<s>[INST] How to find all polynomials satisfying $P(x^2+x-4)=P^2(x)+P(x)$? 
Find all polynomials with real coefficients $P(x)$ such that $$P(x^2+x-4)=P^2(x)+P(x).$$

$P=0$ is a solution. For non-constant polynomial $P$ comparing both sides gives the leading coefficient as $1$.
We have $$P(x^2+x-4)=P(x) \big(P(x)+1 \big)$$ If $z$ is a complex root of $P$ then $w=z^2+z-4$ is also a root of $P$. Now if we let $x=w$ then $w^2+w-4$ is a root too and etc. But then $P$ has infinitely many roots. And this is a contradiction. Thus we must have $z=z^2+z-4$ and $z= \pm 2$ and $P$ is in the form $$P(x)=(x-2)^n(x+2)^m$$ 
I know that this solution is not complete and above arguments have some flaws! I just wrote it to let you know how I thought about it! I would love to see an elementary solution to this problem.
 [/INST] For $x=2$ the relation gives $\require{cancel}\,\bcancel{P(2)}=P^2(2)+\bcancel{P(2)}\,$, therefore $P(2)=0\,$. Assuming $P \not \equiv 0$, there must exist a multiplicity $n \ge 1$ and some polynomial $Q(x)$ such that $P(x)=(x-2)^n\cdot Q(x)$ and $Q(2) \ne 0$. Substituting back into the relation:
$$
\left(x^2+x-6\right)^n \cdot Q(x^2+x-4)=(x-2)^{2n} \cdot Q^2(x)+(x-2)^n \cdot Q(x)
$$
Since $x^2+x-6=(x-2)(x+3)\,$:
$$
\bcancel{(x-2)^n} \cdot (x+3)^n \cdot Q(x^2+x-4)=\bcancel{(x-2)^n} \cdot (x-2)^n \cdot Q^2(x)+\bcancel{(x-2)^n} \cdot Q(x) \\[5px]
\iff (x+3)^n \cdot Q(x^2+x-4)=(x-2)^n \cdot Q^2(x)+Q(x)
$$
For $x=2\,$, the above reduces to $5^n \cdot Q(2) = Q(2)\,$, but the equality cannot hold for $n \ge 1$ and $Q(2) \ne 0\,$. Therefore, the only solution is the zero polynomial $P \equiv 0\,$.
</s>