be an exponential polynomial over a field of zero characteristic. Assume that for each pair

Hans Peter Schlickewei · Wolfgang M. Schmidt · Michel Waldschmidt

Zeros of linear recurrence sequences

Received: 7 August 1998

Abstract. Letf (x) = P0(x)α^x₀+ · · · +P_k(x)α^x_k

be an exponential polynomial over a field of zero characteristic. Assume that for each pair

with

,

is not a root of unity. Define

j=0(degP_j+1). We introduce a partition of

α0, . . . , α_k

into subsets

(1

into

, so that, for 1

,

∈P_ki(Q)

, while for 1

, the number

either is transcendental or else is algebraic with not too small a height. Then we show that for all but at most exp

5

solutions

of

0, we have

0

1. Introduction

Let K be a field of zero characteristic, α

, . . . , α

be non-zero elements of K and P

, . . . , P

non-zero polynomials with coefficients in K . Consider an exponential polynomial

f (x) = X

P

(x)α

.

We study the equation

f (x) = 0 ( 1 . 1 )

in x ∈ Z . We suppose that for each pair i, j with i 6= j ,

α

/α

is not a root of unity. (1.2)

H. P. Schlickewei: Fachbereich Mathematik, Universität Marburg, Hans-Meerwein-Strasse, Lahnberge, D-35032 Marburg, Germany.

e-mail: hps@mathematik.Uni-Marburg.de

W. M. Schmidt: Department of Mathematics, University of Colorado, Campus Box 395, Boulder, CO 80309-0395, USA. e-mail: schmidt@euclid.colorado.edu

M. Waldschmidt: Université P. et M. Curie (Paris VI), Institut Mathématique de Jussieu, Problèmes Diophantiens, Case 247, 4, Place Jussieu, F-75252 Paris Cedex 05, France. e- mail: miw@math.jussieu.fr

Mathematics Subject Classification (1991): 11B37, 11D61, 11J13

We set

1(f ) = X

( deg P

+ 1 ).

It is well known that f ( 0 ), f ( 1 ), . . .

is a linear recurrence sequence of order 1(f ), which is “non-degenerate”. Vice versa, any non-degenerate linear recurrence sequence u

, u

, . . .

of elements of K of order q has some representation u

= f (n) , where f is an exponential polynomial as above satisfying (1.2) and with 1(f ) = q . For more details, cf. e.g. [8].

So in studying (1.1) we study the zeros of linear recurrence sequences. An old conjecture says that the number of solutions x ∈ Z of equation (1.1) is bounded above by a function that depends only upon 1(f ) . Let us briefly review what is known so far in this context

.

For q = 1, equation (1.1) reduces to a

α

= 0 and clearly there is no solution x at all.

For q = 2, we have one of the following two equations (a

+ a

x)α

= 0 or a

α

+ a

α

= 0.

In either case, in view of our assumption (1.2) on non-degeneracy, we clearly do not have more than one solution x .

The first non-trivial case is q = 3. Here, Schlickewei [4] proved the con- jecture to be true. His bound has been improved by Beukers and Schlickewei [1]. They showed that for q = 3 equation (1.1) does not have more than 61 solutions.

Now suppose q ≥ 4. In a recent paper [3], Evertse, Schlickewei and Schmidt proved the following: Suppose that in (1.1) the polynomials f

for i = 0 , . . . , k are constant. Then equation (1.1) does not have more than exp ( 7 k)

solutions. As in this situation q = k + 1, we see that when the polynomials f

in (1.1) are all constant, the conjecture is true.

There remains the case when q ≥ 4 and when not all f

’s are constant.

Now obviously in (1.1) we may suppose without loss of generality that α

= 1. With this normalization, Schlickewei [5] proved the following:

Suppose that α

, . . . , α

are algebraic and that [Q(α

, . . . , α

) : Q] ≤ d . Then the number of solutions of equation (1.1) is bounded in terms of q and d only. (A bound was given explicitly). Schlickewei and Schmidt [6] later on established the bound ( 2 q)

d

.

1

After the present paper was written, the second author [7] settled this conjecture.

We denote by Q the algebraic closure of Q in K (this is the field of algebraic elements in K ). We define an equivalence relation on the set K

of non-zero elements of K by the condition

z

∼ z

⇐⇒ z

/z

is algebraic . This relation induces a partition of

α

, . . . , α

: α

, . . . , α

=

[

α

, . . . , α

,

where, for 1 ≤ i ≤ m ,

α

: · · · : α

∈ P

(Q),

while for 1 ≤ i 6= u ≤ m , the number α

/α

is transcendental. Accord- ingly, f is decomposed into

f = f

+ · · · + f

, (1.3) with

f

(x) = P

(x)α

+ · · · + P

(x)α

(1 ≤ i ≤ m).

We prove

Theorem 1.1. Suppose we have (1.2). Define 1 = 1(f ) and F (1) = exp 1( 5 1)

.

Then for all but at most F (1) solutions x ∈ Z of (1.1), we have

f

(x) = · · · = f

(x) = 0 . ( 1 . 4 ) Our result, in other words, says that the only case when the conjec- ture possibly could fail to be true arises from the algebraic case, i.e. when α

, . . . , α

are in Q . Moreover we shall see that the conjecture would follow from the special case where α

, . . . , α

are algebraic and each α

/α

has a small height. Actually our method of proof gives a result of the type stated in the Theorem also under the assumption that the quotients α

/α

are not transcendental but have logarithmic height bounded away from zero (for more details, see the final remark in Sect. 6).

We mention that our proof was inspired by a similar result for q = 3 by Beukers and Tijdeman [2]. They showed:

Let α and β be non-zero elements of K . Suppose that α , β and α/β are not roots of unity. Let a and b be non-zero elements of K . Suppose that the equation

aα

+ bβ

= 1

has at least 4 solutions x ∈ Z . Then α and β are algebraic.

Our proof uses a recent result of Schlickewei and Schmidt [6] on polynomial

exponential equations.

2. Heights

Let K be a number field of degree d . Write M(K) for the set of places of K . For v ∈ M(K) , let | |

be the valuation which extends either the standard absolute value of Q , or if v|p for a rational prime p , let | |

be the valuation with |p|

= p

. Write d

for the local degree [K

: Q

] and define the absolute value k k

by

k k

= | |

.

Let n ≥ 1 and let α = (α

, . . . , α

) 6= (0, . . . , 0) be a point in K

. We then put

kαk

= max

kα

k

, . . . , kα

k

and we define the homogeneous height as H (α) = Y

v∈M(K)

kαk

.

Since it depends only on the class α = (α

: · · · : α

) of α in P

(Q), we also denote it by H (α) . Let

h(α) = h(α

: · · · : α

) = log H(α)

be the homogeneous logarithmic absolute height of α ∈ P

(Q) We shall also need the inhomogeneous absolute heights

H

(x) = H(1 : x

: · · · : x

) and

h

(x) = h(1 : x

: · · · : x

) = log H

(x) of x = (x

, . . . , x

) ∈ Q

. Further, for x ∈ Q, we set

H

(x) = H ( 1 : x) and h

(x) = h ( 1 : x) = log H

(x).

Given D ∈ N and h / > 0, we will use the fact that the set of elements α ∈ Q

with

deg α ≤ D and h

(α) ≤ h /

is finite.

3. Algebraic linear recurrence sequences

The results in this section are consequences of the Subspace Theorem.

Lemma 3.1. Let m ≥ 1 and 0 be a finitely generated subgroup of (Q

)

of rank r ≥ 0. Then the solutions z = x ∗ y = (x

y

, . . . , x

y

) of z

+ · · · + z

= 1 (3.1) with z ∈ 0 , y ∈ Q

and

h

(y) ≤ 1

4m

h

(x) (3.2)

are contained in the union of at most

exp ( 4 m)

(r + 1 ) proper subspaces of Q

.

Proof. This is a variation on Proposition A of [6]. In that proposition there was a distinction between three kinds of solutions:

i) Solutions where some y

= 0, i.e., some z

= 0. These clearly lie in m subspaces.

ii) Solutions where each y

6= 0 and where h

(x) > 2m log m. These were called large solutions in [6] and it was shown in (10.4) of that paper that they lie in the union of fewer than

2

(21m

)

proper subspaces.

iii) Solutions where each y

6= 0 and where h

(x) ≤ 2m log m. These were called small solutions in [6]. Here we argue as follows. We have h

(y) ≤ (2m log m)/(4m

) < log 2 by (3.2). Then each component has h

(y

) < log 2, which is H

(y

) < 2. Since y

∈ Q

, we have y

= ± 1. The equation (3.1) now becomes

±x

± x

± · · · ± x

= 1. (3.3) The group 0

generated by 0 and the vectors (± 1 , · · · , ± 1 ) contains no more than r multiplicatively independent elements. By Proposition 2.1 of [6], the solutions of (3.3) lie in the union of not more than

exp ( 4 m)

· 2 (r + 1 )

proper subspaces of Q

.

Combining our estimates we obtain m + 2

( 21 m

)

+ exp ( 4 m)

· 2 (r + 1 )

< exp ( 4 m)

(r + 1 ) . ut Corollary. Let q > 1 and let 0 be a finitely generated subgroup of (Q

)

of rank r ≥ 0. Then the solutions of

z

+ · · · + z

= 0 (3.4) where z = x ∗ y with x ∈ 0, y ∈ Q

and

h (y) ≤ 1 4q

h (x) are contained in the union of fewer than

exp ( 4 q)

(r + 1 )

( 3 . 5 ) proper subspaces of the space given by (3.4).

Proof. This is just the homogeneous version of Lemma 3.1. We apply Lemma 3.1 with m = q − 1. One needs also to consider the possible solu- tions with z

= 0. But they lie in one subspace, and 1 is absorbed in (3.5) since q > m. ut

Lemma 3.2. Let α ∈ Q

be given with h

(α) > 0. Let a ∈ Q

. Then there is a u ∈ Z such that

h

(aα

) ≥ 1

4 h

(α)|x|

for x ∈ Z.

Proof. This is the case r = n = 1 of Lemma 15.1 in [6]. ut Agreement. We define the degree of the zero polynomial as −1.

Lemma 3.3. Consider an equation

P

(x)α

+ · · · + P

(x)α

= 0 ( 3 . 6 ) where (α

, . . . , α

) ∈ (Q

)

and, for 0 ≤ j ≤ k, P

is a non-zero polynomial of degree d

≥ 0 with algebraic coefficients. Write

1 = X

j=0

(d

+ 1 ), D = max

0≤j≤k

d

. Suppose that 1 ≥ 3,

0≤i,j

max

h (α

: α

) ≥ h /

where 0 < h / ≤ 1 and set

E = 16 1

D/ h /, t = exp ( 5 1)

+ 5 E log E.

Then there are tuples

P

, . . . , P

6= ( 0 , . . . , 0 ) ( 1 ≤ ` ≤ t)

of polynomials where deg P

≤ d

(0 ≤ j < k , 1 ≤ ` ≤ t ) and deg P

<

d

for ` = 1 , . . . , t , such that every solution x ∈ Z of (3.6) satisfies P

(x)α

+ · · · + P

(x)α

= 0 ( 3 . 7 ) for some ` .

Proof. Suppose u ∈ Z and set y = x + u. Then (3.6) may be rewritten as P

(y − u)α

α

+ · · · + P

(y − u)α

α

= 0 ,

which is the same as

P e

(y)α

+ · · · + P e

(y)α

= 0 , ( 3 . 8 ) with P e

(Y ) = P

(Y − u)α

( 0 ≤ j ≤ k).

Suppose our assertion is true for (3.8), with polynomials P e

, . . . , P e

(1 ≤ ` ≤ t). Thus every solution of (3.8) satisfies

P e

(y)α

+ · · · + P e

(y)α

= 0 for some `. But then x = y − u satisfies (3.7) with

P

(X) = P e

(X + u)α

( 0 ≤ j ≤ k).

We therefore may make a change of variables x 7→ y = x + u.

We may suppose that h (α

: α

) ≥ h / for some ι in the range 1 ≤ ι ≤ k . Write P

(X) = a

+a

X +· · ·+ a

X

. Pick u according to Lemma 3.2 with h (a

α

: a

α

) ≥

h /|y| . Writing P e

(Y ) = P

(Y − u)α

= b

+ b

Y + · · · + b

Y

, we have b

= a

α

, b

= a

α

, so that

h(b

α

: b

α

) ≥ 1

4 h /|y|. (3.9)

The equation (3.8) may be written as b

+ b

y +· · ·+b

y

α

+· · ·+ b

+ b

y +· · ·+ b

y

α

= 0 .

Some coefficients may be zero; omitting the zero coefficients, we rewrite this as

b

y

+ · · · + b

y

α

+ · · · + b

y

+ · · · + b

y

α

= 0.

Let q be the total number of (non-zero) coefficients here, and consider the following vectors in q -space:

x = b

α

, . . . , b

α

, . . . , b

α

, . . . , b

α

w = y

, . . . , y

, . . . , y

, . . . , y

. Our equation becomes

z

+ · · · + z

= 0 ( 3 . 10 ) with z = x ∗ w = (x

w

, . . . , x

w

). Hence x lies in the group 0 of rank r ≤ 2 generated by (α

, . . . , α

, . . . , α

, . . . , α

) and (b

, . . . , b

, . . . , b

, . . . , b

) . Further

h (x) ≥ h (b

α

: b

α

) ≥ 1 4 h /|y|

by (3.9). On the other hand, h (w) ≤ D log |y| . Therefore when

|y | ≥ 2E log E, (3.11)

so that |y| ≥ ( 32 q

D/ h /) log ( 16 q

D/ h /) by q ≤ 1 , then

|y| > ( 16 q

D/ h /) log |y|, and

h(w) ≤ D log |y| < h /

16 q

|y | = 1 4 q

1

4 h /|y| ≤ 1 4 q

h(x).

By the corollary, for such y , we have z contained in the union of exp (4q)

· 3

< exp (51)

proper subspaces of the space (3.10). Consider such a subspace c

z

+

· · · + c

z

= 0 (where (c

, . . . , c

) is not proportional to (1, . . . , 1)).

Taking a linear combination of this and (3.10) we obtain a non-trivial relation c

z

+· · ·+c

z

= 0. But this means exactly that y satisfies a non-trivial equation

Q e

(y)α

+ · · · + Q e

(y)α

= 0, (3.12) where deg Q e

≤ d

(0 ≤ j < k ) and deg Q e

< d

.

There are not more than 5 E log E values of y where (3.11) is violated. For fixed y, and since 1 = P

(d

+ 1) ≥ 3, there will certainly be polynomials Q e

, . . . , Q e

as above (deg Q e

≤ d

(0 ≤ j < k ) and deg Q e

< d

) with (3.12). ut

Remark. When h / ≥ exp −(51)

we have t ≤ exp (51)

.

4. A specialization-type argument

Lemma 4.1. Let K be a number field, D , N , M , L

, . . . , L

non-negative integers, A

, . . . , A

homogeneous polynomials in K [X

, . . . , X

] , each of degree ≤ D and B

(1 ≤ λ ≤ L

, 1 ≤ µ ≤ M) homogeneous polynomials in Q[X

, . . . , X

]. Assume that there exists α ∈ P

(K) such that

(i) A

(α) = · · · = A

(α) = 0 and

(ii) for each µ = 1, . . . , M, there exists λ ∈ {1, . . . , L

} with B

(α) 6= 0.

Then there exist elements e α

, . . . , e α

in K , algebraic over Q , not all of which are zero, which generate an extension K e = K(e α

, . . . ,e α

) of K of degree e K : K

≤ D

and such that the point e α = (e α

: · · · : e α

) ∈ P

( K) e satisfies

(i)

A

(e α) = · · · = A

(e α) = 0 and

(ii)

for each µ = 1 , . . . , M , there exists λ ∈ { 1 , . . . , L

} with B

(e α) 6= 0.

Proof. Given homogeneous polynomials Q

, . . . , Q

in K[X

, . . . , X

], we write

Z(Q

, . . . , Q

) ⊂ P

(K)

for the set of zeros in P

(K) of the ideal (Q

, . . . , Q

) in K[X

, . . . , X

] generated by Q

, . . . , Q

.

Let Y be an absolutely irreducible component of Z(A

, . . . , A

) ⊂ P

(K) containing α. Consider the Zariski closed subset

F =

\

Z(B

, . . . , B

)

of P

(K). By assumption α is not in F . Hence Lemma 4.1 is a consequence of the following statement:

Let A

, . . . , A

be homogeneous polynomials in K[X

, . . . , X

], each of degree ≤ D . Let Y be an irreducible component of dimension δ of

Z(A

, . . . , A

)

and F a Zariski closed subset of P

(K) such that Y \F is not empty. Then there exists an element e α = (e α

: · · · : e α

) in Y \ F whose components e α

, . . . ,e α

are algebraic over Q and such that we have

K(e α

, . . . , e α

) : K

≤ D

.

Since Y is absolutely irreducible and not contained in F , we have dim(Y ∩ F ) ≤ δ − 1. Pick linear forms L

, . . . , L

with coefficients in K and in sufficiently general position such that

Z(L

) ∩ · · · ∩ Z(L

) ∩ F ∩ Y = ∅ and such that moreover

Z(L

) ∩ · · · ∩ Z(L

) ∩ Y

is a non-empty finite set which does not contain more than D

points. Let γ = (γ

: · · · : γ

) be one of its elements. One at least among γ

, . . . , γ

is non-zero, say γ

. Put e α

= γ

/γ

. Then our construction implies that e α = ( 1 : e α

: · · · : e α

) = γ lies in Y \ F . Since our linear forms L

as well as the polynomials A

, . . . , A

have coefficients in K , it follows that for any K -embedding σ of K(e α

, . . . ,e α

) in K we have

( 1 : σe α

: · · · : σe α

) ∈ Z(L

) ∩ · · · ∩ Z(L

) ∩ Y.

Since moreover the right hand side has cardinality ≤ D

, we may conclude that in fact e α

, . . . ,e α

are algebraic over K and that

[K(e α

, . . . , e α

) : K ] ≤ D

. ut Here is a consequence of Lemma 4.1.

Lemma 4.2. Let k be a non-negative integer, p , S , T , d

, . . . , d

positive in- tegers and h / a positive real number. For 1 ≤ s ≤ S, let C

= (C

, . . . , C

) be a p-tuple of homogeneous polynomials in Q[X

, . . . , X

], each of de- gree d

. For 1 ≤ t ≤ T , let D

= (D

, . . . , D

) be a p -tuple of homo- geneous polynomials in Q[X

, . . . , X

] , with deg D

= · · · = deg D

. Let α

, . . . , α

be non-zero elements of K and α = (α

, . . . , α

) ∈ K

. Denote by V the subspace of K

spanned by C

(α), . . . , C

(α) . Assume that for each t = 1, . . . , T , we have D

(α) 6∈ V .

Then there exist non-zero algebraic elements e α

, . . . ,e α

in K such that e α = (e α

, . . . , e α

) ∈ Q

has the following properties. The subspace V e of K

spanned by C

(e α), . . . , C

(e α) has dim V e = dim V . Further, for each t = 1, . . . , T , we have D

(e α) 6∈ V e . Furthermore, for 0 ≤ i, j ≤ k , we have

( e α

/e α

= α

/α

if α

/α

is algebraic,

h (e α

: e α

) ≥ h / if α

/α

is transcendental .

Proof. Let K denote a number field containing all coefficients of C

(1 ≤ i ≤ p , 1 ≤ s ≤ S ) and all algebraic elements of K which belong to the set α

/α

; 0 ≤ i, j ≤ k . We shall prove the existence of e α = (e α

, . . . , e α

) ∈ K

satisfying the desired properties together with an upper bound for the degree of the number field K e = K(e α

, . . . ,e α

) , namely

e K : K

≤ D

with D = p max

1≤s≤S

d

.

Define r = dim V . Since D

(α) is not in V , we have V 6= K

, hence r < p . Denote by

A

, . . . , A

the set of (r + 1 ) × (r + 1 ) minors of the p × S

matrix

C

, . . . , C

.

Each of these polynomials A

, . . . , A

is homogeneous of degree

≤ (r + 1 ) max

1≤s≤S

d

≤ D.

Also, for 1 ≤ t ≤ T , denote by

B

, . . . , B

the set of (r + 1 )×(r + 1 ) minors of the p × (S + 1) matrix

C

, . . . , C

, D

. Further, let

A

, . . . , A

denote the set of polynomials α

X

− α

X

where (i, j) runs over the set of pairs with 0 ≤ i, j ≤ k for which α

/α

is algebraic. Furthermore, denote by

B

, . . . , B

the set of polynomials X

, . . . , X

, and βX

− X

, where (i, j) runs over the set of pairs with 0 ≤ i, j ≤ k for which α

/α

is transcendental, while β runs over the (finite) set of algebraic elements of K for which

[K(β) : K] ≤ D

and h

(β) ≤ h /.

By assumption the point α ∈ K

satisfies

A

(α) = · · · = A

(α) = 0 , B

(α) 6= 0 for T + 1 ≤ µ ≤ M,

and for each µ = 1, . . . , T , there exists λ ∈ {1, . . . , L} such that B

(α) 6=

0.

From Lemma 4.1 we deduce that there exists e α ∈ Q

such that K(e α

, . . . ,e α

) : K

≤ D

, A

(e α) = · · · = A

(e α) = 0 , B

(e α) 6= 0 for T + 1 ≤ µ ≤ M,

and for each µ = 1, . . . , T , there exists λ ∈ {1, . . . , L} such that B

(e α) 6=

0. This e α then satisfies all desired properties. ut

We apply Lemma 4.2 to exponential polynomials.

Lemma 4.3. Let k ≥ 1 be an integer, h / a positive real number, d

, . . . , d

non-negative integers and α

, . . . , α

non-zero elements of K satisfying (1.2). For 0 ≤ j ≤ k , let

P

(X) =

dj

X

i=0

a

X

be a non-zero polynomial in K[X] of degree d

. Define

f (x) = X

P

(x)α

and denote by N the set of solutions x ∈ Z of the equation f (x) = 0. Let E be a finite subset of Z . Assume that for each x ∈ E we are given a subset I (x) of

(i, j) ; 0 ≤ i ≤ d

, 0 ≤ j ≤ k for which X

(i,j)∈I (x)

a

x

α

6= 0. (4.1)

Then there exist non-zero algebraic elementse α

, . . . , e α

of K and there exist polynomials P e

, . . . , P e

which are not all zero,

P e

(X) =

dj

X

i=0

e a

X

(0 ≤ j ≤ k),

with algebraic coefficients e a

, and with the following properties:

deg P e

≤ d

( 0 ≤ j ≤ k) (4.2) X

j=0

P e

(x)e α

= 0 for all x ∈ N , (4.3) X

(i,j)∈I (x)

e a

x

e α

6= 0 for each x ∈ E, (4.4)

and, for 0 ≤ i, j ≤ k,

( e α

/e α

= α

/α

if α

/α

is algebraic,

h (e α

: e α

) ≥ h / if α

/α

is transcendental . (4.5)

Proof. We fix an ordering of the set I =

(i, j) ; 0 ≤ i ≤ d

, 0 ≤ j ≤ k, a

6= 0 and we denote by p the number of elements in this set. Also we write N = {n

, . . . , n

} (recall that N is finite) and E = {x

, . . . , x

} . For 1 ≤ s ≤ S, we define C

as the p-tuple composed of the polynomials n

X

for (i, j) ∈ I . For 1 ≤ t ≤ T , let D

be the p -tuple composed of the

polynomials (

x

X

for (i, j) ∈ I ∩ I (x

) 0 for (i, j) ∈ I \ I (x

).

From the definition of N we deduce that the dimension r of the vector space V spanned by C

(α), . . . , C

(α) satisfies r < p. According to (4.1), for each t = 1 , . . . , T we have D

(α) 6∈ V . Therefore Lemma 4.3 follows from Lemma 4.2. ut

Remark. Let K denote the field generated over Q by all algebraic elements which belong to the set

α

/α

; 0 ≤ i, j ≤ k . The proof of Lemma 4.3 also yields an upper bound for the degree of the number field K e = K(e α

, . . . , e α

) , namely

e K : K

≤ 1 max

x∈N

|x|

with 1 = d

+· · ·+d

+k+1. One may prove a variant of Lemma 4.3 where (4.3) holds only for some subset N

of N with Card N

/ Card N ≥ 1 /(k+ 1 ) but with the estimate

e K : K

≤ 1 min

x∈N⁰

|x|

.

5. Dividing exponential polynomials

Let α

, . . . , α

be given non-zero elements of K satisfying (1.2) and P

, . . . , P

be polynomials with coefficients in K , possibly zero. Consider the exponential polynomial

f (x) = X

P

(x)α

.

We set

1(f ) = X

j=0 Pj6=0

( deg P

+ 1 ).

Thus 1(f ) = 0 precisely when P

= · · · = P

= 0. When g(x) =

X

Q

(x)α

is another exponential polynomial with the same frequencies (α

, · · · , α

), we write g ≺ f if deg Q

≤ deg P

for 0 ≤ j ≤ k . We write g f if g ≺ f and 1(g) < 1(f ).

Lemma 5.1. Suppose g ≺ f and g 6= 0. Then there is an exponential polynomial

r(x) = R

(x)α

+ · · · + R

(x)α

with r f such that

f (x) = r(x) + cx

g(x) for some c in K

and some n ≥ 0.

Proof. With f and g written as above, set n = min

0≤j≤k Qj6=0

( deg P

− deg Q

).

We may suppose n = deg P

− deg Q

. When

P

= c

X

+ c

X

+ · · · , Q

= d

X

+ d

X

+ · · · , where now a = b + n , set c = c

/d

and

r(x) = f (x) − cx

g(x).

If again r(x) = R

(x)α

+ · · · + R

(x)α

, we have R

(X) = P

(X) − (c

/d

)x

Q

(X),

so that deg R

< deg P

. Also deg R

≤ max ( deg P

, n+ deg Q

) ≤ deg P

, so that r f . ut

Consider an exponential polynomial f (x) =

X

P

(x)α

where α

, . . . , α

are non-zero algebraic elements in K satisfying (1.2).

Assume

α

, . . . , α

= [

α

: · · · : α

is a partition of

α

, . . . , α

and define

f

(x) = P

(x)α

+ · · · + P

(x)α

( 1 ≤ i ≤ m) so that

f (x) = f

(x) + · · · + f

(x).

Suppose further, for 1 ≤ i 6= u ≤ m, 0 ≤ j ≤ k

and 0 ≤ v ≤ k

,

h

(α

/α

) ≥ 1 . ( 5 . 1 ) From (1.2) we deduce

1(f ) = 1(f

) + · · · + 1(f

).

Set

1 = 1(f ) Lemma 5.2. Define

F (1) = exp 1( 5 1)

.

Then for all but at most F (1) solutions x ∈ Z of f (x) = 0, we have f

(x) = · · · = f

(x) = 0 . ( 5 . 2 ) Proof. The lemma is non-trivial only when m ≥ 2 and at least two of f

, . . . , f

are non-zero, so that 1 ≥ 2. We now proceed by induction on 1 . When 1 = 2 and m ≥ 2, we have in fact f (x) = aα

+ bα

with ab 6= 0 and h

(α

/α

) ≥ 1, so that α

/α

is not a root of 1.

There can be at most one zero x of f , for if f (x) = f (y) = 0, then (α

/α

)

= (α

/α

)

= −b/a, so that (α

/α

)

= 1 hence x = y since α

/α

is not a root of 1.

Now assume 1 ≥ 3. In the induction step we apply Lemma 3.3 with h / = 1. The condition max

h (α

: α

) ≥ 1 is satisfied because m ≥ 2.

Any x ∈ Z with f (x) = 0 satisfies a relation f

(x) = 0

for some ` in the range 1 ≤ ` ≤ t where t = exp ( 5 1)

and each f

6= 0 has f

f . By Lemma 5.1 we have, for 1 ≤ ` ≤ t

f (x) = r

(x) + c

x

f

(x) with r

f . Write out

f

(x) = f

(x) + · · · + f

(x), r

(x) = r

(x) + · · · + r

(x) with

f

(x) = P

(x)α

+ · · · + P

(x)α

, r

(x) = R

(x)α

+ · · · + R

(x)α

and

f

(x) = r

(x) + c

x

f

(x). (5.3)

By induction, and since f

f and r

f , hence 1(f

) < 1(f ), 1(r

) < 1(f ) , we see that all but at most F (1− 1 ) solutions of f

(x) = 0 have

f

(x) = · · · = f

(x) = 0 , ( 5 . 4 ) and similarly all but at most F (1 − 1 ) solutions of r

(x) = 0 have

r

(x) = · · · = r

(x) = 0 . ( 5 . 5 ) But (5.3), (5.4) and (5.5) imply (5.2). Taking the sum over ` in 1 ≤ ` ≤ t , we see that all but at most

2 tF (1 − 1 ) ≤ exp 1 + ( 5 1)

+ (1 − 1 )( 5 1)

≤ F (1) solutions of f (x) = 0 have (5.2). ut

6. Proof of Theorem 1.1

Assume that the assumptions of Theorem 1.1 are satisfied. Let E be a set of more than F (1) solutions of (1.1). Assume that for each x in E there is an index i = i(x) in the range 1 ≤ i ≤ m such that f

(x) 6= 0.

We apply Lemma 4.3 with h / = 1. We produce algebraic elements e α

, . . . ,e α

and polynomials with algebraic coefficients P e

, . . . , P e

satis- fying (4.2), (4.3), (4.4) and (4.5). The exponential polynomial

f (x) e = X

P e

(x)e α

can be written

f (x) e = f e

(x) + · · · + f e

(x) where, for 1 ≤ i ≤ m ,

f e

(x) =

ki

X

j=0

P e

(x)e α

and, for 1 ≤ i 6= u ≤ m, 0 ≤ j ≤ k

and 0 ≤ v ≤ k

, h

(e α

/ e α

) ≥ 1 .

We apply Lemma 5.2 and deduce that one at least of x in E satisfies f e

(x) = 0, which is a contradiction with (4.4). ut

Final remark. The proof of Theorem 1.1 yields a stronger result. Fix h / with

0 < h / ≤ 1. If we replace the assumption that α

/α

is transcendental by

the assumption that either it is transcendental, or else has height ≥ h /, then

we get the same conclusion but with F (1) replaced by a function of 1 and

h /, which is equal to F (1) when h / = 1.

References

[1] Beukers, F. and Schlickewei, H.P.: The equation

1 in finitely generated groups.

Acta Arith. 78, 189–199 (1996)

[2] Beukers, F. and Tijdeman, R.: On the multiplicities of binary complex recurrences.

Compositio Math. 51, 193–213 (1984)

[3] Evertse, J.H., Schlickewei, H.P. and Schmidt, W.M.: Linear equations with variables which lie in a multiplicative group. In preparation

[4] Schlickewei, H.P.: The multiplicity of binary recurrences. Invent. Math. 129, 11–36 (1997)

[5] Schlickewei, H.P.: Multiplicities of recurrence sequences. Acta Math. 176, 171–243 (1996)

[6] Schlickewei, H.P. and Schmidt, W.M.: Linear equations with variables which lie in a multiplicative group. Preprint

[7] Schmidt, W.M.: The zero multiplicity of linear recurrence sequences. Acta Arith., to appear

[8] Shorey, T.N. and Tijdeman, R.: Exponential Diophantine Equations. Cambridge Tracts

in Mathematics 87, Cambridge Univ. Press, 1986, pp. 240