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?
Best Answer
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.