\section*{Problem 2 - 121222}

All ordered pairs \((a, b)\) of real numbers \(a, b\) are to be determined for which the polynomial \(f(x) = x^{2} + a x + b\) is a divisor of the polynomial \(g(x) = x^{4} + a x^{2} + b\).

Definition: A polynomial \(f(x)\) is called a divisor of a polynomial \(g(x)\) if and only if there exists a polynomial \(h(x)\) such that \(f(x) \cdot h(x) = g(x)\).
According to the given, for certain \(c, d \in \mathbb{R}\), the equation

\[
\left(x^{2}+a x+b\right)\left(x^{2}+c x+d\right)=x^{4}+a x^{2}+b
\]

must hold. By expanding and comparing coefficients, we obtain the following nonlinear system of equations for the variables \(a, b, c, d\):
(1) \(a+c=0\).
(2) \(b+a c+d=a\)
(3) \(a d+b c=0\).
(4) \(b d=b\).
To solve this, we proceed with the following case distinction:

1. Case: \(b=0\).

(3) then becomes \(a d=0\). If \(a=0\) here, it immediately follows that \(c=d=0\). Otherwise, we can divide by \(a\), leading to \(d=0, c=1, a=-1, b=0\) in this order. In total, these two cases correspond to the factorizations

\[
x^{2} x^{2}=x^{4} \quad \text{or} \quad\left(x^{2}-x\right)\left(x^{2}+x\right)=x^{4}-x^{2}
\]

2. Case: \(b \neq 0\).

Then, due to (4), it must be that \(d=1\), and (3) can also be written in the form \(a=a b\) because \(c=-a\). Here, either \(b=1\), in which case (2) implies \(a^{2}+a-2=(a-1)(a+2)=0\), leading to the solutions

\[
(a, b, c, d)=(1,1,-1,1) \quad \text{or} \quad(a, b, c, d)=(-2,1,2,1)
\]

or \(a=0\), which leads to the solution

\[
(a, b, c, d)=(0,-1,0,1)
\]

These additional solutions correspond to the following factorizations

\[
\begin{aligned}
\left(x^{2}+x+1\right)\left(x^{2}-x+1\right) & =x^{4}+x^{2}+1 \\
\left(x^{2}-2 x+1\right)\left(x^{2}+2 x+1\right) & =x^{4}-2 x^{2}+1 \\
\left(x^{2}-1\right)\left(x^{2}+1\right) & =x^{4}-1
\end{aligned}
\]

which clearly satisfy the conditions of the problem.