Senderov V.A.

For which natural numbers $n$ can one find such integers $a, b, c$ that their sum is zero, and the number $a^{n}+b^{n}+c^{n}$ is prime?
If $n$ is even, then $1^{n}+(-1)^{n}+0^{n}=2-$ a prime number.

Let's prove that for odd $n$, the number $A=a^{n}+b^{n}+c^{n}$ is not a prime number.

First method. For an integer $x$, $x^{n}-x$ is divisible by $x\left(x^{2}-1\right)=x(x-1)(x+1)$, and this number is divisible by 6 (see problem 30359). Therefore,

$A=\left(a^{n}-a\right)+\left(b^{n}-b\right)+\left(c^{n}-c\right)$ is divisible by 6.

Second method. Suppose $A$ is a prime number, then $n>1$ and $a, b, c$ are different from 0.

Since $b^{n}+c^{n}$ is divisible by $b+c=-a$, the number $A$ is divisible by $a$. Similarly, $A$ is divisible by $b$ and by $c$. Therefore, each of the numbers $a, b, c$ is one of the numbers $\pm 1, \pm A$. Since there are no two opposite numbers among $a, b, c$ (otherwise the third would be zero), there will be two equal numbers among them. Let these numbers be equal to $d$, then the third number is $-2 d$, and $A=2 d^{n}-2^{n} d^{n}$ - a number divisible by $2^{n}-2>2$. Contradiction.

## Answer

For all even $n$.