Pintér, Ákos

Rakaczki, Csaba

Symmetry

English

In the present paper we prove that under certain conditions the linear combination of two Euler polynomials with odd degrees      P  n , m    ( x )  =  E n   ( x )  + c  E m   ( x )      is always indecomposable over    C   , where c denotes a rational number

Ákos Pintér

Multidisciplinary Digital Publishing Institute

On the Decomposability of the Linear Combinations of Euler Polynomials with Odd Degrees

University of Debrecen Electronic Archive

symmetryS S
Article
On the Decomposability of the Linear Combinations of
Euler Polynomials with Odd Degrees
Ákos Pintér 1,* and Csaba Rakaczki 2
1 Institute of Mathematics, MTA-DE Research Group “Equations, Functions and Curves”, Hungarian Academy of
Sciences and University of Debrecen, P. O. Box 12, H-4010 Debrecen, Hungary
2 Institute of Mathematics, University of Miskolc, H-3515 Miskolc Campus, Hungary; matrcs@uni-miskolc.hu
* Correspondence: apinter@science.unideb.hu
Received: 2 April 2019; Accepted: 18 May 2019; Published: 31 May 2019
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Abstract: In the present paper we prove that under certain conditions the linear combination of two
Euler polynomials with odd degrees Pn,m(x) = En(x) + cEm(x) is always indecomposable over C, where
c denotes a rational number.
Keywords: Euler polynomials; higher degree equations
MSC: 11B68; 11B83; 11R09
1. Introduction
Let K be a field. If f (x) ∈ K[x] has degree at least 2, we say that f (x) is decomposable over the field K
if we can write f (x) = f1( f2(x)) for some nonlinear polynomial f1(x), f2(x) ∈ K[x]. Otherwise, we say
that f (x) is indecomposable over K. Two decompositions f (x) = f1( f2(x)) and f (x) = F1(F2(x)) are said
to be equivalent over the field K, written f1 ◦ f2 ∼K F1 ◦ F2, if there exists a linear polynomial l(x) ∈ K[x]
such that
f1(x) = F1(l(x)) and F2(x) = l( f2(x)).
For a given f (x) ∈ K[x] with degree at least 2, a complete decomposition of f (x) over K is a
decomposition f = f1 ◦ · · · ◦ fk, where the polynomials fi ∈ K[x] are indecomposable over K for
i = 1, . . . , k. A polynomial of degree greater than 1 always has a complete decomposition, but it does not
need to be unique even up to equivalence.
Euler polynomials are defined by the following generating function
∞
∑
k=0
Ek(x)
tk
k!
=
2etx
(et + 1)
.
These polynomials play a central role in various branches of mathematics; for example, in various
approximation and expansion formulas in discrete mathematics and in number theory (see for
instance [1,2]), in p-adic analyis (see [3], Chapter 2), in statistical physics as well as in semi-classical
approximations to quantum probability distributions (see [4–7]).
There are several results connected to the decomposability of an infinite family of polynomials, see
for instance [8–12]. Bilu, Brindza, Kirschenhofer, Pintér and Tichy [13] gave all the decompositions of
Bernoulli polynomials. Kreso and Rakaczki [14] characterized the all possible decomposations of Euler
Symmetry 2019, 11, 739; doi:10.3390/sym11060739 www.mdpi.com/journal/symmetry
Symmetry 2019, 11, 739 2 of 8
polynomials with degree even, moreover they showed that every Euler polynomial with odd degree is
indecomposable. It is harder to obtain similar results for the sum of polynomials. Pintér and Rakaczki [15]
describe the complete decomposition of linear combinations of the form
Rn(x) = Bn(x) + cBn−2(x)
of Bernoulli polynomials, where c is an arbitrary rational number. Later, Pintér and Rakaczki in [16]
proved that for all odd n > 1 integer and for all rational number c the polynomials En(x) + cEn−2(x)
are indecomposable.
The main purpose of this paper is to prove that under certain conditions a linear combination with
rational coefficients of two Euler polynomials with odd degrees is always indecomposable. We have
Theorem 1. Let Pn,m(x) = En(x) + cEm(x), where c = A/B is an arbitrary rational number, where B 6= 2a,
a ∈ N∪ {0}, n, m are odd integers with n > m > n/3. Then the polynomials Pn,m(x) are indecomposable over C.
2. Auxiliary Results
In the first lemma we collect some well known properties of the Euler polynomials which will be
used in the sequel, sometimes without particular reference.
Lemma 1. (a) En(x) = (−1)nEn(1− x);
(b) En(x + 1) + En(x) = 2xn;
(c) E′n(x) = nEn−1(x);
(d) E2n−1(1/2) = E2n(0) = E2n(1) = 0 for n ∈ N;
(e) En(x) = ∑nk=0 (
n
k)Ek(0)x
n−k;
Proof. See [2].
The following result is a general theorem from the theory of decomposability.
Lemma 2 (Kreso and Rakaczki [14]). Let F(x) ∈ K[x] be a monic polynomial such that deg F is not divisible by
the characteristic of the fieldK. Then for every nontrivial decomposition F = F1 ◦ F2 over any field extension L ofK,
there exists a decomposition F = F˜1 ◦ F˜2 such that the following conditions are satisfied
• F˜1 ◦ F˜2 and F1 ◦ F2 are equivalent over L,
• F˜1(x) and F˜2(x) are monic polynomials with coefficients in K,
• coeff (xdeg F˜1−1, F˜1(x)) = 0.
Moreover, such decomposition F˜1 ◦ F˜2 is unique.
Lemma 3. Let h(x) ∈ Q[x] with deg h(x) ≥ 4. If h(x) is decomposable over Q then we can write the polynomial
h(x) in the form h(x) = uv f (g(x)), where u and v 6= 0 are relative prime integers, f (x) and g(x) ∈ Z[x] are
primitive polynomials. Moreover, if h(x) is a monic polynomial, then u = 1.
Proof. Suppose that h(x) = F(G(x)), where F(x), G(x) ∈ Q[x]. Let
F(x) = bkxk + bk−1xk−1 + · · ·+ b1x + b0,
G(x) = ctxt + ct−1xt−1 + · · ·+ c1x + c0.
Symmetry 2019, 11, 739 3 of 8
Every polynomial with rational coefficients can be written uniquely as a product of a rational number and
a primitive polynomial. Hence, we can assume that
G(x) =
c
d
g(x), where g(x) is a primitive polynomial, c, d 6= 0 ∈ Z
and so
F(G(x)) = bk
( c
d
)k
g(x)k + bk−1
( c
d
)k−1
g(x)k−1 + · · ·+ b1 cd g(x) + b0.
The polynomial
F1(x) = bk
( c
d
)k
xk + bk−1
( c
d
)k−1
xk−1 + · · ·+ b1 cd x + b0 ∈ Q[x]
can be written in the from uv f (x), where f (x) ∈ Z[x] is a primitive polynomial, u > 0, v 6= 0 are relative
prime integers. However, then we have
h(x) = F(G(x)) = F1(g(x)) =
u
v
f (g(x)). (1)
If the polynomial h(x) is monic, then comparing the leading coefficients in (1) one can deduce that
v = u fkgkt , where fk and gt denotes the leading coefficient of the polynomial f (x) and g(x), respectively.
This means that u divides v that is u = 1.
Let
S+ = { f (x) ∈ C[x] | f (x) = f (1− x)}
and
S− = { f (x) ∈ C[x] | f (x) = − f (1− x)} .
From these definitions it is easy to see that S+ and S− are subspaces in the vector space C[x].
Lemma 4. Let P(x) ∈ Q[x] be a monic polynomial. Assume that P(x) ∈ S− and P(x) = f (g(x)), where f (x),
g(x) ∈ Q[x] and deg( f (x)), deg(g(x)) > 1. Then we can assume that f (x), g(x) are monic, g(x) ∈ S− and
f (x) = − f (−x).
Proof. See [16].
The following Lemma is a simple combination of Lemmas 3 and 4.
Lemma 5. Let P(x) ∈ Q[x] be a monic polynomial. Assume that P(x) ∈ S− and P(x) = F(G(x)), where F(x),
G(x) ∈ Q[x] and deg(F(x)) > 1, deg(G(x)) > 1. Then we can assume that P(x) = 1v f (g(x)), where v 6= 0 is
an integer, f (x) and g(x) are primitive polynomials, g(x) ∈ S− and f (x) = − f (−x).
Proof. From Lemma 4 we can assume that G(x) ∈ S− and F(x) = −F(−x). Using the proof of Lemma 3
and the fact that S− is a subspace of C[x] we get the assertion of our Lemma.
Lemma 6. Let g(x) = ctxt + ct−1xt−1 + · · ·+ c1x + c0 ∈ S−. Then
− 2cs =
(
s + 1
s
)
cs+1 +
(
s + 2
s
)
cs+2 + · · ·+
(
t− 1
s
)
ct−1 +
(
t
s
)
ct (2)
for even index 0 ≤ s ≤ t− 1.
Symmetry 2019, 11, 739 4 of 8
Proof. Since g(x) ∈ S− we have that −g(x) = g(1− x). Computing the coefficient of xs on the both sides
we obtain that
−cs = (−1)s
(
cs +
(
s + 1
s
)
cs+1 +
(
s + 2
s
)
cs+2 + · · ·+
(
t− 1
s
)
ct−1 +
(
t
s
)
ct
)
.
Lemma 7. Let
f (x) = bkxk + bk−2xk−2 + bk−3xk−3 + · · ·+ b1x + b0,
g(x) = ctxt + ct−1xt−1 + · · ·+ c1x + c0 ∈ Q[x].
If k, t ≥ 2 then the coefficient of the monomial xkt−2 in the polynomial f (g(x)) is
bk
(
kck−1t ct−2 +
(
k
2
)
ck−2t c
2
t−1
)
.
Proof. It is easy to see that the monomial xkt−2 occurs only in the term bkg(x)k. Expanding g(x)k we
simply get the assertion.
3. Proof of the Theorem
Let n, m be odd positive integers with n− 2 > m > n/3, B is an arbitrary integer which is not a
power of two. The case of m = n− 2 was treated in [16]. Suppose that Pn,m(x) is decomposable over
C. From Lemmas 2 and 5 we can assume that Pn,m(x) = 1v f (g(x)), where v 6= 0 is an integer, f (x),
g(x) ∈ Z[x] are primitive polynomials and g(x) ∈ S−, f (x) = − f (−x). Let
f (x) = bkxk + bk−2xk−2 + bk−4xk−4 + · · ·+ b3x3 + b1x,
g(x) = ctxt + ct−1xt−1 + · · ·+ c1x + c0.
Using (b) of Lemma 1 one can deduce that
1
v
f (g(x + 1)) +
1
v
f (g(x)) = Pn,m(x + 1) + Pn,m(x) = 2xn + 2cxm. (3)
Since Pn,m(x) ∈ S− thus Pn,m(x + 1) = −Pn,m(−x). From (3) we infer that the polynomial g(x)− g(−x)
divides the polynomial Pn,m(x)− Pn,m(−x) = 2xn + 2cxm, that is
g(x)− g(−x) = dxsh(x), (4)
where d ∈ Q, 0 ≤ s ≤ m and the polynomial h(x) divides the polynomial xn−m + c in Q[x]. We know that
g(x)− g(−x) = 2ctxt + 2ct−2xt−2 + · · ·+ 2c3x3 + 2c1x. (5)
If the polynomial h(x) is a constant polynomial then we have t = s and so ct−2 = 0. It follows from
Pn,m(x) = En(x)+ cEm(x) and (d), (e) of Lemma 1 that the coefficient of xn−2 in Pn,m(x) equals 0. Applying
now Lemma 7 we get that
bk
(
k
2
)
ck−2t c
2
t−1 = 0,
which is impossible since −2ct−1 = tct by Lemma 6.
Symmetry 2019, 11, 739 5 of 8
In the case when h(x) = xn−m + c we get s + n−m = t, d = 2ct and g(x)− g(−x) = 2ctxs+n−m +
2ctcxs = 2ctxt + 2ctcxt−(n−m). Since by assumption n−m > 2, we obtain again that ct−2 = 0, which is
not possible.
Next suppose that 1 ≤ deg h(x) < n−m. In this case one can deduce that s is odd and h(x) = h(−x).
Consider first when 1 < s. Then c1 = c3 = · · · = cs−2 = 0 and cs 6= 0. Let G(x) = g(x) − g(0) and
F(x) = f (x + g(0)). Then f (g(x)) = F(G(x)), G(0) = 0 and
G(x)− G(−x) = g(x)− g(−x) = 2ctxt + 2ct−2xt−2 + · · ·+ 2csxs.
Let
F(x) = akxk + ak−1xk−1 + · · ·+ a2x2 + a1x + a0.
Since s < t ≤ n/3 < m we have that s + 4 ≤ m.
Investigate the coefficients of xs and xs+2 in
vPn,m(x) = F(G(x)) = akG(x)k + ak−1G(x)k−1 + · · ·+ a1G(x) + a0. (6)
Since s + 2 < m in the polynomials vPn.m(x) = En(x) + cEm(x) these coefficients are 0. On the other hand,
one can observe that xs occurs only in the term a1G(x) and so a1cs = 0. This means that a1 = 0 and so
vPn,m(x) = F(G(x)) = akG(x)k + · · ·+ a3G(x)3 + a2G(x)2 + a0. (7)
Since xs+2 appears only in the term a2G(x)2 thus 2a2c2cs = 0.
If a2 = 0 we obtain from (7) that the coefficients of x5, x4, x3, x2 and x in F(G(x)) are zero. This yields
that P(i)n,m(0) = 0 for i = 1, . . . , 5. Further, by Lemma 1
P(j)n,m(0) = P
(j)
n,m(1) = 0, if j is odd and j 6= m, n;
P(j)n,m
(
1
2
)
= 0, if j is even.
Applying the above, we can study the number of zeros of the polynomials P(j)n,m(x) in the interval [0,1] for
j = 1, 2, . . . , m + 1. In the following table we use only the Rolle’s theorem.
zeros of P′n,m(x) u u
0 112
zeros of P′′n,m(x) uu u
0 112
zeros of P′′′n,m(x) u uu u
α1 α20 112
zeros of P(4)n,m(x) uu uu u
β1 β20 112
zeros of P(5)n,m(x) u uu u u u
γ1 γ2 γ3 γ40 112
zeros of P(6)n,m(x) uuu uu
δ2δ1 δ4δ30 112
Symmetry 2019, 11, 739 6 of 8
zeros of P(7)n,m(x) u uuu u u
ε1 ε2 ε3 ε40 112
...
P(m−1)n,m (x) uuu uu
ζ2ζ1 ζ4ζ30 112
P(m)n,m (x) uu u u
η1 η2 η3 η40 112
P(m+1)n,m (x) uu u
θ1 θ20 112
But
P(m+1)n,m (x) =
n!
(n−m− 1)! En−m−1(x)
whose the only zero in the interval [0, 1] is 1/2. This contradiction gives that a2 6= 0.
If c2 = 0 then from G(x) = ctxt + · · ·+ c3x3 and (7) one can deduce that
P(j)n,m(0) for j = 1, 2, 3, 4, 5.
The above argument that we used in the case a2 = 0 shows that this impossible.
Finally, consider the case when s = 1. Let c = A/B, where A and B 6= 0 are relatively prime integers.
From (4) we know that
g(x)− g(−x) = 2ctxt + 2ct−2xt−2 + · · ·+ 2c3x3 + 2c1x = dxh(x), (8)
where the polynomial h(x) is even and divides the polynomial Bxn−m + A in Q[x]. If we write h(x) as a
product of a rational number a/b and a primitive polynomial H(x) = hrxr + hr−2xr−2 + · · ·+ h2x2 + h0
we have that
Bxn−m + A = H(x)u(x), (9)
where u(x) = uqxq + uq−2xq−2 + · · ·+ u2x2 + u0 is a primitive polynomial. We obtain from (8) and (9) that
(2ctxt + 2ct−2xt−2 + · · ·+ 2c3x3 + 2c1x)u(x) = ab d(Bx
n−m+1 + Ax). (10)
Let ct = wc′t, ct−2 = wc′t−2, . . . , c3 = wc′3 and c1 = wc′1, where w denotes the greatest common divisor of
the integers ct, ct−2, . . . , c3, c1. Then
2w(c′txt + c′t−2xt−2 + · · ·+ c′3x3 + c′1x)u(x) =
a
b
d(Bxn−m+1 + Ax), (11)
which yields that 2w = (a/b)d and
(c′txt + c′t−2xt−2 + · · ·+ c′1x)(uqxq + uq−2xq−2 + · · ·+ u0) = Bxn−m+1 + Ax. (12)
Symmetry 2019, 11, 739 7 of 8
It follows from Lemma 6 that if p is an odd prime which divides w then p divides ct, ct−1, . . . , c2, c1, c0
which is not possible since g(x) is a primitive polynomial. Thus w = 2a for some non-negative integer a.
Now assume that p is a prime which divides c′t and j ≥ 1 is the greatest odd index for which
p|c′t, c′t−2, . . . , c′j+2 and p - c′j. (13)
On the right hand side of (12) the coefficient of xα equals 0 apart from when α = q + t = n−m + 1 or
α = 1. Thus
c′juq + c
′
j+2uq−2 + c
′
j+4uq−4 + · · · = 0
which means that p|uq.
Similarly,
c′j−2uq + c
′
juq−2 + c
′
j+2uq−4 + · · · = 0
from which we get that p|uq−2. Continuing the process one can deduce that
p|uq, uq−2, . . . , u2.
Further, if j > 1 then
c′ju0 + c
′
j−2u2 + . . . = 0
and so p|u0 contradicting that the polynomial u(x) is a primitive polynomial. It follows from the above
that j must be 1 and so
p|c′t, c′t−2, . . . , c′3, uq, uq−2, . . . , u2 and p - c′1, u0. (14)
If p is an odd prime then from the above and Lemma 6 we have that
p|ct, ct−1, ct−2 . . . , c3, c2 and p - c1, c0. (15)
Now let U(x) = ctxt + ct−1xt−1 + · · ·+ c2x2 and V(x) = c1x + c0. Then g(x) = U(x) +V(x) and for
j = 0, 1, . . . , k
g(x)j =
j
∑
i=0
(
j
i
)
U(x)j−iV(x)i ≡ V(x)j mod (p). (16)
We know that m > n/3 ≥ k and so the coefficients of xk, xk−2, . . . , x3, x are zeros in Pn,m(x) and so in
f (g(x)) = bkg(x)k + bk−2g(x)k−2 + · · ·+ b3g(x)3 + b1g(x), too. (17)
Now one can infer from (16) and (17) that 0 ≡ bkck1 mod (p) which yields p|bk. Comparing coefficient of
xk−2 we have that 0 ≡ bk−2ck−21 mod (p) from which we obtain p|bk−2. Continuing the process it is easy to
see that p|bk−4, . . . , b3, b1 which contradicts the fact that f (x) is a primitive polynomial. This means that c′t
and ct must be powers of two.
Now suppose that p is a prime with p|uq and p - c′t. Using again that on the right hand side of (12)
the coefficient of xα equals 0 apart from when α = n−m + 1 or 1. From uqc′t−2 + uq−2c′t = 0 we obtain
p divides uq−2. From uqc′t−4 + uq−2c
′
t−2 + uq−4c′t = 0 we obtain p divides uq−4. It follows similarly that
p|uq−6, . . . , u2. Finally, from c′tu0 + c′t−2u2 + · · · = 0 we get that p divides u0 which contradicts that the
polynomial u(x) is a primitive polynomial. This means that uq must be a power of two. Since B = c′tuq
this contradicts to our assumption that B is not a power of two.
4. Concluding Remarks
It is a very hard problem to characterize the general decomposition of an infinite sequence of
polynomials fn(x). The first theorem was proved for Bernoulli polynomials. For other results see our
Symmetry 2019, 11, 739 8 of 8
Introduction. A harder question is to describe the decomposition of the sum of two polynomials. There are
only a few results in this direction, mainly for the rational linear combination of two Bernoulli and Euler
polynomials in the form Bn(x) + cBn−2 and En(x) + cEn−2(x), respectively. This paper contains the first
theorem concerning the decomposition of the linear combination of two Euler polynomials Emx + cEn(x)
with “almost” independent parameters m and n.
Author Contributions: All the authors contributed equally to the conception of the idea, implementing and analyzing
the experimental results, and writing the manuscript.
Funding: This research Supported in part by the Hungarian Academy of Siences, and NKFIH/OTKA grant K128088.
Conflicts of Interest: The authors declare no conflict of interest.
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On the Decomposability of the Linear Combinations of Euler Polynomials with Odd Degrees

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