Functional equation $P(X)=P(1-X)$ for polynomials I have encountered the following problem :
Find all polynomials $P$ such as $P(X)=P(1-X)$ on $\mathbb{C}$ and then $\mathbb{R}$.
I have found that on $\mathbb{C}$ such polynomials have an even degree. Because for each $a$ root, $1-a$ must be a root too. I struggle to find whether we can do something with polynomials of degree $2$ on $\mathbb{R}$.
I was also wondering if we could find a general solution for $P(X)=P(aX+b)$ with $a$ and $b$ complexe for $P$ defined either on $\mathbb{R}$ or $\mathbb{C}$.

First, if $P(X) = P(1-X)$ holds on $\Bbb R$ then the same relation
holds on $\Bbb C$, due to the identity principle for holomorphic functions.
With $Q(x) := P(x + \frac 12)$, the condition $P(X) = P(1-X)$ is equivalent
to $Q(x) = Q(-x)$, which is satisfied exactly by all polynomials
having only terms with even powers of $x$.
Therefore the general solution is
$$
P(X) = Q(X - \frac 12) = a_0 + a_1 (X - \frac 12)^2 + \ldots + a_n (X - \frac 12)^{2n}
$$
with coefficients $a_0, a_1, \ldots, a_n \in \Bbb R$ or $\Bbb C$,
depending on whether you want $P$ to be real-valued on $\Bbb R$ or not.

For the general case $P(X) = P(aX + b)$ with $a \ne 1$, note that the
fixed point of $X \to aX +b$ is $X = \frac{b}{1-a}$, therefore define
$Q(x) := P(x + \frac{b}{1-a})$ to get $Q(x) = Q(ax)$.
Now there are two cases: If $a$ is a "root of unity", i.e. $a^k=1$
for some positive integer $k$, then the solutions are exactly the polynomials having 
only terms with powers which are a multiple of $k$:
$$
 Q(x) = a_0 + a_1 x^k + \ldots + a_n x^{kn} \, .
$$
If $a$ is not a root of unity then 
$$
 Q(1) = Q(a) = Q(a^2) = Q(a^3) = \dots
$$
with all arguments being different, and the only solution are constant
polynomials $Q(x) = a_0$.
For $a \in \Bbb R$, $a \ne 1$, the first case occurs only for $a= -1$.
