4. On the board, the numbers $a, b$, and $c$ were written. They were erased, and in their place, the numbers $a^{4}-2 b^{2}, b^{4}-2 c^{2}, c^{4}-2 a^{2}$ were written. After this, it turned out that the numbers on the board were the same as those at the beginning (possibly in a different order). Find the numbers $a, b, c$, given that $a+b+c=-3$.
4. Answer: $a=b=c=-1$. Solution. From the condition, it follows that $a^{4}-2 b^{2}+b^{4}-$ $2 c^{2}+c^{4}-2 a^{2}=a+b+c=-3 \Leftrightarrow a^{4}-2 b^{2}+b^{4}-2 c^{2}+c^{4}-2 a^{2}+3=\left(a^{2}-1\right)^{2}+\left(b^{2}-\right.$ $1)^{2}+\left(c^{2}-1\right)^{2}=0$. Therefore, $a^{2}=b^{2}=c^{2}=1$, meaning each of the numbers $a, b, c$ is either 1 or -1. Since by the condition $a+b+c=-3$, the only possible case is $a=b=c=-1$.

Grading criteria. Correct solution - 7 points. Proven that the squares are equal to 1, but cases other than all -1 are not eliminated - 5 points. In all other cases $-\mathbf{0}$ points.