The condition that `x^(n+1)-x^(n)+1` shall be divisible by `x^(2)-x+1` is that

Options

A. `n=6x+1`
B. `n=6k-1`
C. `n=3k+1`
D. `n=3k-1`

Answer

To solve the problem of determining the condition under which \( x^{n+1} - x^n + 1 \) is divisible by \( x^2 - x + 1 \), we can follow these steps:

### Step 1: Identify the roots of the divisor
The polynomial \( x^2 - x + 1 \) has roots given by the complex numbers:
\[ \omega = \frac{1}{2} + \frac{\sqrt{3}}{2} i \quad \text{and} \quad \omega^2 = \frac{1}{2} - \frac{\sqrt{3}}{2} i \]
These roots are the primitive cube roots of unity, satisfying \( \omega^3 = 1 \) and \( 1 + \omega + \omega^2 = 0 \).

### Step 2: Use the roots to check divisibility
For \( x^{n+1} - x^n + 1 \) to be divisible by \( x^2 - x + 1 \), it must be zero when evaluated at the roots \( \omega \) and \( \omega^2 \).

### Step 3: Substitute \( x = \omega \)
Substituting \( x = \omega \) into the expression:
\[ \omega^{n+1} - \omega^n + 1 = 0 \]
This simplifies to:
\[ \omega^{n+1} - \omega^n + 1 = \omega^n (\omega - 1) + 1 = 0 \]
Thus, we have:
\[ \omega^n (\omega - 1) + 1 = 0 \]
Rearranging gives:
\[ \omega^n (\omega - 1) = -1 \]

### Step 4: Analyze the equation
Since \( \omega^3 = 1 \), we can express \( \omega^n \) in terms of \( n \mod 3 \):
- If \( n \equiv 0 \mod 3 \), then \( \omega^n = 1 \)
- If \( n \equiv 1 \mod 3 \), then \( \omega^n = \omega \)
- If \( n \equiv 2 \mod 3 \), then \( \omega^n = \omega^2 \)

### Step 5: Solve for each case
1. **If \( n \equiv 0 \mod 3 \)**:
\[ 1(\omega - 1) = -1 \implies \omega - 1 = -1 \implies \omega = 0 \quad \text{(not possible)} \]

2. **If \( n \equiv 1 \mod 3 \)**:
\[ \omega(\omega - 1) = -1 \implies \omega^2 - \omega = -1 \implies \omega^2 - \omega + 1 = 0 \quad \text{(true)} \]

3. **If \( n \equiv 2 \mod 3 \)**:
\[ \omega^2(\omega - 1) = -1 \implies \omega^3 - \omega^2 = -1 \implies 1 - \omega^2 = -1 \quad \text{(not possible)} \]

### Step 6: Conclusion
The only case that satisfies the condition is when \( n \equiv 1 \mod 3 \). Therefore, the condition that \( x^{n+1} - x^n + 1 \) is divisible by \( x^2 - x + 1 \) is:
\[ n \equiv 1 \mod 3 \]