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> I want find the polynomials which satisfy the condition
>  $$f(x)\mid f(x^2).$$ 
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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. 

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> 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
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> [1] $ad=a,$
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> [2] $ae+bd=0,$
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> [3] $af+be+cd=b,$
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> [4] $bf+ce=0,$
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> [5] $cf=c.$
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> 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 $a=c.$
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> 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.
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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 $
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With **the real number coefficients** which can be factored into 

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> $$(x-c\_1)(x-c\_2)\cdots(x^2-2a\_1x-(a\_1^2+b\_1^2))(x^2-2a\_2x-(a\_2^2+b\_2^2))\cdots$$ 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?
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