# $Q(x)$ is a monic polynomial with integer coefficients of degree $12$ ,there exists $P(x)$ satisfying $Q(x)P(x)=Q(x^2)$

$Q(x)$ is a monic polynomial with integer coefficients of degree $12$ ,such that there exists an integer polynomial $P(x)$ satisfying $Q(x)P(x)=Q(x^2)$ Find the number of all such polynomials $Q(x)$.

The idea which came to my mind was considering $Q(x)=x^{12}+a_1 x^{11}+...a_0$ We get $Q(x^2)$ and then use synthetic division to find $P(x)$ . Now the only restriction would be that the coefficients of $P(x)$ should be integers.

But this method is quite tedious and I have not yet been able to find $Q(x)$ .Please help.

• Since $Q(x)$ is monic, the leading terms of $Q(x)$ and $Q(x^2)$ are $x^{12}$ and $x^{24}$ respectively. So at the very least, we can say that $P(x)$ must have leading term $x^{12}$ i.e. $P(x)$ is itself a monic polynomial of degree 12. – Semiclassical Nov 15 '16 at 16:32
• Yes indeed we can say that – Sukanta Ghosh Nov 15 '16 at 16:39

## 1 Answer

Note that if $r$ is a root of $Q(x)$ or $P(x)$, it must also be a root of $Q(x^2)$, which says that $r^2$ is also a root of $Q(x)$. Moreover, of the $24$ square roots of the $12$ roots of $Q(x)$ (by multiplicity), half are the roots of $Q(x)$ and half are the roots of $P(x)$. This cuts down the possibilities considerably.

EDIT:

Consider a directed graph with vertices the $24$ roots (counted by multiplicity) of $P$ and $Q$; if root $r$ is not $0$ or $1$, so that $r^2 \ne r$, an arc goes from $r$ to (one of the instances of) $r^2$. $12$ of the vertices which have in-degree $0$ are coloured red (corresponding to roots of $P$), the rest are blue.

The only possible terminal vertices have values $0$ and $1$. $0$ is isolated, while $1$ has in-degree $1$, corresponding to $-1^2 = 1$. All other vertices have in-degree $2$, corresponding to their two square roots.

A blue vertex could be on a $k$-cycle, corresponding to a root $r \ne 0,1$ with $r^{2^k} = r$ (and $r^{2^j} \ne r$ for $0 < j < k$). The minimal polynomial of such a root must be a cyclotomic polynomial $\Phi_m(x)$ where $m$ divides $2^k-1$ (and not $2^j-1$ for $0 < j < k$), and this cyclotomic polynomial must divide $Q(x)$. Its degree is $\varphi(m)$ where $\varphi$ is the Euler totient function.

Roots $r$ that are not on a cycle but such that $r^2$ is in a cycle that corresponds to polynomial $\Phi_m(x)$ are roots of $\Phi_m(x^2)/\Phi_m(x)$. Roots $r$ that are $k$ steps removed from a cycle are roots of $\Phi_m(x^{2^k})/\Phi_m(x^{2^{k-1}})$. I'm not sure that these are necessarily irreducible, so you might have only some of the factors in $Q(x)$ and the others in $P(x)$.

The odd integers $m$ with $\varphi(m) \le 12$ are $1, 3, 5, 7, 9, 11, 13, 15, 21$, which have $\varphi(m) = 1, 2, 4, 6, 6, 10, 12, 8, 12$. It should be possible (but a bit tedious) to work out all possibilities.

• I am not getting you , my bad – Sukanta Ghosh Nov 15 '16 at 17:02
• See also this answer, – Bill Dubuque Nov 15 '16 at 17:43
• So, can the number of polynomials $Q(x)$ be guessed? – Sukanta Ghosh Nov 15 '16 at 18:31