1. Determine all pairs of real numbers $a$ and $b$ for which the polynomial $x^{4}+a x^{2}+b$ is divisible by the polynomial $x^{2}+b x+a$.
1. By dividing the polynomial $x^{4}+a x^{2}+b$ by the polynomial $x^{2}+b x+a$, we find that

$$
x^{4}+a x^{2}+b=\left(x^{2}+b x+a\right)\left(x^{2}-b x+b^{2}\right)+\left(a b-b^{3}\right) x+\left(b-a b^{2}\right) .
$$

The polynomial $x^{4}+a x^{2}+b$ is divisible by the polynomial $x^{2}+b x+a$ if and only if the remainder $\left(a b-b^{3}\right) x+\left(b-a b^{2}\right)$ is the zero polynomial, i.e., $a b-b^{3}=b\left(a-b^{2}\right)=0$ and simultaneously $b-a b^{2}=$ $=b(1-a b)=0$. If $b=0$, both conditions are satisfied. For $b \neq 0$, it must hold that $a-b^{2}=0$ and $1-a b=0$. Hence, $a=b^{2}, 1-b^{3}=0$, and thus $a=b=1$.

Conclusion. The polynomial $x^{4}+a x^{2}+b$ is divisible by the polynomial $x^{2}+b x+a$ if and only if $b=0$ (and $a$ is arbitrary) or $a=b=1$.

Alternative solution. The polynomial $x^{4}+a x^{2}+b$ is divisible by the polynomial $x^{2}+b x+a$ if and only if there exist real numbers $p, q$ such that $x^{4}+a x^{2}+b=\left(x^{2}+b x+a\right)\left(x^{2}+p x+q\right)$. By expanding and comparing coefficients, we obtain the system of equations

$$
p+b=0, \quad q+b p+a=a, \quad a p+b q=0, \quad a q=b .
$$

From the first equation, we express $p=-b$ and from the second $q=-b p=b^{2}$. Substituting into the third and fourth equations, we get $-a b+b^{3}=0, a b^{2}=b$. We complete the solution in the same way as in the previous case.

For a complete solution, award 6 points. For correctly finding the remainder upon division, award 2 points. Award another 2 points for the system of equations $a b-b^{3}=0, b-a b^{2}=0$ and 2 points for solving it. Similarly, award 2 points for the equation $x^{4}+a x^{2}+b=\left(x^{2}+b x+a\right)\left(x^{2}+p x+q\right)$, 2 points for the system $p+b=0, q+b p+a=a, a p+b q=0$, $a q=b$, and 2 points for solving it. Award one point for guessing the solution $b=0$. Water.