Find the polynomials which satisfy the condition $f(x)\mid f(x^2)$

Solution 1:

The polynomials with $f(x)\mid f(x^2)$ are closed under multiplication. In fact, if $f$ is any such polynomial and $g(x)\mid f(x^2)/f(x)$, then $f(x)g(x)$ is also such a polynomial.

WLOG assume $x\nmid f(x)$. The relation $f(x)\mid f(x^2)$ implies

$$ \{\alpha:f(\alpha)=0\}\subseteq \{\beta:f(\beta^2)=0\}=\{\pm\sqrt{\beta}:f(\beta)=0\}.$$

Let $\alpha$ be a zero. Then $\alpha=\pm\sqrt{\beta}$ for some other zero $\beta$, or equivalently $\alpha^2=\beta$. Put another way, the square of any zero is also a zero, so the set of zeros is closed under squaring. Therefore we have a sequence of zeros $\alpha,\alpha^2,\alpha^4,\cdots$ which must eventually terminate since $f$ has finitely many zeros, in which case $\alpha^{2^n}=\alpha^{2^m}$ eventually, so $\alpha^{2^r(2^s-1)}=1$ and thus $\alpha$ is a root of unity.

We can restrict our attention to $f$ that cannot be written as a nontrivial product of other polynomials with this property. I don't think there's a very nice characterization of the possible set of zeros of $f$ beyond "start with a root of unity and keep squaring until you get a repeat." For example, over $\mathbb{C}$ we have that $f(x)=(x-i)(x+1)(x-1)$ is such a polynomial; it includes a kind of "cycle" of length two $\{-1,1\}$ in its zero set, but it also has a kind of "hangnail" at the front, namely $i$. If we think about this in terms of integers mod $n$, we can write $n=2^km$ and use the Chinese Remainder Theorem to track what $x\mapsto 2x$ does to an integer mod $n$; the sequence is eventually periodic but at the beginning the $\mathbb{Z}/2^k\mathbb{Z}$ coordinate may be nonzero.

To get the $f$ with real coefficients, just make sure the set $\{\alpha,\alpha^2,\alpha^4\cdots\}$ is closed under conjugation; if it isn't, then adjoin all their conjugates to construct an $f$ with real coefficients.

And to get $f$ with integer coefficients, if $f$ has an $n$th root of unity as a zero then $f$ is divisible by the cyclotomic polynomial $\Phi_n(x)$. If $n$ is even, then squaring primitive $2n$th roots of unity yield primitive $n$th roots of unity, meaning both $\Phi_{n}(x)$ and $\Phi_{n/2}(x)$ are factors. Writing $n=2^km$, this means it is divisible by $\Phi_{2^km}(x)\Phi_{2^{k-1}m}(x)\cdots\Phi_m(x)$. One can check these polynomials satisfy the condition.

Solution 2:

I don't have "the answer" per se, but I think I can provide a new perspective on it.

Let's work over the complex numbers $\mathbb{C}$. That way, we can write every polynomial as the product of its roots, i.e. in the form $f(x) = (x - r_1)^{e_1}...(x - r_n)^{e_n}$ for distinct $r_1, ... , r_n \in \mathbb{C}$.

Let's first deal with the case of $n=1$, where again, $n$ is the number of distinct roots of $f$. Then $f(x^2) = (x^2 - r_1)^{e_1} = (x - \sqrt{r_1})^{e_1}(x + \sqrt{r_1})^{e_1}$. Therefore since $r_1$ has to be a root of $f(x^2)$ if $f(x) | f(x^2)$, then $r_1 = \pm \sqrt{r_1}$ so $r_1 = 0$ or $r_1 = 1$.

How about two roots? Let's call them $r$ and $s$. Then similarly either $r = \pm \sqrt{r}$ or $r = \pm \sqrt{s}$, and either $s = \pm \sqrt{r}$ or $s = \pm \sqrt{s}$. The only interesting case is if $r = \pm \sqrt{s}$ and $s = \pm \sqrt{r}$. But then $r^2 = s = \pm \sqrt{r}$ so $r^4 - r = 0$, so $r \in \{0, 1, \omega, \omega^2\}$, where $\omega$ is a primitive third root of unity. Similarly $s \in \{0, 1, \omega, \omega^2\}$. By only going after the "new" case, we see that the only choice is to have one of the roots be $\omega$ and the other $\omega^2$. So an obvious candidate is $(x - \omega)(x - \omega^2) = x^2 + x + 1$, which holds since $\omega^2 = -\omega$.

I think it's possible to continue this way, and in particular I think it wouldn't be so hard to show any nonzero solution has complex norm 1.