## Task A-4.2.

Let $w=\frac{1}{2}(-1+i \sqrt{3})$. Determine the largest number $n \in \mathbb{N}_{0}$ for which there exist complex numbers $a, b, c$ such that for every $k \in\{0,1, \ldots, n\}$ the following holds

$$
a+b w^{k}+c w^{2 k}=k
$$

For the determined $n$, find all triples $(a, b, c)$ that satisfy the above equalities.
## Solution.

First, note that $w$ is a third root of 1, i.e., $w^{3}=1$.

Now assume that $n \geqslant 3$. From $w^{3}=1$, it follows that

$$
0=a+b+c=a+b w^{3}+c w^{6}=3
$$

which is obviously impossible.

Therefore, we conclude that $n \leqslant 2$.

Let's prove that the largest possible $n$ that satisfies the conditions of the problem is $n=2$. For $n=2$, we get a system of three equations with three unknowns:

$$
\begin{aligned}
a+b+c & =0 \\
a+b w+c w^{2} & =1 \\
a+b w^{2}+c w & =2
\end{aligned}
$$

By adding the above three equations, we get

$$
3 a+b\left(1+w+w^{2}\right)+c\left(1+w^{2}+w\right)=3
$$

1 point

Since $w^{3}-1=(w-1)\left(w^{2}+w+1\right)=0$, it follows that $w^{2}+w+1=0$, so from the above equation we get $a=1$.

Substituting $a=1$ into the first two equations of the system, we get

$$
\begin{aligned}
b+c & =-1 \\
b w+c w^{2} & =0
\end{aligned}
$$

Multiplying the first equation by $-w$ and adding it to the second, we get $c\left(w^{2}-w\right)=w$, i.e., $c=\frac{1}{w-1}$.

Since $b=-1-c$, it follows that $b=-\frac{w}{w-1}$.

Thus, for $n=2$, the required complex numbers are

$$
(a, b, c)=\left(1,-\frac{w}{w-1}, \frac{1}{w-1}\right)
$$

so $n=2$ is indeed the largest number for which the conditions of the problem hold.

Note: The numbers $a, b$, and $c$ can be expressed in various ways in terms of $w$, and any correct way should be recognized. Some examples for $(a, b, c)$ are $\left(1, \frac{w}{1-w}, \frac{1}{w-1}\right),\left(1, \frac{w-1}{3},-\frac{w+2}{3}\right)$. Also, the triplet $(a, b, c)$ can be written without using the number $w$:

$$
(a, b, c)=\left(1, \frac{-3+i \sqrt{3}}{6}, \frac{-3-i \sqrt{3}}{6}\right)
$$