**1. Introduction**

Let K be a field. If *f*(*x*) ∈ <sup>K</sup>[*x*] has degree at least 2, we say that *f*(*x*) is *decomposable* over the field K if we can write *f*(*x*) = *f*1(*f*2(*x*)) for some nonlinear polynomial *f*1(*x*), *f*2(*x*) ∈ <sup>K</sup>[*x*]. Otherwise, we say that *f*(*x*) is *indecomposable* over K. Two decompositions *f*(*x*) = *f*1(*f*2(*x*)) and *f*(*x*) = *<sup>F</sup>*1(*<sup>F</sup>*2(*x*)) are said to be *equivalent* over the field K, written *f*1 ◦ *f*2 ∼K *F*1 ◦ *F*2, if there exists a linear polynomial *l*(*x*) ∈ <sup>K</sup>[*x*] such that

$$f\_1(\mathbf{x}) = F\_1(l(\mathbf{x})) \text{ and } \quad F\_2(\mathbf{x}) = l(f\_2(\mathbf{x})).$$

For a given *f*(*x*) ∈ <sup>K</sup>[*x*] with degree at least 2, a *complete decomposition* of *f*(*x*) over K is a decomposition *f* = *f*1 ◦···◦ *fk*, where the polynomials *fi* ∈ <sup>K</sup>[*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

$$\sum\_{k=0}^{\infty} E\_k(x) \frac{t^k}{k!} = \frac{2e^{tx}}{(e^t + 1)}.$$

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 polynomials with degree even, moreover they showed that every Euler polynomial with odd degree is

*Symmetry* **2019**, *11*, 739; doi:10.3390/sym11060739

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

$$R\_n(\mathbf{x}) = B\_n(\mathbf{x}) + cB\_{n-2}(\mathbf{x})$$

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*−<sup>2</sup>(*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*,*<sup>m</sup>*(*x*) = *En*(*x*) + *cEm*(*x*)*, where c* = *A*/*B is an arbitrary rational number, where B* = 2*a, a* ∈ N ∪ {0}*, n, m are odd integers with n* > *m* > *<sup>n</sup>*/3*. Then the polynomials Pn*,*<sup>m</sup>*(*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*)=(−<sup>1</sup>)*nEn*(<sup>1</sup> − *x*)*; (b) En*(*x* + 1) + *En*(*x*) = 2*xn; (c) <sup>E</sup>n*(*x*) = *nEn*−<sup>1</sup>(*x*)*; (d) <sup>E</sup>*2*n*−<sup>1</sup>(1/2) = *<sup>E</sup>*2*n*(0) = *<sup>E</sup>*2*n*(1) = 0 *for n (e) En*(*x*) = ∑*nk*=<sup>0</sup> (*nk*)*Ek*(0)*xn*−*k;* **Proof.**See

 [2].

> The following result is a general theorem from the theory of decomposability.

∈

N*;*

**Lemma 2** (Kreso and Rakaczki [14])**.** *Let <sup>F</sup>*(*x*) ∈ <sup>K</sup>[*x*] *be a monic polynomial such that* deg *F is not divisible by the characteristic of the field* K*. Then for every nontrivial decomposition F* = *F*1 ◦ *F*2 *over any field extension* L *of* K*, there exists a decomposition F* = *F* ˜ 1 ◦ *F* ˜ 2 *such that the following conditions are satisfied*


*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* = 0 *are relative prime integers, f*(*x*) *and g*(*x*) ∈ <sup>Z</sup>[*x*] *are primitive polynomials. Moreover, if h*(*x*) *is a monic polynomial, then u* = 1*.*

**Proof.** Suppose that *h*(*x*) = *<sup>F</sup>*(*G*(*x*)), where *<sup>F</sup>*(*x*), *<sup>G</sup>*(*x*) ∈ Q[*x*]. Let

$$\begin{aligned} F(\mathbf{x}) &= b\_k \mathbf{x}^k + b\_{k-1} \mathbf{x}^{k-1} + \dots + b\_1 \mathbf{x} + b\_{0\prime}, \\ G(\mathbf{x}) &= c\_l \mathbf{x}^l + c\_{l-1} \mathbf{x}^{l-1} + \dots + c\_1 \mathbf{x} + c\_0. \end{aligned}$$

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

$$G(\mathbf{x}) = \frac{c}{d} \operatorname{g}(\mathbf{x}), \text{ where } \operatorname{g}(\mathbf{x}) \text{ is a primitive polynomial, } c, d \neq 0 \in \mathbb{Z}$$

and so

$$F(G(\mathbf{x})) = b\_k \left(\frac{c}{d}\right)^k g(\mathbf{x})^k + b\_{k-1} \left(\frac{c}{d}\right)^{k-1} g(\mathbf{x})^{k-1} + \dots + b\_1 \frac{c}{d} g(\mathbf{x}) + b\mathbf{u}.$$

The polynomial

$$F\_1(\mathbf{x}) = b\_k \left(\frac{\mathbf{c}}{d}\right)^k \mathbf{x}^k + b\_{k-1} \left(\frac{\mathbf{c}}{d}\right)^{k-1} \mathbf{x}^{k-1} + \dots + b\_1 \frac{\mathbf{c}}{d} \mathbf{x} + b\_0 \in \mathbb{Q}[\mathbf{x}]^k$$

can be written in the from *uv f*(*x*), where *f*(*x*) ∈ <sup>Z</sup>[*x*] is a primitive polynomial, *u* > 0, *v* = 0 are relative prime integers. However, then we have

$$h(\mathbf{x}) = F(\mathbf{G}(\mathbf{x})) = F\_1(\mathbf{g}(\mathbf{x})) = \frac{\mu}{\upsilon} f(\mathbf{g}(\mathbf{x})).\tag{1}$$

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

$$S^{+} = \{ f(\mathbf{x}) \in \mathbb{C}[\mathbf{x}] \mid \quad f(\mathbf{x}) = f(1-\mathbf{x}) \},$$

and

$$S^{--} = \{ f(\mathbf{x}) \in \mathbb{C}[\mathbf{x}] \mid \quad f(\mathbf{x}) = -f(1-\mathbf{x}) \} \dots$$

From these definitions it is easy to see that *S*<sup>+</sup> and *S*− are subspaces in the vector space C[*x*].

**Lemma 4.** *Let <sup>P</sup>*(*x*) ∈ Q[*x*] *be a monic polynomial. Assume that <sup>P</sup>*(*x*) ∈ *S*− *and <sup>P</sup>*(*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*) = <sup>−</sup>*f*(−*<sup>x</sup>*)*.*

### **Proof.** See [16].

The following Lemma is a simple combination of Lemmas 3 and 4.

**Lemma 5.** *Let <sup>P</sup>*(*x*) ∈ Q[*x*] *be a monic polynomial. Assume that <sup>P</sup>*(*x*) ∈ *S*− *and <sup>P</sup>*(*x*) = *<sup>F</sup>*(*G*(*x*))*, where <sup>F</sup>*(*x*)*, <sup>G</sup>*(*x*) ∈ Q[*x*] *and* deg(*F*(*x*)) > 1*,* deg(*G*(*x*)) > 1*. Then we can assume that <sup>P</sup>*(*x*) = 1*v f*(*g*(*x*))*, where v* = 0 *is an integer, f*(*x*) *and g*(*x*) *are primitive polynomials, g*(*x*) ∈ *S*− *and f*(*x*) = <sup>−</sup>*f*(−*<sup>x</sup>*)*.*

**Proof.** From Lemma 4 we can assume that *<sup>G</sup>*(*x*) ∈ *S*− and *<sup>F</sup>*(*x*) = <sup>−</sup>*<sup>F</sup>*(−*<sup>x</sup>*). Using the proof of Lemma 3 and the fact that *S*− is a subspace of C[*x*] we ge<sup>t</sup> the assertion of our Lemma.

**Lemma 6.** *Let g*(*x*) = *ctx<sup>t</sup>* + *ct*−1*xt*−<sup>1</sup> + ··· + *c*1*x* + *c*0 ∈ *S*<sup>−</sup>*. Then*

$$\mathbf{c} - 2\mathbf{c}\_s = \binom{s+1}{s}\mathbf{c}\_{s+1} + \binom{s+2}{s}\mathbf{c}\_{s+2} + \dots + \binom{t-1}{s}\mathbf{c}\_{t-1} + \binom{t}{s}\mathbf{c}\_t \tag{2}$$

*for even index* 0 ≤ *s* ≤ *t* − 1*.*

**Proof.** Since *g*(*x*) ∈ *S*− we have that −*g*(*x*) = *g*(<sup>1</sup> − *<sup>x</sup>*). Computing the coefficient of *x<sup>s</sup>* on the both sides we obtain that

$$-\mathfrak{c}\_{\mathfrak{s}} = (-1)^{\mathfrak{s}} \left( \mathfrak{c}\_{\mathfrak{s}} + \binom{\mathfrak{s}+1}{\mathfrak{s}} \mathfrak{c}\_{\mathfrak{s}+1} + \binom{\mathfrak{s}+2}{\mathfrak{s}} \mathfrak{c}\_{\mathfrak{s}+2} + \dots + \binom{t-1}{\mathfrak{s}} \mathfrak{c}\_{t-1} + \binom{t}{\mathfrak{s}} \mathfrak{c}\_{t} \right) .$$

**Lemma 7.** *Let*

$$f(\mathbf{x}) = b\_k \mathbf{x}^k + b\_{k-2} \mathbf{x}^{k-2} + b\_{k-3} \mathbf{x}^{k-3} + \dots + b\_1 \mathbf{x} + b\_0 \mathbf{x}$$

$$g(\mathbf{x}) = c\_t \mathbf{x}^t + c\_{t-1} \mathbf{x}^{t-1} + \dots + c\_1 \mathbf{x} + c\_0 \in \mathbb{Q}[\mathbf{x}].$$

*If k, t* ≥ 2 *then the coefficient of the monomial xkt*−<sup>2</sup> *in the polynomial f*(*g*(*x*)) *is*

$$b\_k \left( kc\_t^{k-1} c\_{t-2} + \binom{k}{2} c\_t^{k-2} c\_{t-1}^2 \right).$$

**Proof.** It is easy to see that the monomial *xkt*−<sup>2</sup> occurs only in the term *bkg*(*x*)*<sup>k</sup>*. Expanding *g*(*x*)*<sup>k</sup>* we simply ge<sup>t</sup> 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*,*<sup>m</sup>*(*x*) is decomposable over C. From Lemmas 2 and 5 we can assume that *Pn*,*<sup>m</sup>*(*x*) = 1*v f*(*g*(*x*)), where *v* = 0 is an integer, *f*(*x*), *g*(*x*) ∈ <sup>Z</sup>[*x*] are primitive polynomials and *g*(*x*) ∈ *S*<sup>−</sup>, *f*(*x*) = <sup>−</sup>*f*(−*<sup>x</sup>*). Let

$$\begin{aligned} f(\mathbf{x}) &= b\_k \mathbf{x}^k + b\_{k-2} \mathbf{x}^{k-2} + b\_{k-4} \mathbf{x}^{k-4} + \dots + b\_3 \mathbf{x}^3 + b\_1 \mathbf{x}, \\ g(\mathbf{x}) &= c\_l \mathbf{x}^l + c\_{l-1} \mathbf{x}^{l-1} + \dots + c\_1 \mathbf{x} + c\_0. \end{aligned}$$

Using (b) of Lemma 1 one can deduce that

$$\frac{1}{\upsilon}f(\operatorname{g}(\operatorname{x}+1)) + \frac{1}{\upsilon}f(\operatorname{g}(\operatorname{x})) = P\_{\operatorname{h},\operatorname{\mathfrak{W}}}(\operatorname{x}+1) + P\_{\operatorname{h},\operatorname{\mathfrak{W}}}(\operatorname{x}) = 2\operatorname{x}^{\mathfrak{n}} + 2c\operatorname{x}^{\mathfrak{m}}.\tag{3}$$

Since *Pn*,*<sup>m</sup>*(*x*) ∈ *S*− thus *Pn*,*<sup>m</sup>*(*<sup>x</sup>* + 1) = <sup>−</sup>*Pn*,*<sup>m</sup>*(−*<sup>x</sup>*). From (3) we infer that the polynomial *g*(*x*) − *g*(−*<sup>x</sup>*) divides the polynomial *Pn*,*<sup>m</sup>*(*x*) − *Pn*,*<sup>m</sup>*(−*<sup>x</sup>*) = 2*x<sup>n</sup>* + 2*cxm*, that is

$$
\log(\mathbf{x}) - \mathbf{g}(-\mathbf{x}) = d\mathbf{x}^{\delta} h(\mathbf{x}),
\tag{4}
$$

where *d* ∈ Q, 0 ≤ *s* ≤ *m* and the polynomial *h*(*x*) divides the polynomial *xn*−*<sup>m</sup>* + *c* in Q[*x*]. We know that

$$\log(\mathbf{x}) - \mathbf{g}(-\mathbf{x}) = 2c\mathbf{x}^t + 2c\mathbf{x}^{t-2}\mathbf{x}^{t-2} + \dots + 2c\mathbf{x}^3 + 2c\mathbf{x}.\tag{5}$$

If the polynomial *h*(*x*) is a constant polynomial then we have *t* = *s* and so *ct*−2 = 0. It follows from *Pn*,*<sup>m</sup>*(*x*) = *En*(*x*) + *cEm*(*x*) and (d), (e) of Lemma 1 that the coefficient of *xn*−<sup>2</sup> in *Pn*,*<sup>m</sup>*(*x*) equals 0. Applying now Lemma 7 we ge<sup>t</sup> that

$$b\_k \binom{k}{2} c\_t^{k-2} c\_{t-1}^2 = 0\_r$$

which is impossible since <sup>−</sup>2*ct*−<sup>1</sup> = *tct* by Lemma 6.

In the case when *h*(*x*) = *xn*−*<sup>m</sup>* + *c* we ge<sup>t</sup> *s* + *n* − *m* = *t*, *d* = 2*ct* and *g*(*x*) − *g*(−*<sup>x</sup>*) = 2*ctx<sup>s</sup>*+*n*−*<sup>m</sup>* + 2*ctcx<sup>s</sup>* = <sup>2</sup>*ctx<sup>t</sup>* + <sup>2</sup>*ctcxt*−(*<sup>n</sup>*−*<sup>m</sup>*). 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*(−*<sup>x</sup>*). Consider first when **1** < **s**. Then *c*1 = *c*3 = ··· = *cs*−2 = 0 and *cs* = 0. Let *<sup>G</sup>*(*x*) = *g*(*x*) − *g*(0) and *<sup>F</sup>*(*x*) = *f*(*x* + *g*(0)). Then *f*(*g*(*x*)) = *<sup>F</sup>*(*G*(*x*)), *G*(0) = 0 and

$$G(\mathbf{x}) - G(-\mathbf{x}) = \mathbf{g}(\mathbf{x}) - \mathbf{g}(-\mathbf{x}) = 2c\_1 \mathbf{x}^t + 2c\_{t-2} \mathbf{x}^{t-2} + \dots + 2c\_s \mathbf{x}^s.$$

Let

$$F(\mathbf{x}) = a\_k \mathbf{x}^k + a\_{k-1} \mathbf{x}^{k-1} + \dots + a\_2 \mathbf{x}^2 + a\_1 \mathbf{x} + a\_0 \dots$$

Since *s* < *t* ≤ *n*/3 < *m* we have that *s* + 4 ≤ *m*.

Investigate the coefficients of *x<sup>s</sup>* and *xs*+<sup>2</sup> in

$$
\omega P\_{n,m}(\mathbf{x}) = F(G(\mathbf{x})) = a\_k G(\mathbf{x})^k + a\_{k-1} G(\mathbf{x})^{k-1} + \dots + a\_1 G(\mathbf{x}) + a\_0. \tag{6}
$$

Since *s* + 2 < *m* in the polynomials *vPn*.*<sup>m</sup>*(*x*) = *En*(*x*) + *cEm*(*x*) these coefficients are 0. On the other hand, one can observe that *x<sup>s</sup>* occurs only in the term *<sup>a</sup>*1*<sup>G</sup>*(*x*) and so *a*1*cs* = 0. This means that *a*1 = 0 and so

$$wP\_{n,m}(\mathbf{x}) = F(G(\mathbf{x})) = a\_k G(\mathbf{x})^k + \dots + a\_3 G(\mathbf{x})^3 + a\_2 G(\mathbf{x})^2 + a\_0. \tag{7}$$

Since *xs*+<sup>2</sup> appears only in the term *<sup>a</sup>*2*<sup>G</sup>*(*x*)<sup>2</sup> thus 2*a*2*c*2*cs* = 0.

If **a2** = **0** we obtain from (7) that the coefficients of *x*5, *x*4, *x*3, *x*2 and *x* in *<sup>F</sup>*(*G*(*x*)) are zero. This yields that *P*(*i*) *<sup>n</sup>*,*<sup>m</sup>*(0) = 0 for *i* = 1, . . . , 5. Further, by Lemma 1

$$P\_{\mathfrak{n},\mathfrak{m}}^{(j)}(0) = P\_{\mathfrak{n},\mathfrak{m}}^{(j)}(1) = 0,\text{ if } j \text{ is odd and } j \neq \mathfrak{m},\mathfrak{n};$$

$$P\_{\mathfrak{n},\mathfrak{m}}^{(j)}\left(\frac{1}{2}\right) = 0,\text{ if } j \text{ is even.}$$

Applying the above, we can study the number of zeros of the polynomials *P*(*j*) *<sup>n</sup>*,*<sup>m</sup>*(*x*) in the interval [0,1] for *j* = 1, 2, . . . , *m* + 1. In the following table we use only the Rolle's theorem.

But

$$P\_{n,m}^{(m+1)}(\mathbf{x}) = \frac{n!}{(n-m-1)!} E\_{n-m-1}(\mathbf{x})$$

whose the only zero in the interval [0, 1] is 1/2. This contradiction gives that *a*2 = 0.

If **c2** = **0** then from *<sup>G</sup>*(*x*) = *ctx<sup>t</sup>* + ··· + *<sup>c</sup>*3*x*<sup>3</sup> and (7) one can deduce that

$$P\_{\mathfrak{n},\mathfrak{m}}^{(j)}(0)\text{ for }j=1,2,3,4,5.$$

The above argumen<sup>t</sup> that we used in the case *a*2 = 0 shows that this impossible.

Finally, consider the case when **s** = **1**. Let *c* = *A*/*B*, where *A* and *B* = 0 are relatively prime integers. From (4) we know that

$$\mathbf{g}(\mathbf{x}) - \mathbf{g}(-\mathbf{x}) = 2\mathbf{c}\_1 \mathbf{x}^t + 2\mathbf{c}\_{1-2} \mathbf{x}^{t-2} + \dots + 2\mathbf{c}\_3 \mathbf{x}^3 + 2\mathbf{c}\_1 \mathbf{x} = d\mathbf{x}h(\mathbf{x}),\tag{8}$$

where the polynomial *h*(*x*) is even and divides the polynomial *Bxn*−*<sup>m</sup>* + *A* in Q[*x*]. If we write *h*(*x*) as a product of a rational number *a*/*b* and a primitive polynomial *<sup>H</sup>*(*x*) = *hrxr* + *hr*−2*xr*−<sup>2</sup> + ··· + *<sup>h</sup>*2*x*<sup>2</sup> + *h*0 we have that

$$B\mathbf{x}^{n-m} + A = H(\mathbf{x})u(\mathbf{x}),\tag{9}$$

where *u*(*x*) = *uqx<sup>q</sup>* + *uq*−2*xq*−<sup>2</sup> + ··· + *<sup>u</sup>*2*x*<sup>2</sup> + *u*0 is a primitive polynomial. We obtain from (8) and (9) that

$$d(2c\_l \mathbf{x}^t + 2c\_{l-2} \mathbf{x}^{t-2} + \dots + 2c\_3 \mathbf{x}^3 + 2c\_1 \mathbf{x}) u(\mathbf{x}) = \frac{a}{b} d(B \mathbf{x}^{u-m+1} + A \mathbf{x}).\tag{10}$$

Let *ct* = *wc<sup>t</sup>*, *ct*−2 = *wc<sup>t</sup>*−2, ... , *c*3 = *wc*3 and *c*1 = *wc*1, where *w* denotes the greatest common divisor of the integers *ct*, *ct*−2,..., *c*3, *c*1. Then

$$2\operatorname{w}(c\_t'\mathbf{x}^t + c\_{t-2}'\mathbf{x}^{t-2} + \dots + c\_3'\mathbf{x}^3 + c\_1'\mathbf{x})\mathbf{u}(\mathbf{x}) = \frac{a}{b}d(\mathcal{B}\mathbf{x}^{n-m+1} + A\mathbf{x}),\tag{11}$$

which yields that 2*w* = (*a*/*b*)*d* and

$$(c\_t' \mathbf{x}^t + c\_{t-2}' \mathbf{x}^{t-2} + \dots + c\_1' \mathbf{x})(\mathbf{u}\_\emptyset \mathbf{x}^\emptyset + \mathbf{u}\_{q-2} \mathbf{x}^{q-2} + \dots + \mathbf{u}\_0) = B \mathbf{x}^{n-m+1} + A \mathbf{x}.\tag{12}$$

It follows from Lemma 6 that if *p* is an odd prime which divides *w* then *p* divides *ct*, *ct*−1, ... , *c*2, *c*1, *c*0 which is not possible since *g*(*x*) is a primitive polynomial. Thus *w* = 2*a* for some non-negative integer *a*. Now assume that *p* is a prime which divides *ct* and *j* ≥ 1 is the greatest odd index for which

$$p|c\_{t}',c\_{t-2}',\ldots,c\_{j+2}' \quad \text{and } p \nmid c\_{j}'.\tag{13}$$

On the right hand side of (12) the coefficient of *x<sup>α</sup>* equals 0 apart from when *α* = *q* + *t* = *n* − *m* + 1 or *α* = 1. Thus

$$c'\_j \mu\_q + c'\_{j+2} \mu\_{q-2} + c'\_{j+4} \mu\_{q-4} + \dots = 0$$

which means that *<sup>p</sup>*|*uq*.

> Similarly,

$$c'\_{j-2}u\_q + c'\_j u\_{q-2} + c'\_{j+2}u\_{q-4} + \dots = 0$$

from which we ge<sup>t</sup> that *<sup>p</sup>*|*uq*−2. Continuing the process one can deduce that

> *<sup>c</sup>j<sup>u</sup>*<sup>0</sup> + *<sup>c</sup><sup>j</sup>*−2*u*<sup>2</sup>

$$p \mid u\_{\emptyset}, u\_{\emptyset}, \dots, u\_{2}, \dots$$

Further, if *j* > 1 then

and so  $p|u\_0$  contradicting that the polynomial  $u(x)$  is a primitive polynomial. It follows from the above that  $j$  must be 1 and so

+ ... = 0

$$p\left|c\_{t}^{l},c\_{t-2}^{l},\ldots,c\_{3}^{l},u\_{q},u\_{q-2},\ldots,u\_{2}\quad\text{and }p\nmid c\_{1}^{l},u\_{0}.\tag{14}$$

If *p* is an odd prime then from the above and Lemma 6 we have that

$$p\left|c\_{1},c\_{1-1},c\_{1-2},\ldots,c\_{3},c\_{2}\text{ and }p\nmid c\_{1},c\_{0}.\right.\tag{15}$$

Now let *<sup>U</sup>*(*x*) = *ctx<sup>t</sup>* + *ct*−1*xt*−<sup>1</sup> + ··· + *<sup>c</sup>*2*x*<sup>2</sup> and *<sup>V</sup>*(*x*) = *c*1*x* + *c*0. Then *g*(*x*) = *<sup>U</sup>*(*x*) + *<sup>V</sup>*(*x*) and for *j* = 0, 1, . . . , *k j*

$$\log(\mathbf{x})^{\dot{j}} = \sum\_{i=0}^{l} \binom{\dot{j}}{\dot{i}} \mathcal{U}(\mathbf{x})^{\dot{j}-\dot{i}} \mathcal{V}(\mathbf{x})^{\dot{i}} \equiv \mathcal{V}(\mathbf{x})^{\dot{j}} \bmod (p). \tag{16}$$

We know that *m* > *n*/3 ≥ *k* and so the coefficients of *<sup>x</sup>k*, *xk*−2,..., *x*3, *x* are zeros in *Pn*,*<sup>m</sup>*(*x*) and so in

$$f(\mathbf{g}(\mathbf{x})) = b\_k \mathbf{g}(\mathbf{x})^k + b\_{k-2} \mathbf{g}(\mathbf{x})^{k-2} + \dots + b\_3 \mathbf{g}(\mathbf{x})^3 + b\_1 \mathbf{g}(\mathbf{x}), \text{ too.} \tag{17}$$

Now one can infer from (16) and (17) that 0 ≡ *bkc<sup>k</sup>*1 mod (*p*) which yields *p*|*bk*. Comparing coefficient of *xk*−<sup>2</sup> we have that 0 ≡ *bk*−2*ck*−<sup>2</sup> 1 mod (*p*) from which we obtain *p*|*bk*−2. Continuing the process it is easy to see that *p*|*bk*−4, ... , *b*3, *b*1 which contradicts the fact that *f*(*x*) is a primitive polynomial. This means that *ct* and *ct* must be powers of two.

Now suppose that *p* is a prime with *p*|*uq* and *p* - *ct*. Using again that on the right hand side of (12) the coefficient of *x<sup>α</sup>* equals 0 apart from when *α* = *n* − *m* + 1 or 1. From *uqc<sup>t</sup>*−<sup>2</sup> + *uq*−2*<sup>c</sup><sup>t</sup>* = 0 we obtain *p* divides *uq*−2. From *uqc<sup>t</sup>*−<sup>4</sup> + *uq*−2*<sup>c</sup><sup>t</sup>*−<sup>2</sup> + *uq*−4*<sup>c</sup><sup>t</sup>* = 0 we obtain *p* divides *uq*−4. It follows similarly that *<sup>p</sup>*|*uq*−6, ... , *u*2. Finally, from *<sup>c</sup>tu*<sup>0</sup> + *<sup>c</sup><sup>t</sup>*−2*u*<sup>2</sup> + ··· = 0 we ge<sup>t</sup> that *p* divides *u*0 which contradicts that the polynomial *u*(*x*) is a primitive polynomial. This means that *uq* must be a power of two. Since *B* = *<sup>c</sup>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

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*−<sup>2</sup> and *En*(*x*) + *cEn*−<sup>2</sup>(*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 gran<sup>t</sup> K128088. **Conflicts of Interest:** The authors declare no conflict of interest.
