23.21. For which natural numbers $n$ is the expression $a^{n}(b-c)+$ $+b^{n}(c-a)+c^{n}(a-b)$ divisible by $a^{2}+b^{2}+c^{2}+a b+b c+c a$?
23.21. Answer: only for $n=4$. It is easy to verify that for $n=4$ the result of the division is $(a-b)(b-c)(a-c)$. We will show that for all other natural $n$ the expression $a^{n}(b-c)+b^{n}(c-a)+c^{n}(a-b)$ does not divide $a^{2}+b^{2}+c^{2}+ab+bc+ca$. It is sufficient to check that the first expression does not divide the second when $b=2$ and $c=1$, i.e., $a^{n}-(2^{n}-1)a+(2^{n}-2)$ does not divide $a^{2}+3a+7$. For $n=2$ and 3, this is verified directly. Now let $n \geqslant 5$. The quadratic equation $a^{2}+3a+7=0$ has roots $\frac{-3 \pm i \sqrt{19}}{2}$; the modulus of each root is $\sqrt{7}$. Therefore, it is sufficient to check that if $z$ is a complex number and $|z|=\sqrt{7}$, then $z^{n}-(2^{n}-1)z+(2^{n}-2) \neq 0$. Clearly, $\left|z^{n}-(2^{n}-1)z+(2^{n}-2)\right| \geqslant \left|z^{n}\right|-(2^{n}-1)|z|-(2^{n}-2)=7^{n/2}-(2^{n}-1)\sqrt{7}-2^{n}+2>7^{n/2}-2^{n}(1+\sqrt{7})>7^{n/2}-4 \cdot 2^{n}$, since $\sqrt{7}>2$, it is sufficient to consider the case $n=5$. In this case, we need to check that $7^{5}>(4 \cdot 2^{5})^{2}$.