It isn't easy for f(x) to divide f(2x^3+x)

<p> To Anubhav Singh: I think it is not enough! There could be two polynomials $p,q$ s.t. $p(x) \nmid p(2x^3+x), q(x) \nmid q(2x^3+x)$ , but $p(x) \cdot q(x) \mid p(2x^3+x) \cdot q(2x^3+x)$ , or am I wrong? </p>

<p> Factorize $p(x)$ in $\mathbb{R}[x]$ : it has some factors of the form $(x-s_n)$ and others of the form $(x^2+a_nx + b_n)$ . If $p(x) \mid p(2x^3+x)$ , for every solution $s_n$ it must exists some $s_m$ such that $s_m^3+s_m-s_n = 0$ (*). </p>

<p><p>Let $f(x) : \mathbb{R} \rightarrow \mathbb{R}$ such that $f(s_m) = s_n$ iff $s_m^3+s_m-s_n = 0$ (that is $f(x) = 2x^3+x$ ). We stated that every $s_i$ has an image: for pidgeonhole there exists $n,k$ s.t. $f^k(s_n)=s_n$ . But since $f(x) \begin{cases} >x, & \mbox{ iff } x>0 \\ =x, &\mbox{ iff } x=0 \\ < x, & \mbox{ iff } x <0 \end{cases}$ it is $s_n=0$ . Now we want to show by contradiction that all the solutions equal to 0. Suppose there are some nonzero solutions: one of them must be the counterimage of 0, because there cannot be "cycles" among nonzero solutions. So it satisfies $2s^3+s=0$ which has just 0 as real solution, absurd. On balance, $p(x)$ has just 0 as real solution.</p></p>

<p>First, we notice that $x^2+ax+b \mid (2x^3+x)^2+a(2x^3+x)+b$ iff $a=0,b=1$ (by doing the division). Now we have to get into $\mathbb{C}[x]$ to deal with the problem I stated at the beginning: we want to show that there are no quadratic polynomials $q_1, \ldots, q_m$ s.t. $q_i(x) \nmid q_i(2x^3+x)$ but $q_1(x) \cdot \ldots \cdot q_m(x) \mid q_1(2x^3+x) \cdot \ldots \cdot q_m(2x^3+x)$ . I got hard difficulties here, but I eventually came up with a (very ugly) conclusion. With some calculation, one can notices that $|f(a+bi)|^2 = [4a^4+a^2(8b^2+4) +(1-2b)^2 ] \cdot |a+bi|^2$ . Assume that the ugly factor is $> 1$ : there must be by pidgeonhole some $s, k$ s.t. $f^k(s) = s$ , but this is a contradiction because $|s|^2 = |f^k(s)|^2 > |s|^2$ . If $|a| \ge 1$ , the factor $\ge 4$ , thus $a=0$ (because it is an integer). Then $(1-2b)^2 = 1$ yields $b=0,1$ ; $b=0$ gives the zero real solution, $b=1$ gives $x=i$ , which has to appear with its conjugate $-i$ in factorization of $p(x)$ . So the polynomial has the form $x^{2n} (x^2+1)^{500-n}$ , with $n$ varying from $0$ to $500$ .</p>

<p>(*) It is not possible that $(x-s_n)$ divides a quadratic factor because it has no real solutions. </p>

</p>Some (rough) generalizations.</p> <p>1. We can state this general fact looking at the first part.Let $p,q$ be two polynomials in $\mathbb{R}[x]$ , and $S_p, S_q, S_q^*$ respectively the set of real solutions of $p(x), q(x), q(x)-s$ with $s \in S_q$ . Thus, If $p(x) \mid p(q(x))$ , and $S_q^* \subseteq S_q$ then $S_p \subseteq S_q$ . For example, one can show in this way that if $p(x) \mid p( 1+q^2(x) )$ then $p(x)$ has no real solution, or that if $p(x) \mid p( (x^{2n+1}+1)^2 )$ then $p(x)$ has only the $-1$ real solution, and so on. </p> <p>2. Even the second part offers a generalization. Let $p,q$ be two polynomials in $\mathbb{Z}[x]$ such that $p(x) \mid p(x\cdot q(x))$ . We notice that $| xq(x) |^2 = |x|^2 |q(x)|^2$ . Recall that if $x$ is a solution of both $p(x \cdot q(x) )$ and $p(x)$ then $|q(x)|^2 \le 1$ , because of the considerations we did before. Furthermore $|q(x)|^2 =c^2+d^2 \ge 0$ for some $c,d \in \mathbb{Z}$ . Thus, the only solutions of $p(x)$ are the solutions in $\mathbb{Z}[i]$ of $\begin{cases}\Im[q(a+bi)] & = & 0, \\ \Re[q(a+bi)] & = & 0 \end{cases}$ or $\begin{cases}\Im[q(a+bi)] & = & 1, \\ \Re[q(a+bi)] & = & 0 \end{cases}$ or $\begin{cases}\Im[q(a+bi)] & = & 0, \\ \Re[q(a+bi)] & = & 1 \end{cases}$