Let's determine the necessary and sufficient condition for the fourth-degree polynomial

$$
f(x)=A x^{4}+B x^{3}+C x^{2}+D x+E \quad(A \neq 0)
$$

to be expressible in the form

$$
f(x)=p(q(x))
$$

where $p(x)$ and $q(x)$ are quadratic polynomials.
Let's find the second-degree polynomials $p(x)$ and $q(x)$ in the form $p(x)=a x^{2}+b x+c$ and $q(x)=r x^{2}+s x+t$. If the composite function $p(q(x))$ is identical to the polynomial $f(x)$, then the corresponding coefficients must be equal, and thus by comparing the coefficients, we arrive at the system of equations:

$$
\begin{aligned}
a r^{2} & =A \\
2 a r s & =B \\
2 a r t+a s^{2}+b r & =C, \\
2 a s t+b s & =D \\
a t^{2}+b t+c & =E
\end{aligned}
$$

The necessary and sufficient condition for the desired representation is that this system is solvable for the unknowns $a, b, c, r, s, t$. It is clear that the last equation can be ignored, as $c$ only appears in it. If we have already determined the values of $a, b$, and $t$, we can always choose $c$ such that the fifth equation is satisfied. From the condition $A \neq 0$, it follows from the first equation that $a \neq 0$ and $r \neq 0$. Dividing the remaining three equations by the first equation, we get:

$$
\begin{aligned}
& \frac{B}{A}=2 \frac{s}{r} \\
& \frac{C}{A}=2 \frac{t}{r}+\frac{s^{2}}{r^{2}}+\frac{b}{a r} \\
& \frac{D}{A}=\left(2 \frac{t}{r}+\frac{b}{a r}\right) \frac{s}{r}
\end{aligned}
$$

From these, we obtain:

$$
\left(\frac{C}{A}-\frac{1}{4} \frac{B^{2}}{A^{2}}\right) \frac{1}{2} \cdot \frac{B}{A}=\frac{D}{A}
$$

or

$$
D=\frac{B C}{2 A}-\frac{B^{3}}{8 A^{2}}
$$

For the representation $f(x)=p(g(x))$ to be possible, the coefficients of $f$ must satisfy the above equality, which is a consequence of the system of equations for the coefficients. On the other hand, if the equation $(*)$ is satisfied, then the system is solvable: for example, let $r=1, t=0$, then from the first equation $a=A$, from the second $s=\frac{B}{2 A}$, from the third $b=C-\frac{B^{2}}{4 A}$, and $(*)$ ensures that the fourth equation is not in contradiction with this, and finally, $c$ can be expressed from the fifth equation.

Tamás Hornung (Győr, Czuczor G. Bencés Gymnasium, 4th grade)