At most one degree-$n$ polynomial commutes with $X^2+\alpha$ I took an exam a few hours ago and there was a question I couldn't do.
For those who speak French, here is the document http://www.concours-centrale-supelec.fr/CentraleSupelec/2014/MP/sujets/2012-006.pdf (precisely III.B.3).

Let $\alpha \in \mathbb C$.
Let $P_\alpha = X^2+\alpha$
Prove that for any $n \in \mathbb N$, there exists at most one polynomial $Q$ with degree $n$ such that $P_\alpha \circ Q=Q \circ P_\alpha $

If such $Q$ exists it must have $1$ as leading coefficient.
I tried rewriting $P_\alpha \circ Q=Q \circ P_\alpha$ in terms of coefficients of $Q$ and find a recurrence relation on these coefficients (this would grant the "at most") but computations got awful.
 A: If you let $Q_ = \sum_{0 \le k \le n} b_k X^k$ and write out the coefficients of $P_\alpha \circ Q$ and $Q \circ P_\alpha$, you should notice that, in front of $X^{2n-k}$, in $P_\alpha \circ Q$ you obtain a linear combination of $b_{n-i}b_{n-j}$ for $i+j=k$, and in particular, you get $2b_nb_{n-k} = 2b_{n-k}$. In $Q \circ P_\alpha$, you obtain an expression involving $b_n \ldots b_{n-\lfloor k/2 \rfloor}$ (because $2k$ is the greatest power of $X$ appearing with $b_k$). 
Hence by taking $k=1, \ldots, n$ you obtain equations relating each $b_{n-k}$ to some expression in $b_{n-l}$ with $0\le l<k$. Since $b_n$ has a unique possible value, so does each coefficient. So by looking at the $n+1$ most dominant coefficients of the equality, you obtain $n+1$ equations that determine $Q$ uniquely.
Also note that this doesn't work for degree $0$ polynomials. 
This still works if you change the degree of $P$ to $d>1$, the $\lfloor k/2 \rfloor$ turns into $\lfloor k/d \rfloor$, which is still strictly less than $k$ for $k > 0$. Only the first coefficient of $Q$ can vary a bit, bust the rest is determined by that coefficient.
A: I feel very dumb now...
Suppose for contradiction sake that $Q$ and $R$ are distinct degree $n\geq 2$ polynomials that commute with $P_\alpha$.
Then Let $Z=Q-R$.
Since $Q$ and $R$ are monic, $\deg Z < n$
Furthermore, $Z \circ P_\alpha =P_\alpha\circ Q -P_\alpha\circ R=Q^2-R^2=Z(Q+R) $
Note that $\deg Z \circ P_\alpha=2\deg Z$ and $\deg Z(Q+R)=n+\deg Z$
Hence $\deg Z =n$
Contradiction.
