Find all polynomials $f(x)$ satisfying $f(x^2 ) + f(x) \cdot f(x + 1) = 0$ ∀ $x ∈ R$ Find all polynomials $f(x)$ satisfying $f(x^2 ) + f(x) \cdot f(x + 1) = 0$  ∀  $x ∈ R$
Obvious polynomials are $f(x)=0$ and $f(x) = -1$
There are no linear polynomials satisfying the equation.
I took $f(x) = a_0+a_1x+a_2x^2+a_3x^3+a_4x^4+...a_nx^n$
For $f(x)$ to satisfy the condition
$a_0(1+a_0+a_1+a_2+...a_n) = 0$
$a_n + {a_n}^2=0$
Taking simple values of $x$ like $x=1,x=0$ did not help and I am unable to proceed further.
Can anybody help?
 A: Hint: Basic Complex Analysis tells us that the given equation holds for complex $x$ also. [I am referring to the Identity Theorem].  If $f$ has  a root $x$ then the hypothesis shows that $x^{2}, x^{4},..$ are all zeros. But  a polynomial can have only finite number of zeros (unless it is the zero polynomial). So the only possible roots are roots of unity and $0$.
A: I would not consider this as a complete or rigourous answer since I've given some statements which I've not given the proof to. Consider this more like a hint.
You've noticed that there were no possible linear solutions, in fact, we can proof that any polynomial with an odd degree cannot be solution to that equation.
Let $ n\in 2\mathbb{N} $. Denoting $ f\left(x\right)=\sum\limits_{k=0}^{n}{a_{k}x^{k}} $. We have :
\begin{aligned} f\left(x\right)f\left(x+1\right)&=\left(\sum_{i=0}^{n}{a_{i}x^{i}}\right)\left(\sum_{i=0}^{n}{\left(\sum_{j=i}^{n}{a_{j}\binom{j}{i}}\right)x^{i}}\right)\\&=\sum_{k=0}^{2n}{\left(\sum_{p=\max\left(0,k-n\right)}^{\min\left(k,n\right)}{a_{k-p}\sum_{j=p}^{n}{a_{j}\binom{j}{p}}}\right)x^{k}} \end{aligned}
Suppose $ f\left(x\right)\neq 0 $. Let's try to find the first coefficients $ a_{n-k} $, $ k\in\mathbb{N} $ of our polynomial.
\begin{aligned}a_{n}+\sum_{p=n}^{n}{a_{2n-p}\sum_{j=p}^{n}{a_{j}\binom{j}{p}}}=0 &\iff a_{n}+a_{n}^{2}=0 \\&\ \ \Longrightarrow \ \ \ \ \ \ \ \ \ \ \ a_{n}=-1\end{aligned}
\begin{aligned} \sum_{p=n-1}^{n}{a_{2n-1-p}\sum_{j=p}^{n}{a_{j}\binom{j}{p}}}=0&\iff a_{n-1}a_{n}+a_{n}\left(a_{n-1}+n a_{n}\right)=0\\ &\ \ \Longrightarrow \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ a_{n-1}=\frac{n}{2} \end{aligned}
\begin{aligned} a_{n-1}+\sum_{p=n-2}^{n}{a_{2n-2-p}\sum_{j=p}^{n}{a_{j}\binom{j}{p}}}=0 \Longrightarrow a_{n-2}=-\frac{\frac{n}{2}\left(\frac{n}{2}-1\right)}{2} \end{aligned}
$$ \sum_{p=n-3}^{n}{a_{2n-3-p}\sum_{j=p}^{n}{a_{j}\binom{j}{p}}}=0\Longrightarrow a_{n-3}=\frac{\frac{n}{2}\left(\frac{n}{2}-1\right)\left(\frac{n}{2}-2\right)}{6}$$
So on and so forth.
We can proof by strong induction that $ \left(\forall k\in\mathbb{N}\right),\ a_{n-k}=\left(-1\right)^{k+1}\binom{\frac{n}{2}}{k} $.
Then we can conclude that : $$ \fbox{$\begin{array}{rcl}\displaystyle f\left(x\right)=\sum_{k=0}^{n}{\left(-1\right)^{k+1}\binom{\frac{n}{2}}{k}x^{n-k}}\end{array}$} $$
Since $ n $ must be even, then all the polynomials satisfying to the equation are $ 0 $ and $ \sum\limits_{k=0}^{n}{\left(-1\right)^{k+1}\binom{n}{k}x^{2n-k}}=-x^{n}\left(1-x\right)^{n} $, $ n\in\mathbb{N} $.
The set of solutions would, thus, be : $$\fbox{$\begin{array}{rcl} \displaystyle\mathbb{S}=\left\lbrace 0,-x^{n}\left(1-x\right)^{n}\mid n\in\mathbb{N}\right\rbrace\end{array}$} $$
