When will $f(x)$ be a factor of $f(x^2)$? Given that $f(x)$ is a polynomial, when will it be a factor of $f(x^2)$?
I am thinking that the polynomial could be set as monic, since the multiplying coefficient would not affect the result.
For low degree like one or two, the results can be easily obtained by using undetermined coefficient and division. When it comes to higher degree, however, the approach seemed to be complicated. Are their any other ways to think about this problem? 
 A: Suppose $f(x)$ is a monic polynomial of degree $n$, so that $$f(x)=(x-r_1)\cdots(x-r_n),$$ where the $r_i$ are the roots of $f$. Then $$f(x^2)=(x^2-r_1)\cdots(x^2-r_n)=(x+\sqrt{r_1})(x-\sqrt{r_1})\cdots(x+\sqrt{r_1})(x-\sqrt{r_n}).$$ For $f(x)$ to be a factor of $f(x^2)$, every root of $f(x)$ must also be a root of $f(x^2)$. This means that for all $i$, there is a $j$ such that $$r_i=\pm\sqrt{r_j}\iff r_i^2=r_j.$$ 
So if $f(x)$ has the root $r$, it must also have the roots $r^2, r^4, r^8,\cdots.$ Since it is a polynomial, this list must be finite. This can only happen if the elements of the list eventually repeat, which requires $r^{2^n}=r^{2^m}$ for some $m<n$. This condition is equivalent to $$r^{2^n}-r^{2^m}=0\implies r^{2^m}\left(r^{2^{n-m}}-1\right)=0.$$ 
Thus we either have $r=0$, or since $j=n-m$ can be any integer, $r$ is a $2^{j}$th root of unity. The first case gives the family of polynomials $x^n$ for any positive integer $n$; the second case gives the family $$(x-z)(x-z^2)\cdots\left(x-z^{2^j}\right),$$ where $z$ is a $2^j$th root of unity. We can check that both of these families are solutions. 
