1. Determine the real numbers $a, b, c$ such that the polynomial $x^{4}+a x^{2}+b x+c$ is divisible by the polynomial $x^{2}+x+1$ and the sum $a^{2}+b^{2}+c^{2}$ is as small as possible.
1. By dividing the polynomial $x^{4}+a x^{2}+b x+c$ by the polynomial $x^{2}+x+1$, we find that

$$
x^{4}+a x^{2}+b x+c=\left(x^{2}+x+1\right)\left(x^{2}-x+a\right)+(b-a+1) x+(c-a).
$$

The polynomial $x^{4}+a x^{2}+b x+c$ is divisible by the polynomial $x^{2}+x+1$ if and only if the remainder upon division is the zero polynomial, i.e., $b-a+1=0$ and simultaneously $c-a=0$; hence $b=a-1, c=a$.

Then

$$
a^{2}+b^{2}+c^{2}=a^{2}+(a-1)^{2}+a^{2}=3 a^{2}-2 a+1=3\left(a-\frac{1}{3}\right)^{2}+\frac{2}{3}.
$$

This expression has its minimum value for $a=\frac{1}{3}$; we easily compute $b=a-1=-\frac{2}{3}$, $c=a=\frac{1}{3}$.

Another solution. The polynomial $x^{4}+a x^{2}+b x+c$ is divisible by the polynomial $x^{2}+x+1$ if and only if there exist real numbers $p, q$ such that

$$
x^{4}+a x^{2}+b x+c=\left(x^{2}+x+1\right)\left(x^{2}+p x+q\right).
$$

By expanding the right-hand side and comparing the corresponding coefficients, we obtain four equations $p+1=0, q+p+1=a, q+p=b, q=c$. From these, we express $p=-1, q=a, c=a$, $b=a-1$ and proceed as in the first solution.

For a complete solution, award 6 points. Award two points for the equality $x^{4}+a x^{2}+b x+c=\left(x^{2}+x+1\right)\left(x^{2}-x+a\right)+(b-a+1) x+(c-a)$, one point for expressing two of the unknowns $a, b, c$ in terms of the third, two points for transforming the expression $a^{2}+b^{2}+c^{2}$ into the form $3 a^{2}-2 a+1=3\left(a-\frac{1}{3}\right)^{2}+\frac{2}{3}$ (or $3\left(b+\frac{2}{3}\right)^{2}+\frac{2}{3}$), and one point for correctly determining the numbers $a, b, c$. Similarly, award one point for the equality $x^{4}+a x^{2}+b x+c=\left(x^{2}+x+1\right)\left(x^{2}+p x+q\right)$, one point for the system of equations $p+1=0, q+p+1=a, q+p=b, q=c$, one point for expressing $c=a$, $b=a-1$, and proceed as in the first solution.