Problem 1780

1. Given Information and Initial Setup:
We are given that $a \in \mathbb{R}$ and $z_1, z_2, \ldots, z_n$ are complex numbers of modulus 1 satisfying the relation:
$$\sum_{k=1}^n z_k^3 = 4(a + (a-n)i) - 3 \sum_{k=1}^n \overline{z_k}$$
We need to prove that $a \in \{0, 1, \ldots, n\}$ and $z_k \in \{1, i\}$ for all $k$.

2. **Expressing $z_k$ in Polar Form:**
Let $z_k = e^{i\theta_k} = \cos(\theta_k) + i\sin(\theta_k)$. Since $|z_k| = 1$, we have:
$$z_k^3 = (\cos(\theta_k) + i\sin(\theta_k))^3 = \cos(3\theta_k) + i\sin(3\theta_k)$$
and
$$\overline{z_k} = \cos(\theta_k) - i\sin(\theta_k)$$

3. Substituting into the Given Equation:
Substitute $z_k^3$ and $\overline{z_k}$ into the given equation:
$$\sum_{k=1}^n (\cos(3\theta_k) + i\sin(3\theta_k)) = 4(a + (a-n)i) - 3 \sum_{k=1}^n (\cos(\theta_k) - i\sin(\theta_k))$$

4. Separating Real and Imaginary Parts:
Separate the real and imaginary parts of the equation:
$$\sum_{k=1}^n \cos(3\theta_k) + i \sum_{k=1}^n \sin(3\theta_k) = 4a + 4(a-n)i - 3 \sum_{k=1}^n \cos(\theta_k) + 3i \sum_{k=1}^n \sin(\theta_k)$$
Equate the real and imaginary parts:
$$\sum_{k=1}^n \cos(3\theta_k) = 4a - 3 \sum_{k=1}^n \cos(\theta_k)$$
$$\sum_{k=1}^n \sin(3\theta_k) = 4(a-n) + 3 \sum_{k=1}^n \sin(\theta_k)$$

5. Using Trigonometric Identities:
Recall the trigonometric identities:
$$\cos(3\theta) = 4\cos^3(\theta) - 3\cos(\theta)$$
$$\sin(3\theta) = 3\sin(\theta) - 4\sin^3(\theta)$$
Substitute these into the equations:
$$\sum_{k=1}^n (4\cos^3(\theta_k) - 3\cos(\theta_k)) = 4a - 3 \sum_{k=1}^n \cos(\theta_k)$$
$$\sum_{k=1}^n (3\sin(\theta_k) - 4\sin^3(\theta_k)) = 4(a-n) + 3 \sum_{k=1}^n \sin(\theta_k)$$

6. Simplifying the Equations:
Simplify the real part equation:
$$4 \sum_{k=1}^n \cos^3(\theta_k) - 3 \sum_{k=1}^n \cos(\theta_k) = 4a - 3 \sum_{k=1}^n \cos(\theta_k)$$
$$4 \sum_{k=1}^n \cos^3(\theta_k) = 4a$$
$$\sum_{k=1}^n \cos^3(\theta_k) = a$$
Simplify the imaginary part equation:
$$3 \sum_{k=1}^n \sin(\theta_k) - 4 \sum_{k=1}^n \sin^3(\theta_k) = 4(a-n) + 3 \sum_{k=1}^n \sin(\theta_k)$$
$$-4 \sum_{k=1}^n \sin^3(\theta_k) = 4(a-n)$$
$$\sum_{k=1}^n \sin^3(\theta_k) = n - a$$

7. Combining the Results:
Combine the results:
$$\sum_{k=1}^n \cos^3(\theta_k) + \sum_{k=1}^n \sin^3(\theta_k) = a + (n - a) = n$$
From the lemma, we know that $\cos^3(\theta_k) + \sin^3(\theta_k) \leq 1$. Therefore:
$$\sum_{k=1}^n (\cos^3(\theta_k) + \sin^3(\theta_k)) \leq n$$
Since the sum equals $n$, it must be that $\cos^3(\theta_k) + \sin^3(\theta_k) = 1$ for all $k$.

8. **Determining $\theta_k$:**
The equality $\cos^3(\theta_k) + \sin^3(\theta_k) = 1$ holds if and only if $\theta_k = 2m\pi$ or $\theta_k = \frac{\pi}{2} + 2m\pi$ for some integer $m$. Thus:
$$z_k = e^{i\theta_k} = 1 \text{ or } i$$

9. Conclusion:
Since $\cos^3(\theta_k)$ can only be 0 or 1, $a$ must be an integer in the range $\{0, 1, \ldots, n\}$.

The final answer is $\boxed{ a \in \{0, 1, \ldots, n\} }$ and $z_k \in \{1, i\}$ for all $k$.