CUBO A Mathematical Journal

Vol.10, N
o
¯ 01, (1–10). March 2008

Prime Factorization of Entire Functions

Xinhou Hua and Rémi Vaillancourt1

Department of Mathematics and Statistics

University of Ottawa, Ottawa, ON, K1N 6N5, Canada

email: hua@mathstat.uottawa.ca, remi@uottawa.ca

ABSTRACT

Let n be a prime number and let f (z) be a transcendental entire function. Then

it is proved that both [f (z) + cz]
n

and [f (z) + cz]
−n

are uniquely factorizable for

any complex number c, except for a countable set in C.

RESUMEN

Sea n un número primo y f (z) una función entera transcendental. Entonces

ambos [f (z) + cz]
n

y [f (z) + cz]
−n

se factorizan de manera única para cualquier

número complejo c, excepto para un conjunto numerable en C.

Key words and phrases: entire function, unique factorization.

Math. Subj. Class.: 30D05.

1This research was partially supported by the Natural Sciences and Engineering Research Council of
Canada and the Centre de recherches mathématiques of the Université de Montréal.



2 Xinhou Hua and Rémi Vaillancourt CUBO
10, 1 (2008)

1 Introduction

The fundamental theorem of elementary number theory states that every integer n ≥ 2 can

be expressed uniquely as the product of primes in the form

n = p
m1
1 · · · p

mk
k , for k ≥ 1,

with distinct prime factors p1, . . . , pk and corresponding exponents m1 ≥ 1, . . . , mk ≥ 1

uniquely determined by n. For example, 2700 = 2
2
3
3
5
2
.

In 1922, Ritt ([14]) generalized this theorem to polynomials. To state his result, we

introduce the following concepts.

Let F (z) be a nonconstant meromorphic function. A decomposition

F (z) = f (g(z)) = f ◦ g(z) (1)

will be called a factorization of F (z) with f (z) and g(z) being the left and right factors of

F (z), respectively, where f (z) is meromorphic and g(z) is entire (g(z) may be meromorphic

when f (z) is rational) (see [2], [4], [19]).

A function F (z) is said to be prime (pseudo-prime) if F (z) is nonlinear and every

factorization of the form (1) implies that either f (z) is fractional linear or g(z) is linear

(either f (z) is rational or g(z) is a polynomial).

Example 1 ez + z is prime.

This is stated by Rosenbloom [15] and proved by Gross [3].

Example 2 (cos z)eaz+b + p(z) is prime, where a (6= 0) and b are constants, and p(z) is a

nonconstant polynomial.

This was conjectured by Gross–Yang [5] and proved by Hua [7].

Suppose that a function F (z) has two prime factorizations

F (z) = f1 ◦ · · · ◦ fm(z) = g1 ◦ · · · ◦ gn(z),

i.e., fi (i = 1, . . . , m) and gj (j = 1, . . . , n) are prime functions. If m = n and if there exist

linear functions Lj (j = 1, . . . , n − 1) such that

f1(z) = g1 ◦ L
−1
1 , f2(z) = L1 ◦ g2 ◦ L

−1
2 , . . . , fn(z) = L

−1
n ◦ gn(z),

then the two factorizations are called equivalent. If any two prime factorizations of F (z) are

equivalent, then F (z) is called uniquely factorizable. In particular, for an entire function



CUBO
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Prime Factorization of Entire Functions 3

F (z), if any two prime entire factorizations of F (z) are equivalent, then F (z) is called

uniquely factorizable in the entire sense.

Ritt [14] proved the following result.

Proposition 1 Let p(z) be a nonlinear polynomial. If p(z) has two prime factorizations

p(z) = p1 ◦ · · · ◦ pm(z) = q1 ◦ · · · ◦ qn(z),

where pi (i = 1, . . . , m) and qj (j = 1, . . . , n) are polynomials, then m = n. Moreover,

one factorization can be changed to another one by a sequence of applications of any of the

following three ways:

1. replace pi and pi+1 by pi ◦ L and L
−1 ◦ pi+1, respectively;

2. alternate pi and pi+1 when both are Chebychev polynomials;

3. replace zk and zsh(zk) by zsh(z)k and zk, respectively, where h(z) is a polynomial,

and s and k are natural numbers.

Example 3 z10 + 1 = (z5 + 1) ◦ z2 = (z2 + 1) ◦ z5.

However, Ritt’s result cannot be extended to rational functions.

Example 4 z3 ◦ z
2−4
z−1

◦ z
2+2
z+1

=
z(z−8)3

(z+1)3
◦ z3.

This example was given by Michael Zieve (see [1]).

For transcendental functions, the diverse cases are very complex. For example, e
z

can

have infinitely many nonlinear factors.

Example 5 For any integer n,

e
z

= z
2 ◦ z3 ◦ · · · ◦ zn ◦ ez/n!.

The following example shows that transcendental entire functions can have non-equivalent

prime factorizations (see [10]).

Example 6

z
2 ◦

(

ze
z2

)

=
(

ze
2z

)

◦ z2.

Of course, there are functions which are uniquely factorizable. The following example

is given by Urabe [17].



4 Xinhou Hua and Rémi Vaillancourt CUBO
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Example 7 For any two nonconstant polynomials p(z) and q(z),

(z + e
p(ez )

) ◦ (z + q(ez))

is uniquely factorizable.

The following result, proved by Hua [6], shows that, for a given function, we can con-

struct uncountably many uniquely factorizable functions.

Proposition 2 Let f (z) be a transcendental entire function and n ≥ 3 be a prime number.

Then both f (zn) − czn and (zn − c)f (zn) are uniquely factorizable for any complex number

c except for a countable set.

In this paper, we prove the following two results.

Theorem 1 Let f (z) be a transcendental entire function and n ≥ 3 be a prime number.

Then [f (z) − cz]n is uniquely factorizable for any complex number c except for a countable

set.

Theorem 2 Let f (z) be a transcendental entire function and n ≥ 3 be a prime number.

Then [f (z) − cz]−n is uniquely factorizable for any complex number c except for a countable

set.

2 Some Lemmas

The following lemmas will be used in the proof of the theorems.

Lemma 1 ([4]) Suppose that p(z) is a nonconstant polynomial and g(z) is entire. Then

p(g(z)) is periodic if and only if g(z) is periodic.

Lemma 2 ([11]) Let f (z) be a transcendental entire function. Then for any complex num-

ber c except for a countable set, f (z) − cz is prime.

Remark. So far, there is no example with countably infinite exceptions. In [13], it

is proved that there is at most one exception for f (z) = g(e
z
), where g(z) is an entire

function satisfying max|z|=r |g(z)| ≤ e
Kr

for a positive constant K. In [8] and [18], some

other functions f (z) are studied.



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Prime Factorization of Entire Functions 5

Lemma 3 ([12]) Let f (z) be a transcendental entire function. We denote by ν(a, f ) the

least order of almost all zeros of f (z) − a, where “almost all” means all with possibly finite

exceptions. Then
∑

a6=∞

(

1 −
1

ν(a, f )

)

≤ 1.

Lemma 4 ([16]) Let f (z) and g(z) be prime entire functions. Assume that both f (z) and

F (z) = f (g(z)) are non-periodic. Then F (z) is uniquely factorizable if and only if F (z) is

uniquely factorizable in the entire sense.

Lemma 5 Let f (z) be a nonconstant meromorphic function. Then f (z)−cz is non-periodic

for any complex number c with at most one exception.

Proof of Lemma 5. Suppose there exist two different numbers c and d such that f (z) − cz

and f (z) − dz are periodic with period u and v, respectively. Then f ′(z) is periodic and

f
′
(z + u) = f

′
(z) = f

′
(z + v). Let w be the period of f

′
(z). Then there exist two nonzero

integers m and k such that u = mw and v = kw. This implies that u =
m
k

v. Hence

f (z) − cz = f (z + ku) − c(z + ku)

= f (z + mv) − c(z + ku)

= f (z + mv) − d(z + mv) + d(z + mv) − c(z + ku)

= f (z) − dz + d(z + mv) − c(z + ku)

= f (z) − cz + dmv − cku.

Therefore dmv = cku, and so, d = c, which is a contradiction. 2

The following lemma is a simple version of the so-called Borel Unicity Theorem which

can be found in [2] and [4].

Lemma 6 Let h0(z), . . . , hn(z) be rational functions and let g1(z), . . . , gn(z) be nonconstant

entire functions such that
n

∑

j=1

hj (z)e
gj (z) = h0(z).

Then h0 = 0.

Lemma 7 Let f (z) be a transcendental entire function. Then

f (z) − cz 6= P (z)ef1(z)

for all c ∈ C with at most one exception, where P (z) is a polynomial and f1(z) is a noncon-

stant entire function.



6 Xinhou Hua and Rémi Vaillancourt CUBO
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Proof of Lemma 7. Suppose to the contrary that there exist two different constants c and

d, two polynomials P1(z) and P2(z), and two nonconstant entire functions f1(z) and f2(z)

such that

f (z) − cz = P1(z)e
f1(z)

and

f (z) − dz = P2(z)e
f2(z).

Then

cz − dz = P2(z)e
f2(z) − P1(z)e

f1(z).

By Lemma 6, cz − dz = 0; thus d = c which is a contradiction. 2

3 Proof of Theorem 1

Let

F (z) = [f (z) − cz]n = zn ◦ (f (z) − cz).

Obviously, z
n

is non-periodic.

Let

Z(f ) = {f (z) : f ′′(z) = 0}.

Then Z(f ) is a countable set, and for any c 6∈ Z(f ), f ′(z) − c has only simple zeros ([9,

Theorem F]). We combine Z(f ) and all the exceptions (if any) in Lemmas 1, 2, 5 and 7

to form an exceptional set E. Then E is a countable set which may be empty. For any

c ∈ C − E, we have the following properties:

(P1) The function F (z) is non-periodic;

(P2) The function f (z) − cz is prime;

(P3) f
′
(z) − c has only simple zeros.

(P4) f (z)−cz 6= P (z)ef1(z) for any polynomial P (z) and nonconstant entire function f1(z).

Next we assume c ∈ C − E.

By Lemma 4, we need only prove that F (z) is uniquely factorizable in the entire sense,

which means, we just need to consider entire factors. Assume that

F (z) = g(z) ◦ h(z), (2)

where g(z) and h(z) are nonconstant entire functions. We consider three cases.



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Prime Factorization of Entire Functions 7

Case 1. g(z) has at least two zeros, z1 and z2, of order m1 and m2, respectively, such that

(n, m1) = (n, m2) = 1, that is, n and mi (i = 1, 2) have no common factors other than 1.

Then by (2) and the fact that n is prime, the order of any zero of h(z) − zi (i = 1, 2) should

be a multiple of n. Hence

ν(zi, h) ≥ n ≥ 3 (i = 1, 2),

which implies that
∑

a6=∞

(

1 −
1

ν(a, f )

)

≥ 1 −
1

3
+ 1 −

1

3
> 1.

This is a contradiction to Lemma 3.

Case 2. g(z) has one zero, z0, of order m such that (n, m) = 1. Then by (2) and the fact

that n is prime, g(z) and h(z) can be written as

g(z) = (z − z0)
r
g1(z)

n
, h(z) = z0 + h1(z)

n
, r = m (mod n), (3)

where g1(z) and h1(z) are entire functions. Obviously, 1 ≤ r < n. Substituting (3) into (2)

we have

F (z) = h1(z)
rn

[g1(z0 + h1(z)
n
)]

n
,

which implies that

f (z) − cz = uh1(z)
r
g1(z0 + h1(z)

n
)

= [uz
r
g1(z0 + z

n
)] ◦ h1(z), (4)

where u is an n-th root of unity. Since f (z) − cz is prime, we have two subcases as follows.

Case 2.1. Since the left factor uzrg1(z0 + z
n
) is linear, then r = 1 and g1 is a constant.

It follows from (3) that g(z) is linear. This is a trivial case.

Case 2.2. The right factor h1(z) is linear. Let h1(z) = az + b (a, b ∈ C, a 6= 0). By (4),

f (z) − cz = u(az + b)rg1[z0 + (az + b)
n
)]. (5)

If g1(z) has a zero, then by differentiating (5) we see that f
′
(z) − c has a zero of order

n − 1 ≥ 2, which is a multiple zero of f ′(z) − c. This contradicts (P3). Therefore g1(z)

has no zero. This implies that there exists a nonconstant entire function g2(z) such that

g1(z) = e
g2(z). By (5),

f (z) − cz = u(az + b)reg2[z0+(az+b)
n)]

,

which contradicts (P4).



8 Xinhou Hua and Rémi Vaillancourt CUBO
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Case 3. The order of any zero of g(z) is a multiple of n. Then there exists an entire function

g2(z) such that

g(z) = g2(z)
n
. (6)

It follows from (2) that

[f (z) − cz]n = [g2 ◦ h(z)]
n
,

and so,

f (z) − cz = ug2(z) ◦ h(z)

for an n-th root of unity, u. Since f (z) − cz is prime, we have two subcases.

Case 3.1. The left factor ug2(z) is linear. It follows from (6) that g(z) = z
n ◦ L(z) for

a linear function L(z). Therefore we get an equivalent factorization.

Case 3.2. The right factor h(z) is linear. This is a trivial case.

The proof is complete. 2

4 Proof of Theorem 2

Assume that

[f (z) − cz]−n = g(z) ◦ h(z),

where g(z) is a nonconstant meromorphic function and h(z) is a nonconstant entire function.

Then we have

[f (z) − cz]n =
1

g(z)
◦ h(z).

Now, since the left-hand side is entire, the conclusion follows from Lemma 4 and Theorem 1.

2

5 Open Questions

Question 1 Can n be 2 in Theorems 1 and 2?

Question 2 What kind of rational functions are uniquely factorizable?

Question 3 Is (z + ee
z

) ◦ (z + ee
z

) uniquely factorizable?

Received: November 2005. Revised: January 2006.



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Prime Factorization of Entire Functions 9

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