<s>[INST] Find the polynomials which satisfy the condition $f(x)\mid f(x^2)$ 
I want find the polynomials which satisfy the condition
  $$f(x)\mid f(x^2).$$ 

I want to find such polynomials with integer coefficients, real number coefficients and complex number coefficients.
For example, $x$ and $x-1$ are the linear polynomials which satisfy this condition. 

Here is one way to find the $2$-degree polynomials with integer coefficients.
  Let the quadratic be $p=ax^2+bx+c$, so its value at $x^2$ is $q=ax^4+bx^2+c$. If $p$ is to be a divisor of $q$ let the other factor be $dx^2+ex+f.$ Equating coefficients gives equations
[1]  $ad=a,$
[2] $ae+bd=0,$
[3] $af+be+cd=b,$
[4] $bf+ce=0,$
[5] $cf=c.$
Now we know $a,c$ are nonzero (else $p$ is not quadratic, or is reducible). So from [1] and [5] we have $d=f=1.$ Then from [2] and [4] we obtain $ae=ce.$ Here $e=0$ leads to $b=0$ from either [2] or [4], and [3] then reads $a+c=0$, so that $p=a(x^2-1)$ which is reducible. So we may assume $e$ is nonzero, and also $a=c.$
At this point, [2] and [4] say the same thing, namely $ae+b=0.$ So we may replace $b=-ae$ in [3] (with its $c$ replaced by $a$) obtaining
  $a+(-ae)e+a=-ae,$ which on factoring gives $a(2-e)(e+1)=0.$ The possibility $e=2$ then leads after some algebra to $2a+b=0$ and $p=a(x-1)^2$ which is reducible, while the possibility $e=-1$ leads to $a=b$ and then $p=ax^2+ax+a$ as claimed.

Should we list out all the irreducible degree polynomials and then check if these polynomials satisfy the condition

  
*
  
*$x$
  
*$x+1$
  
*$x^2 + x + 1$
  
*$x^3 + x^2 + 1$
  
*$x^3 + x + 1$
  
*$ x^4 + x^3 + x^2 + x + 1 $
  
*$ x^4 + x^3 + 1 $
  
*$ x^4 + x + 1 $
  

With the real number coefficients which can be factored into 

$$(x-c_1)(x-c_2)\cdots(x^2-2a_1x-(a_1^2+b_1^2))(x^2-2a_2x-(a_2^2+b_2^2))\cdots$$ If all of these linear terms and quadratic terms satisfy $$f(x)\mid f(x^2),$$ this polynomial satisfy too? So what's pattern in the real number polynomials?

 [/INST] 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. 
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