## Task B-3.2.

Niko says to Jure: "I have three numbers whose sum is 2. The sum of their squares is 6, and the sum of their cubes is 8. What do you think, if I continue to sum the $n$-th powers of these numbers, can I get a sum of $2^{2017}$ for some $n$?" What will Jure answer? Justify your response.
## Solution.

If we denote the initial numbers as $x, y, z$, let's introduce the substitution $a=x+y, b=xy$.

We have: $x^{2}+y^{2}=(x+y)^{2}-2xy$ and $x^{3}+y^{3}=(x+y)^{3}-3xy(x+y)$.

The original system becomes:

$$
\begin{aligned}
& a+z=2 \\
& a^{2}-2b+z^{2}=6 \\
& a^{3}-3ab+z^{3}=8
\end{aligned}
$$

From the first equation, we have $z=2-a$ and substituting into the other two, we get the system:

$$
\begin{aligned}
& a^{2}-2b+4-4a+a^{2}=6 \\
& a^{3}-3ab+8-12a+6a^{2}-a^{3}=8
\end{aligned}
$$

Simplifying, we get:

$$
\begin{aligned}
& a^{2}-2a-b=1 \\
& 3a(2a-4-b)=0
\end{aligned}
$$

The second equation gives us two cases:

1. Case:

$a=0 \Longrightarrow z=2, b=-1$

Now we have the system: $x+y=0, xy=-1$.

Its solutions are $x_{1,2} = \pm 1, y_{1,2}=\mp 1$, and the ordered triples are $(1,-1,2),(-1,1,2)$.

2. Case:

$2a-4-b=0 \Longrightarrow b=2a-4$

Substituting into the first equation, we get $a^{2}-4a+3=0$.

The solutions to this equation are: $a_{1}=1 \Longrightarrow z=1, b=-2$

$$
a_{2}=3 \Longrightarrow z=-1, b=2
$$

We solve the system $x+y=1, xy=-2$ and the system $x+y=3, xy=2$.

The solutions to the first system are $x_{1}=-1, y_{1}=2$ and $x_{2}=2, y_{2}=-1$, so the triples are $(x, y, z):(-1,2,1),(2,-1,1)$.

The solutions to the second system are $x_{1}=2, y_{1}=1$ and $x_{2}=1, y_{2}=2$, so the triples are $(2,1,-1),(1,2,-1)$.

The final solutions to the system are the ordered triples:

$$
(x, y, z) \in\{(1,-1,2),(-1,1,2),(-1,2,1),(2,-1,1),(1,2,-1),(2,1,-1)\}
$$

The sum of the $n$-th powers of $x, y$, and $z$ for each ordered triple $(x, y, z)$ is $1^{n}+(-1)^{n}+2^{n}$, which is always equal to $2^{n}$ for odd $n$, and $2+2^{n}$ for even $n$.

Therefore, Jure will answer affirmatively for $n=2017$, i.e., $x^{2017}+y^{2017}+z^{2017}=2^{2017}$.