣∣∣∣∣ < 3γ−k.

This implies that

0 <

∣∣∣∣∣m logα

log γ
− k +

log
(√

5a/Fn

)
log γ

∣∣∣∣∣ < 7γ−k. (18)

We then apply Lemma 2 with w := k,

τ :=
logα

log γ
, µ :=

log
(√

5a/Fn

)
log γ

, A := 7, B := γ, M := 8.75 · 1018.

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THE FIBONACCI QUARTERLY

With the help of Mathematica, we find that the 41st convergent of τ is

p41
q41

=
237161759629456603958

187281242121494666147
.

It satisfies q41 > 6M and ε > 0.010633 > 0 for all n ∈ [3, 89]. Hence, inequality (18) has no
solution for

k ≥ log(7q41/ε)

log γ
> 110.4.

Thus, we obtain k ≤ 110 and consequently m ≤ 175 by inequality (9). We now check for the
solutions of equation (1) for 3 ≤ n ≤ 89, n ≤ k ≤ 110, and 4 ≤ m ≤ 175. This is done quickly
with a small program in Mathematica and we get

(m, k, n) ∈ {(4, 3, 3), (7, 6, 4)} .
This yields the additional Tribonacci numbers T4 = 4 = F 2

3 and T7 = 24 = F4 · F6, which are
products of two Fibonacci numbers from the statement of Theorem 1. This finishes the proof.

Acknowledgment

We thank the anonymous referee for carefully reading our manuscript and making valuable
suggestions. The second author worked on this paper during a visit to the School of Maths
of Wits University in Johannesburg, South Africa, in August 2022 partially supported by an
EMS (European Mathematical Society) collaboration grant. This author thanks the School
for its hospitality and support.

References

[1] A. Baker and H. Davenport, The equations 3x2−2 = y2 and 8x2−7 = z2, Quart. J. Math. Oxford Ser. (2),
20 (1969), 129–137.

[2] J. J. Bravo and F. Luca, Coincidences in generalized Fibonacci sequences, J. Number Theory, 133.6 (2013),
2121–2137.

[3] Y. Bugeaud, M. Mignotte, and S. Siksek, Classical and modular approaches to exponential Diophantine
equations (I), Fibonacci and Lucas perfect powers, Ann. Math. (2), 163.3 (2006), 969–1018.

[4] A. Dujella and A. Pethö, A generalization of a theorem of Baker and Davenport, Quart. J. Math. Oxford
Ser. (2), 49 (1998), 291–306.

[5] S. Guzmán and F. Luca, Linear combinations of factorials and s-units in a binary recurrence sequence,
Ann. Math. Québec, 38 (2014), 169–188.

[6] E. M. Matveev, An explicit lower bound for a homogeneous rational linear form in logarithms of algebraic
numbers II, Izv. Ross. Akad. Nauk Ser. Mat., 64.6 (2000), 125–180.

[7] J. Odjoumani and V. Ziegler, On prime powers in linear recurrence sequences, Ann. Math. Québec (to
appear).

MSC2020: 11D45, 11B37, 11D61

School of Mathematics, Wits University, Johannesburg, South Africa, and Centro de Ciencias
Matemáticas UNAM, Morelia, Mexico

Email address: florian.luca@wits.ac.za

Institut de Mathématiques et de Sciences Physiques (IMSP), Université d’Abomey-Calavi
(UAC), Dangbo, Benin

Email address: japhet.odjoumani@imsp-uac.org

Department of Mathematics, Statistics, Purdue University Northwest, 1401 S. U.S. 421,
Westville IN 46391 USA

Email address: atogbe@pnw.edu

304 VOLUME 61, NUMBER 4