2. Determine all positive integers $n$, for which

$$
S_{n}=x^{n}+y^{n}+z^{n}
$$

is constant, for any real numbers $x, y, z$ such that, $x y z=1$ and $x+y+z=0$.
## SOLUTION:

Suppose $z$ is known. Then $x, y$ are two real numbers with sum $-z$ and product $\frac{1}{z}$; that is, they are the roots of the equation $\rho^{2}-\rho z+\frac{1}{z}=0$, or: $x, y=\frac{-z \pm \sqrt{z^{2}-\frac{4}{z}}}{2}$.

Thus, one possible solution is, for example, $z=\sqrt[3]{4} \quad$ and $\quad x=y=-\frac{1}{\sqrt[3]{2}}$. However, we can make $z$ arbitrarily large in absolute value, making one of the two numbers $x$ or $y$ also arbitrarily large in absolute value and of opposite sign to $z$. The other will be arbitrarily small in absolute value and negative. It immediately follows that $n$ cannot be even when $S_{n}$ is constant, as on the one hand it takes a finite and well-determined value when $z=\sqrt[3]{4}$, and on the other hand, it can take arbitrarily large and always positive values when $z$ grows arbitrarily.

To complete the problem, we will use the following result.

Lemma: For all $n \geq 3$ odd, given $z$, $x^{n}+y^{n}$ can be written as a polynomial in $z$ of degree $n$ (plus some term possibly with a negative exponent) and with coefficients equal to -1 for the terms of degree $n$, equal to 0 for the terms of degree $n-1, n-2$, and equal to $n$ for the terms of degree $n-3$.

## Proof:

For $n=3$ we have $x^{3}+y^{3}=(x+y)^{3}-3 x y(x+y)=-z^{3}+3$.

For $n=5$ we have $x^{5}+y^{5}=(x+y)^{5}-5 x y\left(x^{3}+y^{3}\right)-10 x^{2} y^{2}(x+y)=-z^{5}+5 z^{2}-\frac{5}{z}$.

If the result is true for an odd integer $n$, then for $n+2$ we have

$$
\begin{aligned}
& x^{n+2}+y^{n+2}=(x+y)^{n+2}-\sum_{u=1}^{n+1}\left(\begin{array}{c}
n+2 \\
u
\end{array}\right) x^{u} y^{u}\left(x^{n+2-2 u}+y^{n+2-2 u}\right)= \\
& -z^{n+2}-\frac{n+2}{z}\left(x^{n}+y^{n}\right)+\sum_{u=2}^{\frac{n+1}{2}}\left(\begin{array}{c}
n+2 \\
u
\end{array}\right) \frac{x^{n+2-2 u}+y^{n+2-2 u}}{z^{u}}
\end{aligned}
$$

By the induction hypothesis, the sum for $u \geq 2$ does not contain any term of degree higher than $n-4$, and also by the induction hypothesis, the term of highest degree in the second summand on the right-hand side is $-\frac{n+2}{z}\left(-z^{n}\right)=(n+2) z^{n-1}$.

Thus, the lemma is proven by complete induction.

From the initial reasoning, it follows that if $S_{n}$ is constant, $n$ cannot be even, and by the lemma, when $z \rightarrow \infty$ for a given $n \geq 5$ odd, we have

$$
x^{n}+y^{n}+z^{n} \rightarrow n z^{n-3} \rightarrow \infty
$$

with a positive sign as well, since $n-3$ is even in this case. Therefore, only $S_{1}$ and $S_{3}$ could be constant: $S_{1}=0$ is constant by hypothesis, and $S_{3}=3$ as a result of a finding during the proof of the lemma.

Thus, it is concluded that $S_{n}$ is constant under the conditions of the problem if and only if $n=1$ or $n=3$.