Find all $P$ with $P(x^2)=P(x)^2$ The following problem is from Golan's book on linear algebra, chapter 4. I have posted a proposed answer below. 
Problem: Let $F$ be a field. Find all nonzero polynomials $P\in F[x]$ satisfying 
$$P(x^2)=[P(x)]^2.$$

Assume first that $F$ is a field with characteristic not equal to 2. The only ones are 1 and $x^n$, $n\in \mathbb{N}$.
Let $a_n$ denote the coefficient of $x^n$ in $P$. We see immediately that all $a_n=0$ for odd $n>0$. Examining the constant coefficient, we see $a_0=a_0^2\Rightarrow a_0=1$ or $a_0=0$. 
Now proceed by induction. Consider the case where $a_0=1$. Assume we have shown $a_n=0$ for all $n<k$, $n\neq 1$. We will show $a_k=0$. If $k$ is odd, we are done. If $k$ is even, the coefficient of $x^{k}$ in $P(x^2)$ is $a_{k/2}$, so it is 0. We evaluate $[P(x)]^2$ and ignore higher order terms, and see
$$(a_kx^{k}+1)^2=a_k^2x^{2k}+2a_kx^k+1$$
and the only way for the coefficient of $x^k$ to vanish here is for $a_k$ to be 0.
The case with $a_0=0$ is similar. Assume we have shown $a_n=0$ for all $n<k$. The coefficient of $x^{2k}$ in $P(x^2)$ is $a_{k}$. If evaluate $[P(x)]^2=[...a_kx^k]^2$ and ignore higher order terms again, we get $a_k^2x^{2k}$. So $a_k=1$ or $a_k=0$. If $a_k=0$, we continue the induction. If $a_k=1$, we factor $x^k$ out of the original polynomial and are reduced to the first case. 
In a field of characteristic 2 however, I believe that any polynomial with all coefficients equal to 0 or 1 works. Just use the "freshman's dream." Further, because equating constants on both sides gives $a_n^2=a_n$, these are the only ones that work.
