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

Options

A. `n=6k+1`
B. `n=6k-1`
C. `n=3k+1`
D. none of these

Answer

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

### Step 1: Factor the divisor
The polynomial \( x^2 - x + 1 \) can be factored using its roots. The roots are given by:
\[ \omega = e^{2\pi i / 3} \quad \text{and} \quad \omega^2 = e^{-2\pi i / 3} \]
Thus, we can express \( x^2 - x + 1 \) as:
\[ x^2 - x + 1 = (x - \omega)(x - \omega^2) \]

### Step 2: Apply the divisibility condition
For \( x^{n+1} - x^n + 1 \) to be divisible by \( x^2 - x + 1 \), it must equal zero when \( x = \omega \) and \( x = \omega^2 \). Therefore, we need to check:
1. \( f(\omega) = \omega^{n+1} - \omega^n + 1 = 0 \)
2. \( f(\omega^2) = (\omega^2)^{n+1} - (\omega^2)^n + 1 = 0 \)

### Step 3: Evaluate \( f(\omega) \)
Calculating \( f(\omega) \):
\[ f(\omega) = \omega^{n+1} - \omega^n + 1 \]
Factoring out \( \omega^n \):
\[ f(\omega) = \omega^n(\omega - 1) + 1 \]
Setting this equal to zero gives:
\[ \omega^n(\omega - 1) + 1 = 0 \implies \omega^n(\omega - 1) = -1 \]

### Step 4: Evaluate \( f(\omega^2) \)
Calculating \( f(\omega^2) \):
\[ f(\omega^2) = (\omega^2)^{n+1} - (\omega^2)^n + 1 \]
Factoring out \( (\omega^2)^n \):
\[ f(\omega^2) = (\omega^2)^n(\omega^2 - 1) + 1 \]
Setting this equal to zero gives:
\[ (\omega^2)^n(\omega^2 - 1) + 1 = 0 \implies (\omega^2)^n(\omega^2 - 1) = -1 \]

### Step 5: Analyze the conditions
From both evaluations, we have two conditions:
1. \( \omega^n(\omega - 1) = -1 \)
2. \( (\omega^2)^n(\omega^2 - 1) = -1 \)

### Step 6: Determine values of \( n \)
To satisfy these equations, we observe the periodic nature of powers of \( \omega \):
- \( \omega^3 = 1 \)
- Thus, \( \omega^n \) will repeat every 3 values.

We find that:
- For \( n \equiv 0 \mod 3 \): \( \omega^n = 1 \)
- For \( n \equiv 1 \mod 3 \): \( \omega^n = \omega \)
- For \( n \equiv 2 \mod 3 \): \( \omega^n = \omega^2 \)

By substituting these into the conditions, we find that \( n \) must satisfy:
- \( n \equiv 1 \mod 3 \)

### Conclusion
Thus, the condition that \( x^{n+1} - x^n + 1 \) is divisible by \( x^2 - x + 1 \) is:
\[ \boxed{n \equiv 1 \mod 3} \]