# 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

 $ad=a,$

 $ae+bd=0,$

 $af+be+cd=b,$

 $bf+ce=0,$

 $cf=c.$

Now we know $a,c$ are nonzero (else $p$ is not quadratic, or is reducible). So from  and  we have $d=f=1.$ Then from  and  we obtain $ae=ce.$ Here $e=0$ leads to $b=0$ from either  or , and  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,  and  say the same thing, namely $ae+b=0.$ So we may replace $b=-ae$ in  (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?

• I imagine you want to represent polynomials by their sets of roots rather than by their coefficients. Incidentally, if $f(x) | f(x^2)$, then $(-1)^n a f(x^2) = f(x) f(-x)$, where $a$ is the leading coefficient of $f$ and $n$ its degree. – user14972 Feb 4 '17 at 3:09
• can you give me a proof? I can hardly understand you – Mclalalala Feb 4 '17 at 3:12
• Hrm. Actually, the easy argument I had in mind only concludes $\operatorname{lcm}(f(x), f(-x)) \mid f(x^2)$. I still think what I wrote is true, but am no longer certain. – user14972 Feb 4 '17 at 3:34

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.

• Comments are not for extended discussion; this conversation has been moved to chat. Pertinent information should be added to the post itself via an edit. – user642796 Mar 3 '17 at 5:54
• $\alpha^{2^n}=\alpha^{2^m}$ then $\alpha^{2^n-2^m}=1$ and we can factor $2^n-2^m=2^r(2^s-1)$ for some $r,s$. – Mclalalala Mar 3 '17 at 6:36
• @arctic tern why from this we can conclude that $\alpha$ was a root of unity – Mclalalala Mar 3 '17 at 6:38
• If $\alpha$ raised to an integer power equals $1$, then $\alpha$ is a root of unity. That's, like, the definition of "root of unity." – Mclalalala Mar 3 '17 at 13:37
• I already gave every possible example. The only monic irreducible integer-coefficient polynomials $f(x)$ such that $f(x)\mid f(x^2)$ are cyclotomic polynomials $\Phi_m(x)$ where $m$ is odd. Since $\deg\Phi_m(x)=\phi(m)$ and $\phi(m)$ is never $3$, there are no examples of degree $3$. As for degree $4$, the only odd solution to $\phi(m)=4$ is $m=5$, so the unique polynomial $f(x)$ is $\Phi_5(x)=x^4+x^3+x^2+x+1$. – Mclalalala Mar 3 '17 at 14:37

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.