For which polynomials $p(x)$ is $p(p(x))+p(x)$=$x^4+3x^2+3$, for all $x \in \mathbb{R}$

Question

For which polynomials $p(x)$ is $p(p(x))+p(x)$=$x^4+3x^2+3$, for all $x \in \mathbb{R}$

Since the power of the right hand side is 4, $p(x)$ has to be 2. So I assumed a solution of: $p(x)=ax^2+bx+c$ and then i put it in $p(p(x))+p(x)$ and got:
$a(ax^2+bc+c)^2+b(ax^2+bx+c)+c+ax^2+bx+c=x^4+3x^2+3$

To find the coefficients I tried $x=0$ and got $ac^2+bc+2c=3$. But here I'm stuck, how do I go from here?

Answer 1

Identifying the coefficient of $x^4$ one gets $a=1$ and then the coefficient of $x^3$ you get $b=0$ hence $c=1$ or $c=-3$. Then check that $x^2+1$ is indeed a solution while $x^2-3$ is not.

Answer 2

The polynomials

$$Ax^5+Bx^4+Cx^3+Dx^2+Ex+F$$ and $$ax^5+bx^4+cx^3+dx^2+ex+f$$

are equal for all $x$ if and only if

$$A=a\\B=b\\C=c\\D=d\\E=e\\F=f$$