18. Find all real polynomials $p(x)$, for all real numbers $x$, that satisfy the equation $p(2 p(x))=2 p(p(x))+2(p(x))^{2}$.
(19th Iranian Mathematical Olympiad (First Round))
Prove: Let $p(x)=a_{n} x^{n}+a_{n-1} x^{n-1}+\cdots+a_{1} x+a_{0}$, then the leading terms of $p(2 p(x)), 2 p(p(x)), 2(p(x))^{2}$ are $2^{n} a_{n}^{n+1} x^{n^{2}}, 2 a_{n}^{n+1} x^{n^{2}}, 2 a_{n}^{2} x^{2 n}$, respectively.

If $n \geqslant 3$, then $n^{2}>2 n$, so $2^{n} a_{n}^{n+1}=2 a_{n}^{n+1}$. Since $a_{n} \neq 0$ is impossible, therefore, $n=0,1,2$.
If $n=0$, let $p(x)=a$, then
$a=2 a+2 a^{2}, a=0$ or $a=-\frac{1}{2}$.
If $n=1$, let $p(x)=a x+b$, we get $a=0$, which is not consistent with the problem.
If $n=2$, let $p(x)=a x^{2}+b x+c$, comparing the coefficient of $x^{4}$ we get $a=1$, comparing the constant term we get $c=0, b$ is any real number.
In summary, we have
$$
p(x) \equiv 0, p(x) \equiv-\frac{1}{2}, p(x)=x^{2}+b x, b \in \mathbf{R} \text {. }
$$