Kamal Boussaf

Alain Escassut

Jacqueline Ojeda
email: jacqojeda@udec.cljacqueline.ojeda@math.univ-bpclermont.fr

Complex meromorphic functions f P (f ), g P (g) sharing a small function

Keywords: 2000 Mathematics Subject Classification: 30D35, 30D45, 12E05. 0 Meromorphic, Nevanlinna, Sharing Value, Unicity, Distribution of values. 1

Let f, g be two transcendental meromorphic functions in C, let P be a polynomial of uniqueness for meromorphic functions in C and let α be a small meromorphic function with regards to f and g. If f P (f ) and g P (g) share α counting multiplicity, then we show that f = g provided that the multiplicity order of zeroes of P satisfy certain inequalities. There is no additional condition on α. We consider the particular case of entire functions.

Introduction and Main Results

Following several previous studies, the problem we consider consists of determining polynomials P such that, if f and g are complex meromorphic functions and if α is a small meromorphic function such that f P (f ) and g P (g) share α counting multiplicity, then we can conclude f = g. The first (and hardest) step consists of showing that f P (f ) = g P (g), so P (f ) -P (g) is a constant. The second step consists of showing that the constant is zero. And then, if we take for P a polynomial of uniqueness for complex meromorphic functions, we can get f = g.

Problems of uniqueness on meromorphic functions were examined first in C [START_REF] Fang | Entire functions that share one value[END_REF], [START_REF] Fang | A unicity theorem for entire functions concerning differential polynomials[END_REF], [START_REF] Fujimoto | On uniqueness of Meromorphic Functions sharing finite sets[END_REF], [START_REF] Hua | Uniqueness and value-sharing of meromorphic functions[END_REF], [START_REF] Lahiri | Uniqueness of nonlinear differential polynomials sharing simple and double 1-points[END_REF], [START_REF] Li | Some further results on the unique range sets of meromorphic functions[END_REF], [START_REF] Lin | Uniqueness theorems for meromorphic functions concerning fixed-points[END_REF], [START_REF] Xu | Entire functions sharing one value I[END_REF], [START_REF] Yang | Unique polynomials of entire and meromorphic functions[END_REF], [START_REF] Yi | Uniqueness theorems of meromorphic functions[END_REF] and next in a p-adic field [START_REF] An | Strong uniqueness polynomials: the complex case[END_REF], [START_REF] Boutabaa | URS and URSIMS for p-adic meromorphic functions inside a disc[END_REF], [5], [START_REF] Escassut | Meromorphic functions of uniqueness[END_REF], [START_REF] Escassut | Functional equations in a p-adic context[END_REF], [START_REF] Hoa | On the functional equation P (f ) = Q(g) in non-archimedean field[END_REF], [START_REF] Hu | Meromorphic functions over non archimedean fields[END_REF], [START_REF] Ojeda | Applications of the p-adic Nevanlinna theory to problems of uniqueness, Advances in p-adic and Non-Archimedean analysis[END_REF], [START_REF] Ojeda | zeros of ultrametric meromorphic functions f f n (f -a) k -α[END_REF], [START_REF] Ojeda | Uniqueness for ultrametric analytic functions[END_REF], [START_REF] Wang | Uniqueness polynomials and bi-unique range sets[END_REF]. After examining problems of the form P (f ) = P (g), where P is a polynomial, several studies were made on the equality f P (f ) = g P (g), or value sharing questions: if P (f ) and P (g) share a value, or a small function, do we have f = g?

We denote by A(C) the C-algebra of entire functions in C, by M(C) the field of meromorphic functions in C, i.e. the field of fractions of A(C) and by C(x) the field of rational functions.

The problem of value sharing a small function for functions of the form f P (f ) was examined first when P was just of the form x n [START_REF] Fang | Entire functions that share one value[END_REF], [START_REF] Ojeda | Applications of the p-adic Nevanlinna theory to problems of uniqueness, Advances in p-adic and Non-Archimedean analysis[END_REF] in both complex and p-adic contexts. More recently it was examined when P was a polynomial such that P had exactly two distinct zeros [START_REF] Lahiri | Uniqueness of nonlinear differential polynomials sharing simple and double 1-points[END_REF], [START_REF] Lin | Uniqueness theorems for meromorphic functions concerning fixed-points[END_REF], [START_REF] Ojeda | Uniqueness for ultrametric analytic functions[END_REF], both in complex analysis and in p-adic analysis. In [START_REF] Lahiri | Uniqueness of nonlinear differential polynomials sharing simple and double 1-points[END_REF], [START_REF] Lin | Uniqueness theorems for meromorphic functions concerning fixed-points[END_REF] the functions where meromorphic on C, with a small function that was a constant or the identity. In [START_REF] Ojeda | Applications of the p-adic Nevanlinna theory to problems of uniqueness, Advances in p-adic and Non-Archimedean analysis[END_REF], the problem was considered for analytic functions in a complete ultrametric algebraically closed field K of characteristic 0 concerning entire functions and unbounded analytic functions in an open disk, sharing a rational function. In [START_REF] Boussaf | p-adic meromorphic functions f P (f ), g P (g) sharing a small function[END_REF] the three authors considered the same problem for meromorphic functions in the p-adic fileld K. The results obtained in that case have then suggested a similar study in C. But several crucial lemmas obtained on a p-adic field have to be re-examined particularly because the characteristic Nevanlinna function T (r, f ) is quite different for complex meromorphic functions and for p-adic ones, particularly all statements involving functions m(r, f ) on M(C). So, we must develop complete proofs for all intermediate and final results.

Here we enjoy polynomials of uniqueness for complex meromorphic functions [START_REF] An | Strong uniqueness polynomials: the complex case[END_REF], [START_REF] Escassut | Meromorphic functions of uniqueness[END_REF] and we can finally obtain results similar to those obtained in p-adic analysis: see Theorem 1 [START_REF] Boussaf | p-adic meromorphic functions f P (f ), g P (g) sharing a small function[END_REF].

The definition of polynomials of uniqueness was introduced in [START_REF] Li | Some further results on the unique range sets of meromorphic functions[END_REF] by P. Li and C. C. Yang and was studied in many papers [START_REF] An | Strong uniqueness polynomials: the complex case[END_REF], [START_REF] Frank | A unique range set for meromorphic functions with 11 elements[END_REF], [START_REF] Fujimoto | On uniqueness of Meromorphic Functions sharing finite sets[END_REF] , [START_REF] Yang | Unique polynomials of entire and meromorphic functions[END_REF] for complex functions and [5], [START_REF] Escassut | Meromorphic functions of uniqueness[END_REF], [START_REF] Wang | Uniqueness polynomials and bi-unique range sets[END_REF] for p-adic functions.

Definition. Recall that a polynomial P ∈ C[x] is called a polynomial of uniqueness for a class of functions F if for any two functions f, g ∈ F the property P (f ) = P (g) implies f = g. Now, in order to define small functions, we have to briefly recall the definitions of the classical Nevanlinna theory in the field C. Here, for convenience, we will use notation long ago used in p-adic analysis in order to denote the counting functions.

Let log be a real logarithm function of base > 1. Given u ∈ R * + , we denote by log + the real function defined as log + (u) = max(log(u), 0).

Let f ∈ M(C). Given r> 0, we set m(r, f ) = 2π 0 log + |f (re iθ )|dθ. Suppose f has no zero and no pole at 0. Let γ ∈ C. If f has a zero of order n at γ, we set ω γ (f ) = n. If f has a pole of order n at γ, we put ω γ (f ) = -n and finally, if f (γ) = 0, ∞, we put ω γ (f ) = 0

We denote by Z(r, f ) the counting function of zeros of f in the ball |z| ≤ r, counting multiplicity, i.e. we set Z(r, f ) = ωγ (f )>0, |γ|≤r ω γ (f )(log r -log |γ|).

Similarly, we denote by Z(r, f ) the counting function of zeros of f in the ball |z| ≤ r, ignoring multiplicity, and set

Z(r, f ) = ωγ (f )>0, |γ|≤r (log r -log |γ|).
In the same way, we set

N (r, f ) = Z r, 1 f resp. N (r, f ) = Z r, 1 f
to denote the counting function of poles of f in the ball |z| ≤ r, counting multiplicity (resp. ignoring multiplicity).

If f admits a zero of order s at 0, we can make a change of origin or count the zero at 0 by adding s log r and similarly, if f admits a pole at 0 of order s, we can make a change of origin or count the pole at 0 by adding -s log r.

For f ∈ M(C) having no zero and no pole at 0, the Nevanlinna function is defined by

T (r, f ) = m(r, f ) + N (r, f ). Definition. Let f ∈ M(C) such that f (0) = 0, ∞. A function α ∈ M(C)
having no zero and no pole at 0 is called a small function with respect to f , if it satisfies lim r→+∞ T (r, α) T (r, f ) = 0. We denote by M f (C) the set of small meromorphic functions with respect to f in C. And we say that two functions f, g ∈ M(C) share a function α C.M. if f -α and g -α admit the same zeros with the same multiplicity.

Remark 1. For simplicity, we have kept the same notation as in p-adic analysis for the counting function of zeros of a meromorphic function.

Now, we can give some sufficient conditions to get polynomials P such that, if f P (f ) and g P (g) share a small meromorphic function, then f = g.

Recall the following Theorems A, B that we can get from [START_REF] Lin | Uniqueness theorems for meromorphic functions concerning fixed-points[END_REF]:

Theorem A. Let f, g ∈ M(C) be transcendental. Let n ∈ N, n ≥ 13 and let a ∈ C \ {0}. If f n (f -a) 2 f and g n (g -a) 2 g share the identical function x C.M. then f = g. Theorem B. Let f, g ∈ M(C) be transcendental and let a ∈ C \ {0} and let ∈ N, n ≥ 12. If f n (f -a)f and g n (g -a)g share the identical function x C.M. then either f = g or there exists h ∈ M(C) such that f = a(n + 2) n + 1 h n+1 -1 h n+2 -1 h and g = a(n + 2) n + 1 h n+1 -1 h n+2 -1 .
In [START_REF] Boussaf | p-adic meromorphic functions f P (f ), g P (g) sharing a small function[END_REF], we have obtained the following Theorem C.

Theorem C. Let K be a complete ultrametric algebraically closed field of characteristic zero. Let P be a polynomial of uniqueness for M(K), let

P = b(x -a 1 ) n l i=2 (x -a i ) ki with b ∈ K * , with l ≥ 2 and k i ≥ k i+1 , 2 ≤ i ≤ l -1, whenever l > 2 and let k = l i=2 k i . Suppose P satisfies the following conditions: n ≥ 10 + l i=3 max(0, 4 -k i ) + max(0, 5 -k 2 ), n ≥ k + 2, if l = 2, then n = 2k, 2k + 1, 3k + 1, if l = 3, then n = 2k + 1, 3k i -k ∀i = 2, 3.
Let f, g ∈ M(K) be transcendental and let α ∈ M f (K) ∩ M g (K) be non-identically zero. If f P (f ) and g P (g) share α C.M., then f = g.

Next, several particular applications were given when the small function is a constant or a Moebius function. Here we can't get similar refinements because the complex Nevanlinna Theory is less accurate than the p-adic Nevanlinna Theory.

However, we aim at obtaining for complex meromorphic functions results almost similar to those obtained for p-adic functions, thanks to theorems of uniqueness in C. Let us recall Theorems D and E that we can extract from Theorem 3 [START_REF] An | Strong uniqueness polynomials: the complex case[END_REF] and Corollary 9.1 [START_REF] Escassut | Meromorphic functions of uniqueness[END_REF] respectively.

Notation: Let L be an algebraically closed field and let P ∈ L[x] \ K and let Ξ(P ) be the set of zeros c of P such that P (c) = P (d) for every zero d of P other than c. We denote by Φ(P ) its cardinal.

In [START_REF] An | Strong uniqueness polynomials: the complex case[END_REF], Theorem 3 is stated under the general condition that the polynomial P we consider is injective on the set of zeros of its derivative. Here we reduce that hypothesis to a hypothesis concerning Φ(P ). Particularly, the following Theorem D (concerning the case when P only has two distinct zeros) comes fromTheorem 3 [START_REF] An | Strong uniqueness polynomials: the complex case[END_REF], but actually, it is useless to assume the general hypothesis that the restriction of P on the set of zeros of P is injective because in that case, such a polynomial P necessarily satisfies P (c 1 ) = P (c 2 ) with the two zeros c 1 , c 2 of P , as shown in Lemma 10 of [START_REF] Escassut | Meromorphic functions of uniqueness[END_REF].

Remark 2. The hypothesis Φ(P ) ≥ 4 is weaker than the hypothesis: "the restriction of P on the set of zeros of P is injective" as soon as the number of distinct zeros of P is at least 6.

Theorem D Let P ∈ C[x] be such that P has exactly two distinct zeros γ 1 of order c 1 and γ 2 of order c 2 with min{c 1 , c 2 } ≥ 2 and max(c 1 , c 2 ) ≥ 3. Then P is a polynomial of uniqueness for M(C).

Theorem E Let P ∈ C[x] be such that P has exactly three distinct zeros γ 1 of order c 1 , γ 2 of order c 2 , γ 3 of order c 3 with max{c 1 , c 2 , c 3 } ≥ 2, satisfying further Φ(P ) = 3. Then P is a polynomial of uniqueness for M(C).

We can find Theorem F in [START_REF] Escassut | Meromorphic functions of uniqueness[END_REF] (Corollary 9.1, p. 226):

Theorem F. Let P ∈ C[x]. If Φ(P ) ≥ 4 then P is a polynomial of uniqueness for M(C).
We can now state our main theorems on the problem f P (f ), g P (g) sharing a small function.

Theorem 1. Let P be a polynomial of uniqueness for A(C), let

P = b(x -a 1 ) n l i=2 (x -a i ) ki with b ∈ C * , l ≥ 2, and k i ≥ k i+1 , 2 ≤ i ≤ l -1 when l > 2 and let k = l i=2 k i . Suppose P satisfies the following conditions: n ≥ k + 2 n ≥ 5 + l i=3 max(0, 4 -k i ) + max(0, 5 -k 2 ),
Let f, g ∈ A(C) be transcendental and let α ∈ A f (C)∩A g (C) be non-identically zero. If f P (f ) and g P (g) share α C.M., then f = g.

By Theorem F, we can derive Corollary 1.1:

Corollary 1.1 Let P ∈ C[x] be such that P (x) = b(x -a 1 ) n l i=2 (x -a i ) ki with b ∈ C * , Φ(P ) ≥ 4,
and k i ≥ k i+1 , 2 ≤ i ≤ l -1 and let k = l i=2 k i . Suppose P satisfies the following conditions: n ≥ k + 2, n ≥ 5 + l i=3 max(0, 4 -k i ) + max(0, 5 -k 2 ).
Let f, g ∈ A(C) be transcendental and let α ∈ A f (C)∩A g (C) be non-identically zero. If f P (f ) and g P (g) share α C.M., then f = g.

By Theorem E, we can drive Corollary 1.2:

Corollary 1.2 Let P ∈ C[x] be such that P (x) = b(x -a 1 ) n (x -a 2 ) k2 (x -a 3 ) k3 with b ∈ C * , k 2 ≥ k 3 , Φ(P ) = 3. Suppose P satisfies the following conditions: n ≥ k 2 + k 3 + 2, n ≥ 5 + max(0, 4 -k 3 ) + max(0, 5 -k 2 ).
Let f, g ∈ A(C) be transcendental and let α ∈ A f (C)∩A g (C) be non-identically zero. If f P (f ) and g P (g) share α C.M., then f = g. Theorem 2. Let P be a polynomial of uniqueness for A(C) such that P is of the form

b(x -a 1 ) n l i=2 (x -a i ) with l ≥ 3 , b ∈ C * , satisfying, n ≥ l + 5.
Let f, g ∈ A(C) be transcendental and let α ∈ A f (C) ∩ A g (C) be non-identically zero. If f P (f ) and g P (g) share α C.M., then f = g.

By Theorem F we have Corollary 2.1:

Corollary 2.1 Let P ∈ C[x] be such that P (x) = b(x -a 1 ) n l i=2 (x -a i ) with Φ(P ) ≥ 4 , b ∈ C * , satisfying n ≥ l + 5.
Let f, g ∈ A(C) be transcendental and let α ∈ A f (C) ∩ A g (C) be non-identically zero. If f P (f ) and g P (g) share α C.M., then f = g.

And by Theorem E we have Corollary 2.1:

Corollary 2.2 Let P ∈ C[x] be such that P (x) = b(x -a 1 ) n (x -a 2 ) k2 (x -a 3 ) k3 with Φ(P ) = 3 , b ∈ C * , satisfying n ≥ 8.
Let f, g ∈ A(C) be transcendental and let α ∈ A f (C) ∩ A g (C) be non-identically zero. If f P (f ) and g P (g) share α C.M., then f = g.

Theorem 3. Let P be a polynomial of uniqueness for M(C), let

P = b(x -a 1 ) n l i=2 (x -a i ) ki with b ∈ C * , and l ≥ 2, k i ≥ k i+1 , 2 ≤ i ≤ l -1 when l > 2 and let k = l i=2 k i .
Suppose P satisfies the following conditions:

n ≥ 10 + l i=3 max(0, 4 -k i ) + max(0, 5 -k 2 ), n ≥ k + 3, if l = 2, then n = 2k, 2k + 1, 3k + 1, if l = 3, then n = 2k + 1, 3k i -k, ∀i = 2, 3.
Let f, g ∈ M(C) be transcendental and let α ∈ M f (C) ∩ M g (C) be non-identically zero. If f P (f ) and g P (g) share α C.M., then f = g. By Theorem F, we have Corollary 3.1:

Corollary 3.1 Let P ∈ C[x] be such that P (x) = b(x -a 1 ) n l i=2 (x -a i ) ki with b ∈ C * , Φ(P ) ≥ 4, k i ≥ k i+1 and 2 ≤ i ≤ l -1 and let k = l i=2 k i . Suppose P satisfies the following conditions: n ≥ 10 + l i=3 max(0, 4 -k i ) + max(0, 5 -k 2 ), n ≥ k + 3. Let f, g ∈ M(C) be transcendental and let α ∈ M f (C) ∩ M g (C) be non-identically zero. If f P (f ) and g P (g) share α C.M., then f = g.
And by Theorem D we also have Corollary 3.2

Corollary 3.2 Let P ∈ C[x] be such that P is of the form b(x -a 1 ) n (x -a 2 ) k with k ≥ 2.
Suppose that P satisfies the further conditions:

n ≥ 10 + max(0, 5 -k), n ≥ k + 3, n = 2k, 2k + 1, 3k + 1.
Let f, g ∈ M(C) be transcendental and let α ∈ M f (C) ∩ M g (C) be non-identically zero. If f P (f ) and g P (g) share α C.M., then f = g.

Example: Let P (x) =

x 16 16 -2x 15 15 + x 14 14 . Then P (x) = x 13 (x -1) 2 . Given f, g ∈ M(C)

transcendental such that f P (f ) and g P (g) share a small function α ∈ M(C) C.M., we have f = g.

By Corollary 3.2 we can also derive Corollary 3.3:

Corollary 3.3 Let f, g ∈ M(C) be transcendental and let α ∈ M f (C)∩M g (C) be non-identically zero. Let a ∈ C * . If f f n (f -a) 2 and g g n (g -a) 2 share the function α C.M. and if n ≥ 13, then f = g.
By Theorem E, we can also derive Corollary 3.4:

Corollary 3.4 Let P ∈ C[x] be such that P (x) = b(x -a 1 ) n (x -a 2 ) k2 (x -a 3 ) k3 with b ∈ C * , Φ(P ) = 3, k i ≥ k i+1 and i = 2, 3
. Suppose P satisfies the following conditions:

n ≥ 10 + l i=3 max(0, 4 -k 2 ) + max(0, 5 -k 3 ), n ≥ k 2 + k 3 + 3. Let f, g ∈ M(C) be transcendental and let α ∈ M f (C) ∩ M g (C) be non-identically zero. If f P (f ) and g P (g) share α C.M., then f = g.
When all k i are equal to 1, we can obtain a better formulation:

Theorem 4. Let P be a polynomial of uniqueness for M(C) such that P is of the form b(x -a 1 ) n l i=2 (x -a i ) with l ≥ 3 , b ∈ C * , with n ≥ l + 10.
Let f, g ∈ M(C) be transcendental and let α ∈ M f (C) ∩ M g (C) be non-identically zero. If f P (f ) and g P (g) share α C.M., then f = g.

By Theorem F, we have Corollary 4.1:

Corollary 4.1 Let P ∈ C[x] be such that P is of the form P (x) = b(x -a 1 ) n l i=2 (x -a i ), with Φ(P ) ≥ 4, b ∈ C * satisfying: n ≥ l + 10. Let f, g ∈ M(C) be transcendental and let α ∈ M f (C) ∩ M g (C) be non-identically zero. If f P (f ) and g P (g) share α C.M., then f = g. Example: Let P (x) = x 18 18 - 2x 17 17 - x 16 16 + 2x 15 15 . Then P (x) = x 17 -2x 16 -x 15 + 2x 14 = x 14 (x -1)(x + 1)(x -2).
We check that: P (0), P (1), P (-1), P (2) are all distinct. Consequently Φ(P ) = 4. So, P is a polynomial of uniqueness for M(C). Moreover, we have n = 14, l = 4, so we can apply Corollary 4.1. Given f, g ∈ M(C) transcendental such that f P (f ) and g P (g) share a small function α ∈ M(C) C.M., we have f = g.

And by Theorem E we have Corollary 4.2

Corollary 4.2 Let P ∈ C[x] be such that P is of the form

P (x) = b(x -a 1 ) n (x -a 2 )(x -a 3 ) with Φ(P ) = 3, b ∈ C * , satisfying n ≥ 13.
Let f, g ∈ M(C) be transcendental and let α ∈ M f (C) ∩ M g (C) be non-identically zero. If f P (f ) and g P (g) share α C.M., then f = g.

Example: Consider the general form of a polynomial P such that P has 3 distinct zeros, one of them being 0 of order n and the two others a and b of order 1. We can obviously suppose that P (0) = 0. And then we have

P (x) = x n+3 (n + 2)(n + 1) -x n+2 (a 1 + a 2 )(n + 3)(n + 1) + x n+1 (a 1 a 2 )(n + 3)(n + 2)
and therefore

P (x) = (n + 1)(n + 2)(n + 3)x n (x -a 1 )(x -a 2 ).
Now we can check that P (a 1 ) = P (0) = 0 if and only if

a 1 a 2 = n + 3 n + 1 and that P (a 1 ) = P (a 2 ) if and only if [a n+3 1 -a n+3 2 ](n + 1) = [a 1 a 2 (a n+1 1 -a n+1 2
)](n + 3). Particularly, the last equality is satisfied when n is odd and

a 2 = -a 1 . Consequently, if a 1 a 2 = n + 3 n + 1 and a 1 a 2 = n + 1 n + 3 and if [a n+3 1 -a n+3 2 ](n + 1) = [a 1 a 2 (a n+1 1 -a n+1 2
)](n + 3) then P is a polynomial of uniqueness for complex meromorphic functions. Moreover, if n ≥ 13, then we can apply Corollary 4.2.

Conversely, suppose for instance n is odd and a 2 = -a 1 . Then we can check that P (f ) = P (-f ) for every function, therefore P is not a polynomial of uniqueness for M(C).

Theorem 5. Let f, g ∈ M(C) be transcendental and let α ∈ M f (C) ∩ M g (C) be non-identically zero. Let a ∈ C * . If f f n (f -a)
h n+1 -1 h n+2 -1 h and g = a(n + 2) n + 1 h n+1 -1 h n+2 -1 .
Remark 3. In Theorem 5, the second conclusion is well known to occur [5], [START_REF] Yang | Unique polynomials of entire and meromorphic functions[END_REF]. Indeed, let h ∈ M(C) and let us define f and g as:

g = ( n + 2 n + 1 ) h n+1 -1 h n+2 -1 and f = hg. We can see that the polynomial P (y) = 1 n + 2 y n+2 - 1 n + 1 y n+1 satisfies P (f ) = P (g), hence f P (f ) = g P (g),
therefore f P (f ) and g P (g) trivially share any function.

Remark 4. On a p-adic field, we obtained conclusions with hypotheses slightly more general when the small function α is either a Moebius function or a constant, thanks to a very accurate Second Main Theorem. On the field C, we cannot do the same.

2 Basic and specifc results:

Notation: Given f ∈ M(C), we will denote by S f (r) a function in r defined in ]0, +∞[ such that lim r→+∞ S f (r)
T (r,f ) = 0 out of a subset J of ]0, +∞[ of finite measure. Let us recall a few classical lemmas [START_REF] Hayman | Meromorphic Functions[END_REF]:

Lemma 1. Let f, g ∈ M(C), let a ∈ C and let P (f ) ∈ C[x] be of degree q. Then T (r, f + g) ≤ T (r, f ) + T (r, g) + O(1), T (r, f g) ≤ T (r, f ) + T (r, g), T (r, f -a) = T (r, f ) + O(1), T (r, 1 f ) = T (r, f ) + O(1), T (r, P (f )) = qT (r, f ) + O(1). Lemma 2. Let f, g ∈ M(C). Then Z(r, f -a) ≤ T (r, f ) + O(1) ∀a ∈ C, m(r, f g) ≤ m(r, f ) + m(r, g), N (r, f ) = N (r, f ) + N (r, f ), Z(r, f ) ≤ Z(r, f ) + N (r, f ) + S f (r). Let Q ∈ C[x] be of degree q. Then Z(r, f Q(f )) ≥ Z(r, Q(f )) and N (r, f Q(f )) = (q + 1)N (r, f ) + N (r, f ).
By Milloux's Theory we have the following lemma (Theorem 3.1 in [START_REF] Hayman | Meromorphic Functions[END_REF]).

Lemma 3. Let f ∈ M(C). Then m(r, f f ) = S f (r) and T (r, f ) ≤ T (r, f ) + N (r, f ) + S f (r). Moreover, if f ∈ A(C) then T (r, f ) ≤ T (r, f ) + S f (r).
The statements of Lemma 4 are known and used in [START_REF] Lin | Uniqueness theorems for meromorphic functions concerning fixed-points[END_REF]:

Lemma 4. Let f ∈ M(C). Then m(r, 1 f ) ≥ m(r, 1 f ) + S f (r). Moreover, T (r, f ) -Z(r, f ) ≤ T (r, f ) -Z(r, f ) + S f (r). Notation: Let f ∈ M(C) satisfy f (0) = 0, ∞.
Let S be a finite subset of C and r ∈]0, +∞[. We denote by Z S 0 (r, f ) the counting function of zeros of f in C which are not zeros of any f -s for s ∈ S. This is, if (γ n ) n∈N is the finite or infinite sequence of zeros of f in C that are not zeros of f -s for s ∈ S, with multiplicy order q n respectively, we set

Z S 0 (r, f ) = |γn|≤r q n (log r -log |γ n |).
Theorem N. [START_REF] Hayman | Meromorphic Functions[END_REF] Let a 1 , ..., a n ∈ C with n ≥ 2, n ∈ N, and let f ∈ M(C). Let S = {a 1 , ..., a n }.

Assume that none of f, f and f -a j with 1 ≤ j ≤ n, equals 0 or ∞ at the origin. Then, for r > 0 we have

(n -1)T (r, f ) ≤ n j=1 Z(r, f -a j ) + N (r, f ) -Z S 0 (r, f ) + S f (r).
We will also need the following Lemma 5 which is specific to the complex case.

Lemma 5. Let Q ∈ C[x] and let f ∈ M(C) be transcendental. Then T (r, Q(f )) ≤ T (r, f Q(f )) + m(r, 1 f ) and T (r, Q(f )) ≤ T (r, f ) + N (r, f ) + T (r, f Q(f )) + S f (r), r ∈]0, +∞[. Moreover, if deg(Q) ≥ 3, then M f (C) ⊂ M f Q(f ) (C). Proof : We have T (r, Q(f )) = N (r, Q(f ))+m(r, f Q(f ) f ) ≤ N (r, Q(f ))+N (r, f )+m(r, f Q(f ))+ m(r, 1 f ). Now, since the poles of f are those of Q(f ) (with different multiplicities), we have N (r, Q(f ))+N (r, f ) = N (r, f Q(f )) and therefore N (r, Q(f ))+N (r, f )+m(r, f Q(f ))+m(r, 1 f ) = N (r, f Q(f )) + m(r, f Q(f )) + m(r, 1 f ) = T (r, f Q(f )) + m(r, 1 f ), which proves the first inequality. Next, m(r, 1 f ) ≤ T (r, f ) + S f (r) = m(r, f ) + N (r, f ) + S f (r). But by Lemma 3 m(r, f ) ≤ m(r, f ) + S f (r), hence m(r, 1 f ) ≤ m(r, f ) + N (r, f ) + S f (r) = m(r, f ) + N (r, f ) + N (r, f ) + S f (r) and therefore T (r, Q(f )) ≤ T (r, f Q(f ))+T (r, f )+N (r, f )+S f (r) ≤ T (r, f Q(f ))+2T (r, f )+S f (r),
which finishes proving the second inequality. Particularly, if deg(Q) ≥ 3, the last statement is an immediate consequence of both inequalities.

On a p-adic field, the following lemma 6 is only proven for meromorphic functions [START_REF] Boussaf | p-adic meromorphic functions f P (f ), g P (g) sharing a small function[END_REF]. Here we must consider the particular case of entire functions which will be useful in the proof of Theorem 1.

Lemma 6. Let Q ∈ C[x] and let f, g ∈ M(C) (resp. f, g ∈ A(C)) be transcendental. Let P (x) = x n+1 Q(x) be such that n ≥ deg(Q) + 3 (resp. n ≥ deg(Q) + 2). If P (f )f = P (g)g then P (f ) = P (g).
Proof : Set k = deg(Q). Since P (f )f = P (g)g , there exists c ∈ C such that P (f ) = P (g) + c. Suppose that c = 0. Then by Theorem N, we have

T (r, P (f )) ≤ Z(r, P (f )) + Z(r, P (f ) -c) + N (r, P (f )) + S f (r).
(

) 1
Obviously we see that Z(r, P (f 1) and then Z(r, P (f )) ≤ (k +1)T (r, f )+O [START_REF] An | Strong uniqueness polynomials: the complex case[END_REF]. We also have Z(r, P (f

)) = Z(r, f n+1 Q(f )) = Z(r, f Q(f )) ≤ T (r, f Q(f )). By Lemma 1 we have T (r, f Q(f )) = (k +1)T (r, f )+O(
) -c) = Z(r, P (g)) ≤ Z(r, g) + Z(r, Q(g)) ≤ T (r, g) + T (r, Q(g)). Then by Lemma 1, Z(r, P (f ) -c) ≤ (k + 1)T (r, g) + O(1). Notice that N (r, P (f )) = N (r, f ) ≤ T (r, f ) + O(1) then by (1) we obtain T (r, P (f )) ≤ (k + 2)T (r, f ) + (k + 1)T (r, g) + S f (r) (2)
(resp.

T (r, P (f )) ≤ (k + 1)T (r, f ) + (k + 1)T (r, g) + S f (r) (3))
According to Lemma 1 we have T (r, P (f )) = (n + k + 1)T (r, f ) + O(1). Then by [START_REF] Boussaf | p-adic meromorphic functions f P (f ), g P (g) sharing a small function[END_REF] we have

nT (r, f ) ≤ T (r, f ) + (k + 1)T (r, g) + S f (r). (4)
(resp. nT (r, f ) ≤ T (r, f ) + (k + 1)T (r, g) + S f (r). ( 5))

Similarly we obtain nT (r, g) ≤ T (r, g) + (k + 1)T (r, f ) + S f (r). ( 6))

(resp. nT (r, g) ≤ (k + 1)T (r, f ) + S f (r). ( 7))
Hence adding ( 6) and ( 7) we have

n(T (r, f ) + T (r, g)) ≤ (k + 2) T (r, f ) + T (r, g) + S f (r) and then 0 ≤ (k + 2 -n) T (r, f ) + T (r, g) + S f (r) (resp. 0 ≤ (k + 1 -n) T (r, f ) + T (r, g) + S f (r))
In the general case, that leads to a contradiction because n ≥ k+3 and lim r→+∞ (T (r, f )+T (r, g)) = +∞ outside a subset J of ]0, +∞[ of finite measure. Thus c = 0 and consequently P (f ) = P (g). And when f, g ∈∈ A(C) we have contradiction when n ≥ k + 2.

Notation: Given two meromorphic functions f, g ∈ M(C). We set

Ψ f,g = f f - 2f f -1 - g g + 2g g -1 .
That function was already used in several previous papers, particularly [START_REF] Lin | Uniqueness theorems for meromorphic functions concerning fixed-points[END_REF].

The following Lemma was proved in [START_REF] Yi | Meromorphic functions that share one or two values[END_REF].

Lemma 7. Let F, G ∈ M(C) be non-constant, having no zero and no pole at 0 and sharing the value 1 C.M. Suppose that Ψ f,g = 0 and that

lim sup r→+∞ Z(r, f ) + Z(r, g) + N (r, f ) + N (r, g) max(T (r, f ), T (r, g)) < 1
outside a set of finite measure. Then either f = g or f g = 1.

Lemma 8 is similar to Lemma 9 [START_REF] Boussaf | p-adic meromorphic functions f P (f ), g P (g) sharing a small function[END_REF] for complex meromorphic functions. But here, we can get a better result for entire functions.

Lemma 8. Let P (x) = (x -a 1 ) n l i=2 (x -a i ) ki ∈ C[x] (a i = a j , ∀i = j) with l ≥ 2 and n ≥ max{k 2 , .., k l } and let k = l i=2 k i . Let f, g ∈ M(C) be transcendental and let θ = P (f )f P (g)g . If θ belongs to M f (C) ∩ M g (C), then we have the following : if l = 2 then n belongs to {k, k + 1, 2k, 2k + 1, 3k + 1}, if l = 3 then n belongs to { k 2 , k + 1, 2k + 1, 3k 2 -k, 3k 3 -k}, if l ≥ 4 then n = k + 1.
Moreover, if f, g belong to A(C), then θ does not belong to A f (C).

Proof: Without loss of generality, we can assume a 1 = 0. Suppose f, g ∈ M(C) satisfy

f n ( l i=2 (f -a i ) ki )f g n ( l i=2 (g -a i ) ki )g = θ (1)
Let Σ be the set of zeros and poles of θ. We will first show that P (f ) and P (g) admit zeros or poles out of Σ. Indeed, suppose that none of the zeros and the poles of f P (f ) belong to Σ. So, every zero and every pole of f P (f ) must be either a zero or a pole of θ. Therefore, Z(r, f P (f )) + N (r, f P (f )) ≤ 2T (r, θ). Actually, each zero of f P (f ) is either a zero of f or of P (f ) because the poles of P (f ) are the poles of f . Consequently, Z(r, P (f )) +N (r, f ) ≤ 2T (r, θ) and therefore

l i=1 Z(r, f -a i )) + N (r, f ) ≤ 2T (r, θ) But now by Theorem N we have (l -1)T (r, f ) ≤ l i=1 Z(r, f -a i ) + N (r, f ) + S f (r)
hence (l -1)T (r, f ) ≤ 2T (r, θ) + S f (r), a contradiction. Consequently, P (f ) admits zeros or poles that do not belong to Σ. Similarly, P (g) admits zeros or poles that do not belong to Σ. Now, suppose that f, g belong to A(C). Since both f, g have no pole, all zeros of both must belong to Σ. Then, we have

Z(r, f P (f )g P (g)) = Z(r, f ) + l i=1 k i Z(r, f -a i ) + Z(r, g ) + l i=1 k i Z(r, g -a i ), therefore Z(r, θ) ≥ l i=1 Z(r, f -a i ) ≥ (l -1)T (r, f ) + S f (r) + S g (r)
a contradiction to hypothesis again. Consequently, f P (f )g P (g) cannot belong to A f (C) or A g (C). Now, let us go back to the general case and suppose that f, g are meromorphic functions. Let γ ∈ C \ Σ be a zero of g of order s. Clearly, by (1), γ is a pole of f of order, for example, t. And since γ is neither a zero nor a pole of θ we can derive the following relation

s(n + 1) = t(n + k + 1) + 2 (2)
Now, suppose that for i ∈ {2, .., l}, g -a i has a zero γ ∈ C \ Σ of order s i . It is a pole of f of order t i . So, by (2), we obtain

s i (k i + 1) = t i (n + k + 1) + 2 (3)
By ( 2) and ( 3) it is obvious that s > t and s i > t i . Consider now a pole γ ∈ C \ Σ of f . Either it is a zero of g, or it is a zero of g -a i for some i ∈ {2, .., l}, or it is a zero of g that is neither a zero of g nor a zero of g -a i (∀i ∈ {2, .., l}). Let Z 0 (r, g ) be the counting function of zeros of g that are neither a zero of g nor a zero of g -a i for all i ∈ {2, .., l} (counting multiplicity) and let Z 0 (r, g ) be the counting function of zeros of g that are neither a zero of g nor a zero of g -a i for all i ∈ {2, .., l}, ignoring multiplicity. Since T (r, θ) = S f (r) = S g (r), we have

N (r, f ) ≤ Z(r, g) + l i=2 Z(r, g -a i ) + Z 0 (r, g ) + S f (r) + S g (r)(r) (4)
Now, by Theorem N, we have (l -1)T (r, f ) ≤ Z(r, f ) + l i=2 Z(r, f -a i ) + N (r, f ) -Z 0 (r, f ) + S f (r), hence by (4), we obtain

(l-1)T (r, f ) ≤ Z(r, f )+ l i=2 Z(r, f -a i )+Z(r, g)+ l i=2 Z(r, g-a i )+Z 0 (r, g )-Z 0 (r, f )+S f (r)+S g (r)
(5) And similarly, (l-1)T (r, g) ≤ Z(r, g)+

l i=2 Z(r, g-a i )+Z(r, f )+ l i=2 Z(r, f -a i )+Z 0 (r, f )-Z 0 (r, g )+S f (r)+S g (r) (6)
Hence, adding ( 5) and ( 6), we obtain

(l-1)(T (r, f )+T (r, g)) ≤ 2 Z(r, g)+ l i=2 Z(r, g-a i )+Z(r, f )+ l i=2 Z(r, f -a i ) +S f (r)+S g (r) (7) Case l = 2:
Without loss of generality, we can assume a 2 = 1. Suppose now that all zeros of f, f -1, g, g-1 are at least of order 5, except maybe those lying in Σ: then

Z(r, f ) ≤ 1 5 T (r, f ) + S f (r) + S g (r), Z(r, f -1) ≤ 1 5 T (r, f ) + S f (r) + S g (r), , Z(r, g) ≤ 1 5
T (r, g) + S f (r) + S g (r), Z(r, g -1) ≤ 1 5 T (r, g) + S f (r) + S g (r), a contradiction to [START_REF] Escassut | Functional equations in a p-adic context[END_REF], proving the statement of the Lemma. So, we will look for pairs (n, k) leading to zeros of f or g out of Σ, of order ≤ 4.

Consequently, we will examine all situations leading to zeros of order ≤ 4 for f, f -1, g, g -1. Actually, since f and g play the same role with respect to n and k, it is sufficient to examine the situation, for instance, when g or g -1 has a zero of order s ≤ 4. In each case we denote by t the order of the pole of f which is a zero of g or g -1. Recall that when f has a pole of order 4, g or g -1, if it has a zero, must have a zero of order ≥ 5. Consequently, we only have to examine zeros of g or g -1 that are poles of f of order 1, 2, 3.

Supppose first g has a zero γ / ∈ Σ of order s = 2. Then

2(n + 1) = t(k + n + 1) + 2 (8)
By ( 8) if t = 1 we find a solution:

n = k + 1 (9)
Next, if t ≥ 2, we check that 2n + 2 < t(k + n + 1) + 2, hence ( 9) is the only solution.

Supppose now g has a zero γ / ∈ Σ of order s = 3. Then

3(n + 1) = t(k + n + 1) + 2 (10)
By [START_REF] Frank | A unique range set for meromorphic functions with 11 elements[END_REF] if t = 1 we find no solution because k ≤ n.

If t = 2 we find n = 2k + 1 (11)
If t ≥ 3 we have 3(n + 3) < 3(k + n + 1) + 2 hence [START_REF] Fujimoto | On uniqueness of Meromorphic Functions sharing finite sets[END_REF] is the only solution.

Supppose now g has a zero of order s = 4. Then 4(n + 1) = t(k + n + 1) + 2 [START_REF] Hayman | Meromorphic Functions[END_REF] If t = 1, since k ≤ n, we have 4(n + 1) > t(k + n + 1) + 2. If t = 2, by [START_REF] Hayman | Meromorphic Functions[END_REF] we have a solution

n = k (13)
If t = 3, we have another solution n = 3k + 1 [START_REF] Hu | Meromorphic functions over non archimedean fields[END_REF] Consequently, by ( 9), ( 11), ( 12), ( 14), all possibilities for g to have a zero of order s ≤ 4 are as follows:

n = k + 1, s = 2, n = 2k + 1, s = 3, n = k, s = 4, n = 3k + 1, s = 4. ( 15
)
Now, we will examine zeros of g -1 γ / ∈ Σ of order ≤ 4. So, the order s of g -1 satisfies

s(k + 1) = t(k + n + 1) + 2 (16)
Supppose first g -1 has a zero γ / ∈ Σ of order s = 2. Then by ( 16), we have

2(k + 1) = t(k + n + 1) + 2 (17)
Since k ≤ n, we find no solution neither when t = 1 that would lead to k = n + 1, nor when t ≥ 2 because 2(k + 1) < t(k + n + 1) + 2.

Suppose now that s = 3. If t = 1 we find a solution:

n = 2k ( 18
)
If t ≥ 2, we have no solution with k ≤ n because 3(k + 1) < t(k + n + 1) + 2 Suppose now that s = 4. If t = 1 we find a solution:

n = 3k + 1 ( 19
)
If t = 2 we find another solution:

n = k ( 20
)
If t ≥ 3, we find no solution with k ≤ n because 4(k + 1) < t(k + n + 1) + 2. Consequently, by ( 18), ( 19), ( 20), all possibilities for g -1 to have a zero γ / ∈ Σ of order s ≤ 4 are as follows:

n = 2k, s = 3, n = 3k + 1, s = 4, n = k, s = 4. ( 21
)
Thus, we have proved that when n = k, k + 1, 2k, 2k + 1, 3k + 1, none of the zeros of f , f -1, g, g -1 out of Σ is of order ≤ 4 and therefore the statement of the Lemma is proved in the case l = 2.

Case l = 3 : Suppose that all zeros of f, g, f -a i , g -a i ∀i = 2, 3 are at least of order 4, except maybe those lying in Σ: then

Z(r, f ) ≤ 1 4
T (r, f ) + S f (r) + S g (r), and ∀i = 2, 3, Z(r, f -

a i ) ≤ 1 4 T (r, f ) + S f (r) + S g (r), , Z(r, g) ≤ 1 4
T (r, g) + S f (r) + S g (r), and ∀i = 2, 3 Z(r, g -a i ) ≤ 1 4 T (r, g) + S f (r) + S g (r).

Then by [START_REF] Escassut | Functional equations in a p-adic context[END_REF] we obtain l ≤ 2, a contradiction.

Consequently, we will examine all n and k i (i ∈ {2, 3}) leading to zeros out of Σ of order ≤ 3 for f, g, f -a i , g -a i for i = 2, 3. Actually, since f and g play the same role, it is sufficient to examine the situation, for instance, when g or some g -a i has a zero of order less than 3. In each case we denote by t the order of the pole of f which is a zero of g or g -a i for some i. Recall that when f has a pole of order 3, g or g -a i , if it has a zero, must have a zero of order ≥ 4. Consequently, we only have to examine zeros of g or g -a i (∀i ∈ {2, 3}) that are poles of f of order 1, 2.

Supppose first g has a zero γ / ∈ Σ of order s = 2. By (2) we have

2(n + 1) = t(k + n + 1) + 2 (22)
By [START_REF] Wang | Uniqueness polynomials and bi-unique range sets[END_REF] if t = 1 we find a solution:

n = k + 1 (23)
Next, if t = 2, we check that 2n + 2 < 2(k + n + 1) + 2, hence [START_REF] Xu | Entire functions sharing one value I[END_REF] is the only solution.

Supppose now g has a zero γ / ∈ Σ of order s = 3. Then

3(n + 1) = t(k + n + 1) + 2 (24)
By [START_REF] Yang | Unique polynomials of entire and meromorphic functions[END_REF] if t = 1 we find a solution :

n = k 2 ( 25
)
If t = 2 we find n = 2k + 1 [START_REF] Yi | Uniqueness theorems of meromorphic functions[END_REF] Consequently, by ( 23), ( 25), [START_REF] Yi | Uniqueness theorems of meromorphic functions[END_REF] all possibilities for g to have a zero of order s ≤ 3 are as follows:

n = k + 1, s = 2, t = 1, n = k 2 , s = 3, t = 1, n = 2k + 1, s = 3, t = 2.
Consider now zeros of g -a i γ / ∈ Σ of order s i ≤ 3. So, the order s i of g -a i satisfies

s i (k i + 1) = t(k + n + 1) + 2 (27)
Supppose first g -a i has a zero γ / ∈ Σ of order s i = 2. Then by (28), we have

2(k i + 1) = t(k + n + 1) + 2 (28) Since k i ≤ n and k i ≤ k we have 2(k i + 1) < t(k + n + 1) + 2.
Hence we find no solution for (28).

Suppose now that s = 3. If t = 1 we find a solution:

3k i = n + k (29) If t = 2, we have no solution because 3k i < 2(n + k).
Consequently, the unique possibility for g -a i to have a zero γ / ∈ Σ of order s i ≤ 3 is :

n + k = 3k i , s = 3, t = 1
Thus, we have proved that when n = k + 1, k 2 , 2k + 1, 3k i -k none of the zeros of f , g, f -a i , g -a i (∀i = 2, 3) out of Σ is of order ≤ 3 and therefore the statement of the Lemma is proved in the case l = 3.

Case l ≥ 4 :

Suppose now that all zeros of f, g, f -a i , g -a i ∀i ∈ {2, .., l} are at least of order 3, except maybe those lying in Σ: then

Z(r, f ) ≤ 1 3 T (r, f ) + S f (r) + S g (r), and ∀i ∈ {2, .., l}, Z(r, f -a i ) ≤ 1 3 T (r, f ) + S f (r) + S g (r), Z(r, g) ≤ 1 3 T (r, g) + S f (r) + S g (r)
, and ∀i ∈ {2, .., l}, Z(r, g -

a i ) ≤ 1 3 T (r, g) + S f (r) + S g (r).
Then using [START_REF] Escassut | Functional equations in a p-adic context[END_REF] we obtain l ≤ 3 a contradiction.

Consequently, we will examine all n and k i (i ∈ {2, .., l}) leading to zeros out of Σ of order ≤ 2 for f, g, f -a i , g -a i for all i ∈ {2, .., l}. Actually, since f and g play the same role, it is sufficient to examine the situation, for instance, when g or some g -a i has a zero of order less than 2. In each case we denote by t the order of the pole of f which is a zero of g or g -a i for some i. Recall that when f has a pole of order 2, g or g -a i , if it has a zero, must have a zero of order ≥ 3.

Consequently, we only have to examine zeros of g or g -a i (∀i ∈ {2, .., l}) that are poles of f of order 1.

Supppose first g has a zero γ / ∈ Σ of order s = 2. Then γ is a pole of f of order t = 1. Then by (2) we have 2(n + 1) = (k + n + 1) + 2 (30)

We find a solution :

n = k + 1 (31)
Let i ∈ {2, .., l} and suppose g -a i has a zero γ / ∈ Σ of order s i = 2. Then γ is a pole of f of order t = 1. So by (3) we have :

2(k i + 1) = (n + k + 1) + 2
That means 2k i = n + k + 1. Since k i ≤ n and k i ≤ k, we find no solution when s i = 2 and t = 1. Consequently, by (31), the only possibility for g or some g -a i to have a zero γ / ∈ Σ of order ≤ 3 is :

n = k + 1.
This completes the proof of Lemma 8.

Lemma 9 is known and easily checked (Theorems 2,3 [5] and [START_REF] Yang | Unique polynomials of entire and meromorphic functions[END_REF]):

Lemma 9. Let f, g ∈ M(C) be transcendental satisfy (f -a)f n = (g -a)g n with a ∈ C and let h = f g . If h is not identically 1, then g = h n -1 h n+1 -1 , f = h n+1 -h h n+1 -1 .
Notation: Let f ∈ M(C) be such that f (0) = 0, ∞. We denote by Z [START_REF] Boussaf | p-adic meromorphic functions f P (f ), g P (g) sharing a small function[END_REF] (r, f ) the counting function of the zeros of f each being counted with multiplicity when it is at most 2 and with multiplicity 2 when it is bigger.

We can extract the following Lemma 10 from a result that is proved in several papers and particularly in Lemma 3 [START_REF] Hua | Uniqueness and value-sharing of meromorphic functions[END_REF]. Lemma 10. Let f, g ∈ M(C) be such that f (0) = 0, ∞ and share the value 1 CM. If Ψ f,g is not identically zero, then,

max(T (r, f ), T (r, g)) ≤ N [2] (r, f ) + Z [2] (r, f ) + N [2] (r, g) + Z [2] (r, g) + S f (r) + S g (r)

Proof of Theorems

The polynomial P is the one we considered in theorems 1, 2, 3, 4 and we can assume a 1 = 0. In Theorem 5 we call P the polynomial such that P (0) = 0 and P

(x) = x n (x -a). Set F = f P (f ) α and G = g P (g) α .
Clearly F and G share the value 1 C.M. Since f, g are transcendental, we notice that so are F and G. Recall that

Ψ F,G = F F - 2F F -1 - G G + 2G G -1
We will prove that under the hypotheses of each theorem, Ψ F,G is identically zero. Set F = P (f ), G = P (g). We notice that P (x) is of the form x n+1 Q(x) with Q ∈ C[x] of degree k. Now, by Lemma 4, we have

T (r, F ) -Z(r, F ) ≤ T (r, F ) -Z(r, F ) + S F (r)
Consequently, since ( F ) = αF , we have

T (r, F ) ≤ T (r, F ) + Z(r, F ) -Z(r, F ) + T (r, α) + S f (r), (1)
hence, by (1), we obtain

T (r, F ) ≤ T (r, F )+(n+1)Z(r, f )+Z r, Q(f ) -nZ(r, f )- l i=2 k i Z(r, f -a i )-Z(r, f )+T (r, α)+S f (r) i.e. T (r, F ) ≤ T (r, F ) + Z(r, f ) + Z r, Q(f ) - l i=2 k i Z(r, f -a i ) -Z(r, f ) + T (r, α) + S f (r). ( 2
)
and similarly,

T (r, G) ≤ T (r, G) + Z(r, g) + Z r, Q(g) - l i=2 k i Z(r, g -a i ) -Z(r, g ) + T (r, α) + S f (r). (3)
Now, it follows from the definition of F and G that

Z [2] (r, F ) + N [2] (r, F ) ≤ 2Z(r, f ) + 2 l i=2 Z(r, f -a i ) + Z(r, f ) + 2N (r, f ) + T (r, α) + S f (r) (4)
and similarly

Z [2] (r, G) + N [2] (r, G) ≤ 2Z(r, g) + 2 l i=2 Z(r, g -a i ) + Z(r, g ) + 2N (r, g) + T (r, α) + S g (r) (5)
And particularly, if k i = 1, ∀i ∈ {2, .., l}, then

Z [2] (r, F ) + N [2] (r, F ) ≤ 2Z(r, f ) + l i=2 Z(r, f -a i ) + Z(r, f ) + 2N (r, f ) + T (r, α) + S f (r) (6)
and similarly

Z [2] (r, G) + N [2] (r, G) ≤ 2Z(r, g) + l i=2
Z(r, g -a i ) + Z(r, g ) + 2N (r, g) + T (r, α) + S g (r) [START_REF] Escassut | Functional equations in a p-adic context[END_REF] Suppose now that Ψ F,G is not identically zero. Now, by Lemma 10, we have

T (r, F ) ≤ Z [2] (r, F ) + N [2] (r, F ) + Z [2] (r, G) + N [2] (r, G) + S f (r) + S g (r)
(Z(r, f ) + Z(r, g )) + 4(N (r, f ) + N (r, g)) + S f (r) + S g (r) [START_REF] Fujimoto | On uniqueness of Meromorphic Functions sharing finite sets[END_REF] Now, let us go back to the general case. By Lemma 2, we can write Z(r, f ) + Z(r, g ) ≤ Z(r, f -a 2 ) + Z(r, g -a 2 ) + N (r, f ) + N (r, g) + S f (r). Hence, in general, by [START_REF] Frank | A unique range set for meromorphic functions with 11 elements[END_REF] we obtain

T (r, F ) + T (r, G) ≤ 5(Z(r, f ) + Z(r, g)) + l i=3 (4 -k i ) (Z(r, f -a i ) + Z(r, g -a i )) + (5 -k 2 ) (Z(r, f -a 2 ) + Z(r, g -a 2 )) +5(N (r, f ) + N (r, g)) + (Z(r, Q(f )) + Z(r, Q(g))) + S f (r) + S g (r).
And hence, since T (r, Q(f )) = kT (r, f ) + O(1) and T (r, Q(g)) = kT (r, g) + O(1), we have

T (r, F ) + T (r, G) ≤ 5(T (r, f ) + T (r, g)) + l i=3 (4 -k i ) Z(r, f -a i ) + Z(r, g -a i ) + (5 -k 2 ) Z(r, f -a 2 ) + Z(r, g -a 2 )) +5(N (r, f ) + N (r, g)) + k(T (r, f ) + T (r, g)) + S f (r) + S g (r). ( 12
)
Now, if k i = 1, ∀i ∈ {2, .., l}, by [START_REF] Fujimoto | On uniqueness of Meromorphic Functions sharing finite sets[END_REF] and Lemma 2 we have (T (r, f -a i )+T (r, g-a i ))+(l-1)(T (r, f )+T (r, g))+(T (r, f -a 2 )+T (r, g-a 2 ))

T (r, F ) + T (r, G) ≤ 5(Z(r, f ) + Z(r, g)) + l i=2 (Z(r, f -a i ) + Z(r, g -a i )) + (l -1)(T (r, f ) + T (r, g)) +(Z(r, f -a 2 ) + Z(r
+5(N (r, f ) + N (r, g)) + S f (r) + S g (r)

and hence T (r, F ) + T (r, G) ≤ (9 + 2l)(T (r, f ) + T (r, g)) + S f (r) + S g (r)

Now, let us go back to the general case. Since F is a polynomial in f of degree n + k + 1, we have T (r, F ) = (n + k + 1)T (r, f ) + O(1) and similarly, T (r, G) = (n + k + 1)T (r, g) + O(1), hence by [START_REF] Hayman | Meromorphic Functions[END_REF] we can derive

(n + k + 1)(T (r, f ) + T (r, g)) ≤ 5(T (r, f ) + T (r, g)) + (5 -k 2 )(Z(r, f -a 2 ) + Z(r, g -a 2 )) + l i=3 (4 -k i ) Z(r, f -a i ) + Z(r, g -a i ) +5(N (r, f ) + N (r, g)) + k(T (r, f ) + T (r, g)) + S f (r) + S g (r). (14)
Hence

(n + k + 1)(T (r, f ) + T (r, g)) ≤ (10 + k)(T (r, f ) + T (r, g)) + l i=3 (4 -k i ) Z(r, f -a i ) + Z(r, g -a i ) +(5 -k 2 )(Z(r, f -a 2 ) + Z(r, g -a 2 )) + S f (r) + S g (r)
and hence n(T (r, f ) + T (r, g)) ≤ 9(T (r, f ) + T (r, g))

+ l i=3 (4 -k i ) Z(r, f -a i ) + Z(r, g -a i ) +(5 -k 2 )(Z(r, f -a 2 ) + Z(r, g -a 2 )) + S f (r) + S g (r). ( 15
)
Then at least, for each i = 3, .., l we have (4

-k i )(Z(r, f -a i ) + Z(r, g -a i )) ≤ max(0, 4 -k i )(T (r, f ) + T (r, g)) + O(1) and (5 -k 2 )(Z(r, f -a 2 ) + Z(r, g -a 2 )) ≤ max(0, 5 -k 2 )(T (r, f ) + T (r, g)) + O(1).
Consequently, by [START_REF] Hua | Uniqueness and value-sharing of meromorphic functions[END_REF] we have n(T (r, f ) + T (r, g)) ≤ 9(T (r, f ) + T (r, g))+

+ l i=3 max(0, 4 -k i )(T (r, f ) + T (r, g)) + max(0, 5 -k 2 )(T (r, f ) + T (r, g)) + O(1)
and hence,

n ≤ 9 + l i=3 max(0, 4 -k i ) + max(0, 5 -k 2 ). (16)
Thus, we have proved that Ψ F,G = 0 when n ≥ 10 + l i=3 max(0, 4 -k i ) + max(0, 5 -k 2 ), concerning Theorem 3. And, if k i = 1, ∀i ∈ {2, .., l}, by [START_REF] Hoa | On the functional equation P (f ) = Q(g) in non-archimedean field[END_REF] we have n + k + 1 ≤ 9 + 2l, hence

n ≤ 9 + l. (17)
So, if k i = 1, ∀i ∈ {2, .., l}, when n ≥ l + 10 we have Ψ F,G = 0 which concerns Theorems 4 and 5.

Now, suppose that f, g, α belong to A(C). By ( 14), we have

(n + k + 1)(T (r, f ) + T (r, g)) ≤ (5 + k)(T (r, f ) + T (r, g)) + l i=3 (4 -k i ) (Z(r, f -a i ) + Z(r, g -a i )) +(5 -k 2 )(Z(r, f -a 2 ) + Z(r, g -a 2 )) + S f (r) + S g (r)
and hence n(T (r, f ) + T (r, g)) ≤ 4(T (r, f ) + T (r, g))

+ l i=3 (4 -k i ) (Z(r, f -a i ) + Z(r, g -a i )) +(5 -k 2 )(Z(r, f -a 2 ) + Z(r, g -a 2 )) + S f (r) + S g (r)
and hence n(T (r, f ) + T (r, g)) ≤ 4(T (r, f ) + T (r, g))

+ l i=3 max(0, 4 -k i )(T (r, f ) + T (r, g)) + max(0, 5 -k 2 )(T (r, f ) + T (r, g)) + O(1),
and hence,

n ≤ 4 + l i=3 max(0, 4 -k i ) + max(0, 5 -k 2 ). Consequently, if n ≥ 5 + l i=3 max(0, 4 -k i ) + max(0, 5 -k 2 ) when f, g, α belong to A(C), we have Ψ F,G = 0, which concerns Theorem 1.
Finally, suppose that f, g, α belong to A(C) and k i = 1 ∀i = 1, ..., l. We have n+k +1 ≤ 4+2l, hence n ≤ 4 + l. Consequently, if n ≥ 5 + l when f, g, α belong to A(C), we have Ψ F,G = 0, which concerns Theorem 2.

Thus, henceforth, we can assume that Ψ F,G = 0 in each theorem. Note that we can write

Ψ F,G = φ φ with φ = F (F -1) 2 (G -1) 2 G . Since Ψ F,G = 0, there exist A, B ∈ C such that 1 G -1 = A F -1 + B (18)
and A = 0. We notice Z(r, f

) ≤ T (r, f ), N (r, f ) ≤ T (r, f ), Z(r, f -a i ) ≤ T (r, f -a i ) ≤ T (r, f ) + O(1), i = 2, ..., l and Z(r, f ) ≤ T (r, f ) ≤ 2T (r, f ) + O(1)
. Similarly for g and g . Moreover, by Lemma 5, we have

T (r, F ) ≥ (n + k)T (r, f ) -m(r, 1 f ) + S f (r). (19)
Will consider the following two cases: B = 0 and B = 0.

Case 1: B = 0. Suppose A = 1. Then, by [START_REF] Lin | Uniqueness theorems for meromorphic functions concerning fixed-points[END_REF], we have F = AG + (1 -A). Applying Theorem N to F , we obtain

T (r, F ) ≤ Z(r, F ) + Z r, F -1 -A + N (r, F ) + S F (r)) ≤ Z(r, f ) + l i=2 Z(r, f -a i ) +Z(r, f ) + Z(r, g) + l i=2 Z(r, g -a i ) + Z(r, g ) + N (r, f ) + S f (r) + S g (r). (20)
Consequently, by [START_REF] Ojeda | Applications of the p-adic Nevanlinna theory to problems of uniqueness, Advances in p-adic and Non-Archimedean analysis[END_REF] and [START_REF] Ojeda | zeros of ultrametric meromorphic functions f f n (f -a) k -α[END_REF] we obtain

(n + k)T (r, f ) -m(r, 1 f ) ≤ Z(r, F ) + Z r, F -1 -A + N (r, F ) + S F (r) ≤ Z(r, f ) + l i=2 Z(r, f -a i ) + Z(r, f ) + Z(r, g) + l i=2 Z(r, g -a i ) + Z(r, g ) + N (r, f ) + S f (r) + S g (r) therefore (n+k)T (r, f ) ≤ Z(r, F )+Z r, F -1-A +N (r, F )+m(r, 1 f )+S F (r) ≤ Z(r, f )+ l i=2 Z(r, f -a i ) +Z(r, f ) + m(r, 1 f ) + Z(r, g) + l i=2 Z(r, g -a i ) + Z(r, g ) + N (r, f ) + S f (r) + S g (r).
Here we notice that Z(r, f

) + m(r, 1 f ) ≤ T (r, 1 f ) = T (r, f ) + O(1), hence (n + k)T (r, f ) ≤ ≤ Z(r, f ) + l i=2 Z(r, f -a i ) + Z(r, g) + l i=2 Z(r, g -a i ) + N (r, f ) + Z(r, g ) + T (r, f ) + S f (r) + S g (r).
(21) Then, considering all the previous inequalities in [START_REF] Ojeda | Uniqueness for ultrametric analytic functions[END_REF], by Lemma 3 we can derive

(n + k)T (r, f ) ≤ (l + 3)T (r, f ) + (l + 2)T (r, g) + S f (r) + S g (r). ( 22
)
Since f and g satisfy the same hypothesis, we also have

(n + k)T (r, g) ≤ (l + 3)T (r, g) + (l + 2)T (r, f ) + S f (r) + S g (r). (23)
Hence, adding ( 22) and ( 23), we have

(n + k) T (r, f ) + T (r, g) ≤ (2l + 5) T (r, f ) + T (r, g) + S f (r) + S g (r),
which leads to a contradiction whenever n + k ≥ (2l + 6).

In the hypothesis of Theorems 3, we have

n ≥ 10 + l i=3 max(0, 4 -k i ) + max(0, 5 -k 2 ). That implies n + k ≥ 10 + l i=3 max(0, 4 -k i ) + max(0, 5 -k 2 ) + l i=2 k i ≥ 10 + l i=2 max(k i , 4) ≥ 10 + 4(l -1) hence n + k ≥ 4l + 6. (24)
That contradiction proves that A = 1 in Theorem 3.

In the hypothesis of Theorems 4 and 5 we have n ≥ l + 10 with k = l -1 hence

n + k ≥ 2l + 9 (25)
therefore A = 1 again in Theorems 4 and 5.

Consider now the situation in Theorem 1. First we notice that given a positive number c we have max(0, 4 -c) + c ≥ 4. By [START_REF] Ojeda | Uniqueness for ultrametric analytic functions[END_REF], with help of Lemma 3, we can derive

(n+k)T (r, f ) ≤ Z(r, f )+ l i=2 Z(r, f -a i )+Z(r, g)+ l i=2 Z(r, g-a i )+Z(r, g )+T (r, f )+S f (r)+S g (r), ≤ Z(r, f ) + l i=2 Z(r, f -a i ) + Z(r, g) + l i=2 Z(r, g -a i ) + T (r, g) + T (r, f ) + S f (r) + S g (r),
with a similar inequality for g, thereby

(n + k)(T (r, f ) + T (r, g)) ≤ (2l + 2)(T (r, f ) + T (r, g))
and hence n + k ≤ 2l + 2, which leads to a contradiction as soon as n + k ≥ 2l + 3. Instead of ( 24) we now have n + k ≥ 4l + 1. Consequently, the relation n + k ≥ 2l + 3 is easily satisfied in Theorem 1, proving A = 1 again.

And now, consider the hypothesis of Theorem 2. We have n ≥ l + 5, hence n + k ≥ 2l + 4 assuring that A = 1 again.

Thus, in case B = 0, we have proved that F = G. Now, αF = αG, i.e. ( F ) = ( G) . We assume n ≥ k + 3 in Theorem 3 and this is automatically satisfied in Theorems 4 and 5. Next, we assume n ≥ k + 2 in Theorem 1 and this is automatically satisfied in Theorems 2. Consequently, by Lemma 6, we have F = G, i.e. P (f ) = P (g). But in Theorems 3 and 4, P is a polynomial of uniqueness for the family of meromorphic functions we consider, hence we have f = g. And in Theorem 5, the conclusion comes from Lemma 9. Finally, in Theorems 1 and 2, we assume P is a polynomial of uniqueness for A(C), we have f = g again. Suppose F G = 1, hence f P (f )g P (g) = α 2 . In Theorem 3 we have assumed that n ≥ k + 3 and in Theorems 4 and 5 this hypothesis is automatically satisfied. Next, if l = 2, then n = 2k, 2k + 1, 3k + 1 and if l = 3 then n = 2k + 1, 3k i -k, i = 2, 3. Moreover, these conditions are automatically satisfied in Theorems 4 and 5. So, we have a contradiction to Lemma 8 in Theorems 3, 4, 5. Consequently, F = G and therefore we can conclude as in the case B = 0, that P (f ) = P (g). Now, in Theorems 3 and 4 P is a polynomial of uniqueness for M(C), hence we have f = g. And in Theorem 5 we know that either f = g or f and g are linked by the relation mentionned in Theorems 2, 3 [5] and in [START_REF] Yang | Unique polynomials of entire and meromorphic functions[END_REF].

Consider now the situation in Theorems 1 and 2. By [START_REF] Yi | Uniqueness theorems of meromorphic functions[END_REF] and Lemma 3, here we can write Z(r, F ) + Z(r, G) + N (r, F ) + N (r, G) ≤ Z(r, f ) + l i=2 Z(r, f -a i ) + Z(r, f ) + Z(r, g ) + Z(r, g) On the other hand, in Theorem 1, instead of (24) here we have n + k ≥ 4l + 1 hence 2l+2 n+k < 1 and hence we can apply Lemma 7 proving that either F = G or F G = 1. But since αF, αG and α belong to A(C), by Lemma 8 we know that (αF )(αG) = (α) 2 is impossible. Consequently F = G. Now in Threorem 2, we have n + k ≥ 2l + 4 hence 2l+2 n+k < 1 and hence we can go on as in Theorem 1. So, finally we have F = G in Theorems 1 and 2 again. Next, we can conclude as in the case B = 0.

  and g g n (g -a) share the function α C.M. and if n ≥ 12, then either f = g or there exists h ∈ M(C) such that f = a(n + 2) n + 1

  , g -a 2 )) + 5(N (r, f ) + N (r, g)) + S f (r) + S g (r) hence T (r, F ) + T (r, G) ≤ 5(T (r, f )+T (r, g))+ l i=2

Case 2 :

 2 B = 0. We have Z(r, F ) ≤ Z(r, f ) + l i=2 Z(r, f -a i ) + Z(r, f ) + S f (r) and N (r, F ) ≤ N (r, f ) + S f (r) and similarly for G, so we can deriveZ(r, F ) + Z(r, G) + N (r, F ) + N (r, G) ≤ Z(r, f ) + l i=2 Z(r, f -a i ) + Z(r, f ) +Z(r, g) + l i=2 Z(r, g -a i ) + Z(r, g ) + N (r, f ) + N (r, g) + S f (r) + S g (r). (26)Moreover, by[START_REF] Ojeda | Applications of the p-adic Nevanlinna theory to problems of uniqueness, Advances in p-adic and Non-Archimedean analysis[END_REF] we have(n + k)T (r, f ) ≤ T (r, F ) + m(r, 1 f ) + S f (r) and (n + k)T (r, g) ≤ T (r, G) + m(r,1 g) + S g (r). Consequently, by[START_REF] Yi | Uniqueness theorems of meromorphic functions[END_REF] we obtainZ(r, F ) + Z(r, G) + N (r, F ) + N (r, G) ≤ Z(r, f ) + l i=2 Z(r, f -a i ) + [Z(r, f ) + m(r, g -a i ) + [Z(r, g ) + m(r, 1 g )] -m(r, 1 g ) + N (r, f ) + N (r, g) + S f (r) + S g (r) ≤ (l + 1) T (r, f ) + T (r, g) + T (r, f ) + T (r, g ) -m(r, 1 f ) -m(r, 1 g ) + S f (r) + S g (r)hence by Lemma 3,Z(r, F )+Z(r, G)+N (r, F )+N (r, G) ≤ (l+3)(T (r, f )+T (r, g))-m(r, f (r)+S g (r)(28)Now, in Theorems 3, 4, 5 we have n + k ≥ min(4l + 6, 2l + 9) hence 2l+6 k+n < 1, therefore by (19) we notice that2(n + k)( T (r, f ) + T (r, g) 2 ) -m(r, 1 f ) -m(r, 1 g ) ≤ T (r, F ) + T (r, G) + S f (r) + S g (r)hence (28) yieldsZ(r, F ) + Z(r, G) + N (r, F ) + N (r, G) ≤ ( 2l + 6 n + k )( T (r, F ) + T (r, G) 2 ) + S f (r) + S g (r).Now by Lemma 5 we can deriveZ(r, F ) + Z(r, G) + N (r, F ) + N (r, G) ≤ ( 2l + 6 n + k )( T (r, F ) + T (r, G) 2 ) + S F (r) + S G (r)(29)Consequently, by (29), we obtain lim sup r→+∞ Z(r, F ) + Z(r, G) + N (r, F ) + N (r, G) max(T (r, F ), T (r, G)) < 1 outside a set of finite measure and hence, by Lemma 7, we have F = G or F G = 1.

  , g -a i ) + S f (r) + S g (r)≤ Z(r, f ) + l i=2 Z(r, f -a i ) + T (r, f ) + T (r, g) -m(r, g -a i ) + S f (r) + S g (r)hence finally, Z(r, F ) + Z(r, G) + N (r, F ) + N (r, G) ≤ (l + 1)(T (r, f ) + T (r, g)) -m(r, 29) here, thanks to Lemma 5 we can obtain (30):Z(r, F ) + Z(r, G) + N (r, F ) + N (r, G) ≤ ( 2l + 2 n + k )( T (r, F ) + T (r, G)2) + S F (r) + S G (r) (30)

Partially supported by CONICYT N • 79090014 (Inserción de Capital Humano a la Academia)"

hence by [START_REF] Boussaf | p-adic meromorphic functions f P (f ), g P (g) sharing a small function[END_REF], we obtain T (r, F ) ≤ Z [START_REF] Boussaf | p-adic meromorphic functions f P (f ), g P (g) sharing a small function[END_REF] (r, F ) + N [START_REF] Boussaf | p-adic meromorphic functions f P (f ), g P (g) sharing a small function[END_REF] (r, F ) + Z [START_REF] Boussaf | p-adic meromorphic functions f P (f ), g P (g) sharing a small function[END_REF] (r, G) + N [START_REF] Boussaf | p-adic meromorphic functions f P (f ), g P (g) sharing a small function[END_REF] (r, G) + Z(r, f ) + Z(r, Q(f ))

and hence by ( 4) and (5):

and similarly,

Consequently,

Moreover, if k i = 1, ∀i ∈ {2, .., l}, then by ( 6) and ( 7) we have

and similarly,

Consequently, T (r, F ) + T (r, G) ≤ 5(Z(r, f ) + Z(r, g)) + l i=2 (Z(r, f -a i ) + Z(r, g -a i )) + Z(r, Q(f )) + Z(r, Q(g))+