# Find all polynomials $P(x)$ such that $P(x^2)=P(x)^2$

Find all polynomials $P:\mathbb{C}\rightarrow\mathbb{C}$ such that $$P(x^2)=P(x)^2 .$$

Here is what I tried:

First, it is easy to see the constant solutions, namely $P\equiv 0,P\equiv 1$.

Let $r$ be a root of $P$ (i.e. $P(r)=0$). It follows that all terms in the infinite sequence $r,r^2,r^4,\dots,r^{2n},\dots$ are roots of $P$. To avoid a polynomial with infinite roots, we have to have the sequence be periodic. So it turns out that either $r=0$, or the roots of $P$ are roots of unity of degree $2n$ for some $n\in\mathbb{N}$.

In the first case, we get $P(x)=xQ(x)$. Plugging in, we get $$x^2Q(x^2)=x^2Q(x)^2\implies Q(x^2)=Q(x)^2 ,$$ i.e. that $Q$ satisfies the same condition as $P$.

In the second case, we have $P(x)=(x^{2n}-1)Q(x)$. Plugging in, we get $(x^{2n}-1)Q(x^2)=(x^{2n}-2x^n+1)Q(x)^2$. I was not sure how to proceed from here.

• The roots $e^{\pm 2 \pi i / 3}$ of the polynomial $x^2 + x + 1$ are periodic under the squaring map (of period $2$), but they are third roots of unity. May 8, 2016 at 23:19
• @Travis Glad you said that. Your point of course is that $3$ is not even. The OP and one of the answerers seem to be assuming that if $S$ is a subset of a finite cyclic group which is invariant under squaring then $S$ must be a subgroup. I'd expressed doubts about that, but was wondering whether it might be so for some reason I didn't see - hadn't got to thinking about a counterexample, which you give here. May 8, 2016 at 23:49

Hint If $$P \neq 0$$, then $$P$$ has the form $$P(x) = a_n x^n + a_{n - 1} x^{n - 1} + \cdots + a_1 x + a_0 ,$$ where $$a_n \neq 0$$. Comparing the leading terms of $$P(x^2)$$ and $$P(x)^2$$ gives that $$a_n = 1$$. If $$P(x) \neq x^n$$, then there is some largest index $$m < n$$ such that $$a_m \neq 0$$, and so $$P$$ has the form $$P(x) = x^n + a_m x^m + O(x^{m - 1}) .$$ Now, substitute in both sides of the condition.

Doing so gives $$P(x^2) = x^{2n} + a_m x^{2m} + O(x^{2m - 1})$$ and $$P(x)^2 = x^{2n} + 2 a_m x^{m + n} + O(x^{m + n - 1}) .$$ The second-largest nonzero term in $$P(x^2)$$ has degree $$2m < m + n$$, so comparing the degree $$m + n$$ terms gives $$a_m = 0$$, a contradiction. Thus, $$P(x)$$ must be $$0$$ or $$x^n$$ for some $$n$$. On the other hand, checking directly shows that these polynomials satisfy the condition.

Incidentally, this argument works over any field of characteristic $$\neq 2$$ (in characteristic $$2$$, $$P(x^2) = P(x)^2$$ holds for all polynomials).

• Thanks for the very nice solution. I have a quick question though, what is the notation $O(x^k)$?, does it just denote a polynomial with degree less than or equal to $k$ (it doesn't seem to be a polynomial in $x^k$)? Not that it changes the solution, but I'm just curious.
– Max
May 12, 2016 at 20:37
• You're welcome, I'm glad you found it useful. Yes, it's just shorthand for an unspecified quantity that grows no faster than $x^k$. If a polynomial is $O(x^k)$, then it has degree $\leq k$. May 12, 2016 at 20:39

Well, $P(0)$ must be $0$ or $1$, and as you note, if $P(0)=0$ then $P(x)=xQ(x)$ and it follows that $Q(x^2)=Q(x)^2$ as well.

Great so far. We can repeat this until we get a $Q$ that does not have $Q(0)=0$. So there exists $j\ge0$ and $Q$ so that $P(x)=x^jQ(x)$ and $$Q(x^2)=Q(x)^2,\quad Q(0)=1.$$

But there are problems with your argument about roots of unity. The relevant sequence is $r^{2^n}$, not $r^{2n}$. Yes, that sequence has to be periodic, but that does not imply that $r^{2n}=1$, it implies that $r^{2^j-2^k}=1$. The biggest problem I see is that having shown that each root is a root of unity of some order you jump to the conclusion that all roots of unity of that order are roots of $P$; I don't see why that would follow.

So. We assume that $Q(x^2)=Q(x)^2$ and that $Q(0)=1$.

If $Q\ne 1$ then $$Q(x)=1+a_kx^k+\dots,$$where $k>0$, $a_k\ne0$, and all the missing terms have order greater than $k$. Then $$Q(x^2)=1+a_kx^{2k}+\dots,$$while $$Q(x)^2=1+2a_kx^k+\dots.$$That's impossible unless $a_k=0$ and $k=0$, contradiction.

So $Q=1$; the only such polynomials are $P(x)=x^j$.

Continuation from the OP's wrong idea: You probably meant to write $P(x)=\left(x^n-1\right)\,Q(x)$. Anyway, show that $x^n+1$ divides $Q(x)$, so that $Q(x)=\left(x^n+1\right)\,R(x)$ for some $R(x)\in\mathbb{C}[x]$. Then prove that $x^{2n}+1$ divides $R(x)$. Continue this process and you can show that $x^{2^kn}+1$ divides $P(x)$ for all $k=0,1,2,\ldots$. (Note: This is a false hint. See the comments below.)

Hint: Assume that $P(x)$ is nonconstant. If $r$ is a root of $P(x)$, then it also follows that any $2^n$-th root of $r$ is a root of $P(x)$. The only complex number $r$ that will allow $P(x)$ to have finitely many roots is $r=0$. By the way, this hint also works if $\mathbb{C}$ is replaced by a field of characteristic not equal to $2$ as well (since $x^{2^n}-r$ has distinct roots in the algebraic closure of any such field).

More generally, let $k\in\mathbb{N}$ with $k>1$ and $P(x)\in K[x]$, where $K$ is an integral domain. If $P\left(x^k\right)=\big(P(x)\big)^k$ and $k \neq p^j$, where $p:=\text{char}(K)$, for all $j=1,2,3,\ldots$, then $P(x)=0$ or $P(x)=x^n$ for some $n=0,1,2,\ldots$.

• ??? Regardless of whether it's what he meant to write, why is it true that $P(x)=(x^n-1)Q(x)$? If $r$ is a root of $P$ then $r$ is a root of unity, great; why would it follow that every root of unity of the same order is a root of $P$? May 8, 2016 at 23:37
• You are right. He should have written just $P(x)=(x-r)\,Q(x)$ for some root of unity $r$, but there is an easy fix. See my edit. May 8, 2016 at 23:45
• The Hint gives a very nice solution - I sort of saw something along those lines but didn't quite get it right. But I don't see why you're still including that "Continuation". Since there's simply no reason to think that $P=(x^n-1)Q$ (or at least nobody's given any) what's the point to showing how one could continue from there? May 8, 2016 at 23:56
• I left it there so as to say that "if that were true," you could do this. That wrong hint actually has the same idea as the correct hint. May 8, 2016 at 23:58
• Whatever. It doesn't read that way to me, it looks as though you're saying it is true. You say "the OP's wrong idea" but don't specify what's wrong, exactly. I think the hint gives by far the best of the three (really only two) solutions we've seen, but that section of the Answer makes me almost want to downvote it - you're saying false things without, it seems to me, making it clear that what you're saying is false. May 9, 2016 at 0:05