#### Diophantine Triples and k-Generalized Fibonacci Sequences

Diophantine Triples and k-Generalized Fibonacci Sequences
Clemens Fuchs 0 1 2 3 4
Christoph Hutle 0 1 2 3 4
Florian Luca 0 1 2 3 4
László Szalay 0 1 2 3 4
B Clemens Fuchs 0 1 2 3 4
Florian Luca 0 1 2 3 4
0 Centro de Ciencias Matemáticas UNAM , Morelia , Mexico
1 University of Witwatersrand , Johannesburg , South Africa
2 University of Salzburg , Hellbrunner Str. 34/I, 5020 Salzburg , Austria
3 University of West Hungary , Sopron , Hungary
4 J. Selye University , Hradna ul. 21, 94501 Komarno , Slovakia
We show that if k ≥ 2 is an integer and Fn(k) generalized Fibonacci numbers, then there are only finitelyn≥m0ainsythtreipsleeqsuoefncpeosoiftivkeintegers 1 < a < b < c such that ab + 1, ac + 1, bc + 1 are all members of Fn(k) : n ≥ 1 . This generalizes a previous result where the statement for k = 3 was proved. The result is ineffective since it is based on Schmidt's subspace theorem.
Diophantine triples; Generalized Fibonacci numbers; Diophantine equations; Application of the Subspace theorem
1 Introduction
There are many papers in the literature concerning Diophantine m-tuples which are
sets of m distinct positive integers {a1, . . . , am } such that ai a j + 1 is a square for all
1 ≤ i < j ≤ m (see [4], for example). A variation of this classical problem is obtained
if one changes the set of squares by some different subsets of positive integers like
kpowers for some fixed k ≥ 3, or perfect powers, or primes, or members of some linearly
recurrent sequence, etc. (see [7,12–14,18]). In this paper, we study this problem with
the set of values of k-generalized Fibonacci numbers for some integer k ≥ 2. Recall
that these numbers denoted by Fn(k) satisfy the recurrence
F (k) n+k−1 + · · · + Fn(k)
n+k = F (k)
and start with 0, 0, . . . , 0 (k − 1 times) followed by 1. Notationwise, we assume that
Fi(k) = 0 for i = −(k − 2), − (k − 1), . . . , 0, and F1(k) = 1. For k = 2 we obtain
the Fibonacci numbers and for k = 3 the sequence of Tribonacci numbers. The result
is the following:
Theorem 1 Let k ≥ 2 be fixed. Then there are only finitely many triples of positive
integers 1 < a < b < c such that
ab + 1 = Fx(k), ac + 1 = Fy(k), bc + 1 = Fz(k)
(
1
)
hold for some integers x , y, z.
Our result generalizes the results obtained in [8], [11] and [13], where this problem
was treated for the cases k = 2 and k = 3. In [13] it was shown that there does not
exist a triple of positive integers a, b, c such that ab + 1, ac + 1, bc + 1 are Fibonacci
numbers. In [11] it was shown that there is no Tribonacci Diophantine quadruple, that
is a set of four positive integers {a1, a2, a3, a4} such that ai a j + 1 is a member of the
Tribonacci sequence (3-generalized Fibonacci sequence) for 1 ≤ i < j ≤ 4, and in
[8] it was proved that there are only finitely many Tribonacci Diophantine triples. In
the current paper, we prove the same result for all such triples having values in the
sequence of k-generalized Fibonacci numbers.
For the Proof of Theorem 1, we proceed as follows. In Sect. 2, we recall some
properties of the k-generalized Fibonacci sequence Fn(k) which we will need and we
prove two lemmata. The first lemma shows that any k − 1 roots of the characteristic
polynomial are multiplicatively independent. In the second lemma, the greatest
common divisor of Fx(k) − 1 and Fy(k) − 1 for 2 < y < x is estimated. In Sect. 3, we assume
that the Theorem 1 is false and give, using the Subspace theorem, a finite expansion
of infinitely many solutions. In Sect. 4, we use a parametrization lemma which is
proved by using results about finiteness of the number of non-degenerate solutions to
S-unit equations. Applying it to the finite expansion, this leads us to a condition on
the leading coefficient, which turns out to be wrong. This contradiction is obtained by
showing that a certain Diophantine equation has no solutions; this last Diophantine
equation has been treated in particular cases in [2] and [16].
2 Preliminaries
There are already many results in the literature about (Fn(k))n≥0. We will only use
what we need, which are the following properties. The sequence (Fn(k))n≥0 is linearly
recurrent of characteristic polynomial
Ψk (X ) = X k − X k−1 − · · · − X − 1.
The polynomial Ψk (X ) is separable and irreducible in Q[X ] and the Galois group thus
acts transitively on the roots, which we denote by α1, . . . , αk . If k is even or prime, the
Galois group is certainly Sk (see [16] for these statements). The polynomial Ψk (X )
has only one root, without loss of generality assume that this root is α1 > 1, which
is outside the unit disk (formally, this root depends also on k, but in what follows we
shall omit the dependence on k on this and the other roots of Ψk (X ) in order to avoid
notational clutter). Thus,
k
i=1
Ψk (X ) =
(X − αi ), where |αi | < 1, i = 2, . . . , k.
Observe that α1α2 · · · αk = (−1)k−1. Note also that
Ψk (X ) = X k −
X k−1 + · · · + 1
= X k − XXk −−11 =
X k+1 − 2X k + 1
,
X − 1
a representation which is sometimes useful. Furthermore,
(see Lemma 3 in [3]). The Binet formula of (Fn(k))n≥0 is given by
where
and
(see Theorem 1 in [3]). We have
1
2 − k < α1 < 2
F (k)
n
=
k
i=1
fi αin for all n ≥ 0,
(αi − 1)αi−1
fi = 2 + (k + 1)(αi − 2) , i = 1, . . . , k
|Fn(k) − f1α1k | <
for all n ≥ 1
(see Theorem 2 in [3]). We also need the fact that
(see [1]).
Furthermore, the following property is of importance; it follows from the fact that
there is no non-trivial multiplicative relation between the conjugates of a Pisot number
(cf. [15]). Since in our case it is rather easy to verify, we shall present a proof of what
we need.
Lemma 1 Each set of k − 1 different roots (e.g., {α1, . . . , αk−1}) is multiplicatively
independent.
Proof We shall prove the statement only for the set {α1, . . . , αk−1}. The general
statement follows easily by the transitivity of the Galois Group of the irreducible polynomial
Ψk (X ).
Let us denote by L := Q(α1, . . . , αk ) the splitting field of Ψk over Q and by OL
the ring of integers in L. Note that the roots α1, . . . , αk are certainly in the group of
units OL× which follows from
αi−1 = (−1)k−1(α1 · · · αi−1 · αi+1 · · · αk )
for i = 1, . . . , k. The extension L/Q is Galois. We denote by d its degree and by
G := Gal(L/Q) the Galois group of L over Q. Let G = {σ1, . . . , σd }. We consider
the map λ : OL× → Rd defined by x → (log |σ1(x )|, log |σ2(x )|, . . . , log |σd (x )|).
Observe that by the product formula we have for x ∈ L× that
and for x ∈ OL× this leads to
v∈ML
v∈L
v|∞
|x |v = 1;
|x |v = 1
since for every finite place v we have |x |v = 1. This means that the image of λ lies in
the hyperplane defined by X1 + · · · + Xd = 0 of Rd .
We will use the property that λ is a homomorphism (see e.g., [17]). So if we can
prove that the k − 1 vectors λ(α1), . . . , λ(αk−1) are linearly independent, this will
prove the statement.
Since Ψk is irreducible over Q, the Galois group G of L over Q acts transitively
on {α1, . . . , αk }, i.e., for each i = 1, . . . , k there is some Galois automorphism which
sends αi to α1; without loss of generality let σ1, . . . , σk be such that σi (αi ) = α1.
Observe that σi−1(α1) = {αi } for i = 1, . . . , k. We have
λ(α1) = (log |α1|, log |σ2(α1)|, . . . , log |σk−1(α1)|, . . .),
λ(α2) = (log |σ1(α2)|, log |α1|, . . . , log |σk−1(α2)|, . . .),
.
.
.
λ(αk−1) = (log |σ1(αk−1)|, log |σ2(αk−1)|, . . . , log |α1|, . . .).
We will show that the matrix (log |σi (α j )|)i=1,...,k−1; j=1,...,k−1 consisting of the first
k − 1 entries of these k − 1 vectors has rank k − 1 implying the statement of the lemma.
Observe now that α1 · · · αk = (−1)k−1 and thus |σi (α1 · · · αk )| = 1 for all i =
1, . . . , k − 1. It follows that kj=1 log |σi (α j )| = 0 and thus
k−1
j=1
log |σi (α j )| = − log |σi (αk )| > 0
for j = 1, . . . , k − 1 and all i = 1, . . . , k − 1. Hence the transpose of the matrix above
is strictly diagonal-dominant since the diagonal entries, which are all equal to log |α1|,
are positive, all other entries are negative, and each row-sum (in the transpose matrix)
of all off-diagonal entries is in absolute value less than the corresponding diagonal
entry. It follows that the matrix is regular, which is what we wanted to show.
Finally, we prove the following result, which generalizes Proposition 1 in [8].
Observe that the upper bound depends now on k.
Lemma 2 Let x > y ≥ 3. Then
kx
gcd Fx(k) − 1, Fy(k) − 1 < α1k+1 .
Proof We may assume that y ≥ 4, since for y = 3, we get Fy(k) − 1 = 1, and there
is nothing to prove. We put d = gcd Fx(k) − 1, Fy(k) − 1 . Let κ be a constant to be
determined later. If y ≤ κ x + 1, then
From now on, we assume that y > κ x + 1. Using (
3
) and (
5
), we write
d ≤ Fy(k) − 1 < Fy(k) < α1y−1 ≤ α1κx .
Fx(k) = f1α1x + ζx , |ζx | < 1/2,
Fy(k) = f1α1y + ζy , |ζy | < 1/2.
d | Fx(k) − 1 − α1λ Fy(k) − 1
in K.
dη =
Fx(k) − 1 − α1λ Fy(k) − 1 ,
We put K = Q(α1). We let λ = x − y < (1 − κ)x − 1, and note that
We write (8) (9) (10)
where η is some algebraic integer in K. Note that the right-hand side above is not zero,
for if it were, we would get α1λ = Fx(k) − 1 / Fy(k) − 1 ∈ Q, which is false for λ > 0.
We compute norms from K to Q. Observe that
(see (
5
)) as well as (
2
). Further, let σi be any Galois automorphism that maps α1 to
αi . Then for i ≥ 2, we have
σi
Fx(k) − 1 − α1λ Fy(k) − 1
=
Fx(k) − 1 − αiλ Fy(k) − 1
< Fx(k) − 1 + Fy(k) − 1 < αx−1 + α y−1 − 2
In order to balance (9) and (11), we choose κ such that κ = 1 − κ/ k, giving κ =
k/(k + 1), and the lemma is proved.
3 Parametrizing the Solutions
In order to simplify the notations, we shall from now onwards write Fn instead of
Fn(k); we still mean the nth k-generalized Fibonacci number. The arguments in this
section follow the arguments from [8]. We will show that if there are infinitely many
solutions to (
1
), then all of them can be parametrized by finitely many expressions as
given in (18) for c below.
a =
From and hence
We assume that there are infinitely many solutions to (
1
). Then, for each integer
solution (a, b, c), we have
(Fx − 1)(Fy − 1)
Fz − 1
, b =
(Fx − 1)(Fz − 1)
Fy − 1
, c =
(Fy − 1)(Fz − 1)
Fx − 1
.
α1x+y−2 ≥ Fx Fy > (Fx − 1)(Fy − 1) ≥ Fz − 1 ≥ α1z−2 − 1 > α1z−3
we see that x + y > z − 1 and thus y ≥ z/2. In order to get a similar correspondence
for x and z, we denote d1 := gcd(Fy − 1, Fz − 1) and d2 := gcd(Fx − 1, Fz − 1),
such that Fz − 1 | d1d2. Then we use Lemma 2 to obtain
αx−1 > Fx > Fx − 1 ≥ d2 ≥
1
Fz − 1
d1
≥
α1z−2 − 1 > α1z− kk+z1 −3
kz
α k+1
1
which we can write as x > C1z for some small constant C1 < 1 (depending only on
k), when z is sufficiently large.
Next, we do a Taylor series expansion for c which was given by
k
x > 1 − k + 1 z − 2,
c =
(Fy − 1)(Fz − 1)
Fx − 1
.
Using the power sum representations of Fx , Fy, Fz, we get
c =
f1α1(−x+y+z)/2
· 1 + (−1/ f1)α1−x + ( f2/ f1)α2x α1−x + · · · + ( fk / f1)αkx α1−x −1/2
· 1 + (−1/ f1)α1−y + ( f2/ f1)α2yα1−y + · · · + ( fk / f1)αkyα1−y 1/2
· 1 + (−1/ f1)α1−z + ( f2/ f1)α2zα1−z + · · · + ( fk / f1)αkzα1−z 1/2 .
We then use the binomial expansion to obtain
1 + (−1/ f1)α1−x + ( f2/ f1)α2x α1−x + · · · + ( fk/ f1)αkx α1−x 1/2
(13)
(14)
where O has the usual meaning, using estimates from [9] and where T is some index,
which we will specify later. Since x < z and z < x /C1, the remainder term can also
be written as O α1−T x /C1 , where x = max{x , y, z} = z. Doing the same for y
and z likewise and multiplying those expression gives
c =
f1α1(−x+y+z)/2 ⎝ 1 +
⎛
n−1
j=1
⎞
d j M j ⎠ + O(α1−T x /C1 ),
where the integer n depends only on T and where J is a finite set, d j are non-zero
coefficients in L := Q(α1, . . . , αk ), and M j is a monomial of the form
M j =
k
i=1
in which x = (x , y, z), and Li, j (x) are linear forms in x ∈ R3 with integer coefficients
which are all non-negative if i = 2, . . . , k and non-positive if i = 1. Note that each
monomial M j is “small” that is there exists a constant κ > 0 (which we can even
choose independently of k), such that
This follows directly from
|M j | ≤ e−κx for all j ∈ J.
|M j | = |α1|L1, j (x) · |α2|L2, j (x) · · · |αk |Lk, j (x)
≤ (2 − 1/ k)L1, j (x) · 1 · · · 1
≤ (3/2)−x
≤ e−κx for all j ∈ J.
Our aim of this section is to apply a version of the Subspace theorem given in [5]
to show that there is a finite expansion of c involving terms as in (13); the version we
are going to use can also be found in Sect. 3 of [10]. For the set-up—in particular the
notion of heights—we refer to the mentioned papers.
We work with the field L = Q(α1, . . . , αk ) and let S be the finite set of infinite
places (which are normalized so that the Product Formula holds, cf. [5]). Observe that
α1, . . . , αk are S-units. According to whether −x + y + z is even or odd, we set = 0
or = 1 respectively, such that
α1(−x+y+z− )/2 ∈ L.
By going to a still infinite subset of the solutions, we may assume that is always
either 0 or 1.
Using the fixed integer n (depending on T ) from above, we now define n +1 linearly
independent linear forms in indeterminants (C, Y0, . . . , Yn ). For the standard infinite
place ∞ on C, we set
l0,∞(C, Y0, . . . , Yn−1) := C −
f1α1Y0 −
f1α1
d j Y j ,
(15)
| det(l0,v, . . . , ln,v)|v · H(y)−(n+1)−δ
(16)
where
∈ {0, 1} is as explained above, and
For all other places v in S, we define
li,∞(C, Y0, . . . , Yn−1) := Yi−1 for i ∈ {1, . . . , n}.
l0,v := C, li,v := Yi−1 for i = 1, . . . , n.
We will show that there is some δ > 0, such that the inequality
n
and the double product on the left-hand side can be split up into
|c −
f1α1 y0 −
f1α1
d j y j |∞ ·
|c|v ·
v∈ML,∞,v=∞
v∈S\ML,∞
n−1
j=1 v∈S
|c|v ·
|y j |v.
Now notice that the last double product equals 1 due to the Product Formula and that
v∈S\ML,∞
|c|v ≤ 1,
since c ∈ Z. An upper bound on the number of infinite places in L is k! and hence
v∈ML,∞,v=∞
|c|v <
(Ty − 1)(Tz − 1) k!
Tx − 1
y y k k
≤ f1α1 + · · · + fk αk − 1 ! f1α1z + · · · + fk αkz − 1 !
≤ (1 · 2 x + 1/2)2·k!
using (
6
) and (
5
). And finally the first expression is just
which, by (13), is smaller than some expression of the form C2α1T x /C1 . Therefore,
we have
v∈S i=0
Now we choose T (and the corresponding n) in such a way that
n
|li,v(y)|v < C2α1T x /C1 · (2 x
|y|v
+ 1/2)2·k!.
(17)
and
holds. Then we can write
For the height of our vector y, we estimate
T x
C2α1−T x /C1 < α1− 2C1
(2 x
T x
+ 1/2)2·k! < α14C1
n
with suitable constants C3, C4, C5. For the second estimate, we used that
H(M j ) ≤ H(α1)Cα1 (x)H(α2)Cα2 (x) · · · H(αk )Cαk (x)
and bounded it by the maximum of those expressions. Furthermore we have
−x+y2+z−
H α1
n
≤ α1n x ,
which just changes our constant C4.
Now finally, the estimate
T x
α1− 2C1
is satisfied when we pick δ small enough.
≤ α1−δC5 x
So all the conditions for the Subspace theorem are met. Since we assumed that
there are infinitely many solutions (x , y, z) of (16), we now can conclude that all of
them lie in finitely many proper linear subspaces. Therefore, there must be at least one
proper linear subspace, which contains infinitely many solutions and we see that there
exists a finite set J and (new) coefficients e j for j ∈ J in L such that we have
c = α1(−x+y+z− )/2 ⎝ e0 +
⎛
j∈Jc
e j M j ⎠
⎞
(18)
with (new) non-zero coefficients e j and monomials M j as before.
Likewise, we can find finite expressions of this form for a and b.
4 Proof of the Theorem
We use the following parametrization lemma:
Lemma 3 Suppose, we have infinitely many solutions for (
1
). Then there exists a line
in R3 given by
x (t ) = r1t + s1, y(t ) = r2t + s2, z(t ) = r3t + s3
with rationals r1, r2, r3, s1, s2, s3, such that infinitely many of the solutions (x , y, z)
are of the form (x (n), y(n), z(n)) for some integer n.
Proof Assume that (
1
) has infinitely many solutions. We already deduced in Sect. 3
that c can be written in the form
c = α1(−x+y+z− )/2 ⎝ ec,0 +
ec, j Mc, j ⎠
⎛
⎛
j∈Jc
j∈Jb
⎞ ⎛
⎞
⎞
with Jc being a finite set, ec, j being coefficients in L for j ∈ Jc ∪ {0} and Mc, j =
ik=1 αiLc,i, j (x) with x = (x , y, z). In the same manner, we can write
b = α1(x−y+z− )/2 ⎝ eb,0 +
eb, j Mb, j ⎠ .
Since 1 + bc = Fz = f1α1 + · · · + fk αkz , we get
z
⎛
f1α1z +· · ·+ fk αkz −α1z−ε ⎝ eb,0 +
j∈Jb
eb, j Mb, j ⎠ ⎝ ec,0 +
ec, j Mc, j ⎠
= 1. (19)
j∈Jc
⎞
Substituting
into (19), we obtain an equation of the form
αk = α1 · · · αk−1
(−1)k−1
j∈J
e j α1L1, j (x) · · · αkL−k−11, j (x) = 0,
(20)
where again J is some finite set, e j are non-zero coefficients in L and Li, j are linear
forms in x with integer coefficients.
This is an S-unit equation, where S is the multiplicative group generated by
{α1, . . . , αk , −1}. We may assume that infinitely many of the solutions x are
nondegenerate solutions of (20) by replacing the equation by an equation given by a
suitable vanishing subsum if necessary.
We may assume that (L1,i , . . . , Lk−1,i ) = (L1, j , . . . , Lk−1, j ) for any i = j ,
because otherwise we could just merge these two terms.
Therefore for i = j , the theorem on non-degenerate solutions to S-unit equations
(see [6]) yields that the set of
α1L1,i (x)−L1, j (x) · · · αkL−k−11,i (x)−Lk−1, j (x)
is contained in a finite set of numbers. By Lemma 1, α1, . . . , αk−1 are multiplicatively
independent and thus the exponents (L1,i − L1, j )(x), . . . , (Lk−1,i − Lk−1, j )(x) take
the same value for infinitely many x. Since we assumed that these linear forms are not
all identically zero, this implies that there is some non-trivial linear form L defined
over Q and some c ∈ Q with L(x) = c for infinitely many x. So there exist rationals
ri , si , ti for i = 1, 2, 3 such that we can parametrize
x = r1 p + s1q + t1, y = r2 p + s2q + t2, z = r3 p + s3q + t3
with infinitely many pairs ( p, q) ∈ Z2.
We can assume that ri , si , ti are all integers. If not, we define Δ as the least common
multiple of the denominators of ri , si (i = 1, 2, 3) and let p0, q0 be such that for
infinitely many pairs ( p, q) we have p ≡ p0 mod Δ and q ≡ q0 mod Δ. Then
p = p0 + Δλ, q = q0 + Δμ and
x = (r1Δ)λ + (s1Δ)μ + (r1 p0 + s1q0 + t1)
y = (r2Δ)λ + (s2Δ)μ + (r2 p0 + s2q0 + t2)
z = (r3Δ)λ + (s3Δ)μ + (r3 p0 + s3q0 + t3).
Since ri Δ, si Δ and x , y, z are all integers, ri p0 + si q0 + ti are integers as well.
Replacing ri by ri Δ, si by si Δ and ti by ri p0 + si q0 + ti , we can indeed assume that
all coefficients ri , si , ti in our parametrization are integers.
Using a similar argument as in the beginning of the proof, we get that our equation
is of the form
j∈J
e j α1L1, j (r) · · · αkL−k−11, j (r) = 0,
where r := (λ, μ), J is a finite set of indices, e j are new non-zero coefficients in L
and Li, j (r) are linear forms in r with integer coefficients. Again we may assume that
we have (L1,i (r), . . . , Lk−1,i (r)) = (L1, j (r), . . . , Lk−1, j (r)) for any i = j .
Applying the theorem of non-degenerate solutions to S-unit equations once more,
we obtain a finite set of numbers Λ, such that for some i = j , we have
α(L1,i −L1, j )(r)
1
(Lk−1,i −Lk−1, j )(r)
· · · αk−1
∈ Λ.
So every r lies on a finite collection of lines and since we had infinitely many r, there
must be some line, which contains infinitely many solution, which proves our lemma.
We apply this lemma and define Δ as the least common multiple of the denominators
of r1, r2, r3. Infinitely many of our n will be in the same residue class modulo Δ, which
we shall call r . Writing n = mΔ + r , we get
(x , y, z) = ((r1Δ)m + (rr1 + s1), (r2Δ)m + (rr2 + s2), (r3Δ)m + (rr3 + s3)).
Replacing n by m, ri by ri Δ and si by rri + s, we can even assume that ri , si are
integers. So we have
−x + y + z −
2
=
(−r1 + r2 + r3)m
2
+
−s1 + s2 + s3 −
2
.
This holds for infinitely many m, so we can choose a still infinite subset such that all
of them are in the same residue class δ modulo 2 and we can write m = 2 + δ with
fixed δ ∈ {0, 1}. Thus we have
−x + y + z −
2
= (−r1 + r2 + r3) + η,
where η ∈ Z or η ∈ Z + 1/2.
Using this representation, we can write (18) as
c( ) = α1(−r1+r2+r3) +S ⎝ e0 +
⎛
j∈Jc
⎞
e j M j ⎠ .
(21)
for infinitely many , where
M j =
k
as before and x = x( ) = (x (2 + δ), y(2 + δ), z(2 + δ)). From this, we will now
derive a contradiction.
First we observe that there are only finitely many solutions of (21) with c( ) = 0.
This can be shown by using the fact that a simple non-degenerate linear recurrence
has only finite zero-multiplicity (see [6] for an explicit bound). We will apply this
statement here for the linear recurrence in ; it only remains to check that no quotient
of two distinct roots of the form α1L1,i (x( )) · · · αkLk,i (x( )) is a root of unity or, in other
words, that
α1m1 α2m2 · · · αkmk n = 1
(22)
has no solutions in n ∈ Z/{0}, m1 < 0 and mi > 0 for i = 2, . . . , k. Assume relation
(22) holds. Replacing α1 by (−1)k−1(α2 · · · αk )−1 gives
α22(m2−m1) · · · αk2(mk −m1) = 1.
By squaring this equation and applying Lemma 1 we get 2(m2 − m1) = · · · =
2(mk − m1) = 0 and thus m1 = m2 = · · · = mk , which is impossible because of the
signs of m1 and m2, . . . , mk .
So we have confirmed that c( ) = 0 for still infinitely many solutions. We use (12)
and write
(Fx − 1)c2 = (Fy − 1)(Fz − 1).
(23)
Then we insert the finite expansion (21) in for c into (23). Furthermore, we use
the Binet formula (
3
) and write Fx , Fy , Fz as power sums in x , y and z respectively.
Using the parametrization (x , y, z) = (r1m + s1, r2m + s2, r3m + s3) with m = 2 or
m = 2 + 1 as above, we have expansions in on both sides of (
3
). Since there must
be infinitely many solutions in , the largest terms on both sides have to grow with the
same rate. In order to find the largest terms, we have to distinguish some cases: If we
assume that e0 = 0 for infinitely many of our solutions, then e0α1(−x+y+z− )/2 is the
largest term in the expansion of c and we have
f1α1x e02α1−x+y+z−
y z
= f1α1 f1α1.
It follows that e02 = f1α1. The case e0 = 0 for infinitely many of our solutions is not
possible, because then, the right-hand side of (23) would grow faster than the left-hand
side so that (23) could be true for only finitely many of our . In the other cases, we
have e0 = f1α1, where ∈ {0, 1}. This now contradicts the following lemma, which
turns out to be slightly more involved than in the special case on Tribonacci numbers
(cf. [8]).
Lemma 4 √ f1 ∈/ L and √ f1α1 ∈/ L.
Proof Suppose that f1α1 ∈ L for some ∈ {0,1}. Then there is β ∈ L such that
f1α1 = β2. Using (
4
), we get that
Computing norms over Q, we get that where denotes a rational square. Note that
NL/Q(α1)− NL/Q(α1 − 1)
NL/Q(2 + (k + 1)(α1 − 2)) = NL/Q(β)2 = ,
(24)
and
and finally that
k
i=1
k
NL/Q(α1) =
αi = |(−1)k · (−1)| = 1,
NL/Q(α1 − 1) = (αi − 1) = |Ψk(
1
)| = k − 1,
i=1
Hence, we get that equation (24) leads to This leads to
for some integer w. But this equation has no integer solutions, which is proved in the
theorem below. This concludes the proof.
In order to finish the proof, we have the following result, which might be of
independent interest since particular cases were considered before in [2] and [16].
Theorem 2 The Diophantine equation (25) has no positive integer solutions (k, w)
with k ≥ 2.
Proof The cases k ≡ 1, 2 (mod 4) have already been treated both in [2] and in [16].
We treat the remaining cases. If k ≡ 0 (mod 4), then the left-hand side of (25) is
congruent to −1 (mod 4), and therefore it cannot be a square. Finally, assume that
k ≡ 3 (mod 4). Then k + 1 is even, 2(k+1)/2 | w, and putting w1 = w/2(k+1)/2, we
get
(26)
We then get
kk − ((k + 1)/2)k+1 = w12.
kk = w12 + ((k + 1)/2)k+1.
Note that the two numbers in the right-hand side of (26) are coprime, for if p divides
w1 and (k + 1)/2, then p divides the left-hand side of (26). Thus p divides both k
and (k + 1)/2, so also k − 2((k + 1)/2) = −1, a contradiction. Thus, the right-hand
side is a sum of two coprime squares and therefore all odd prime factors of it must be
1 modulo 4 contradicting the fact that in the left-hand side we have k ≡ 3 (mod 4).
This finishes the proof of this theorem.
Acknowledgments Open access funding provided by University of Salzburg. C.F. and C.H. were supported
by FWF (Austrian Science Fund) Grant No. P24574 and by the Sparkling Science project EMMA Grant
No. SPA 05/172.
Open Access This article is distributed under the terms of the Creative Commons Attribution 4.0
International License (http://creativecommons.org/licenses/by/4.0/), which permits unrestricted use, distribution,
and reproduction in any medium, provided you give appropriate credit to the original author(s) and the
source, provide a link to the Creative Commons license, and indicate if changes were made.
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