Let `n > 1`, be a positive integer. Then the largest integer m, such that `(n^m + 1)` divides `(1 + n + n^2 + n^3 + …..+ n^(127))` is:

Options

A. a. 127
B. b. 63
C. c. 64
D. d. 32

Answer

To solve the problem, we need to find the largest integer \( m \) such that \( n^m + 1 \) divides the sum \( 1 + n + n^2 + n^3 + \ldots + n^{127} \).

### Step-by-Step Solution:

1. **Identify the Sum**:
The expression \( 1 + n + n^2 + n^3 + \ldots + n^{127} \) is a geometric series. The formula for the sum of a geometric series is given by:
\[ S = \frac{a(r^n - 1)}{r - 1} \]
where \( a \) is the first term, \( r \) is the common ratio, and \( n \) is the number of terms. Here, \( a = 1 \), \( r = n \), and there are \( 128 \) terms (from \( n^0 \) to \( n^{127} \)). Thus, we can write:
\[ S = \frac{1(n^{128} - 1)}{n - 1} = \frac{n^{128} - 1}{n - 1} \]

2. **Factor the Numerator**:
We can factor \( n^{128} - 1 \) using the difference of squares:
\[ n^{128} - 1 = (n^{64} - 1)(n^{64} + 1) \]
Continuing to factor \( n^{64} - 1 \):
\[ n^{64} - 1 = (n^{32} - 1)(n^{32} + 1) \]
and so on, until we reach:
\[ n^{128} - 1 = (n - 1)(n + 1)(n^2 + 1)(n^4 + 1)(n^8 + 1)(n^{16} + 1)(n^{32} + 1)(n^{64} + 1) \]

3. **Divisibility by \( n^m + 1 \)**:
We need to check which factors of \( n^{128} - 1 \) can be expressed in the form \( n^m + 1 \). The factor \( n^{64} + 1 \) is of the form \( n^m + 1 \) where \( m = 64 \).

4. **Determine the Largest \( m \)**:
To find the largest \( m \) such that \( n^m + 1 \) divides \( n^{128} - 1 \), we observe that:
- \( n^1 + 1 \) divides \( n^{128} - 1 \)
- \( n^2 + 1 \) divides \( n^{128} - 1 \)
- \( n^4 + 1 \) divides \( n^{128} - 1 \)
- \( n^8 + 1 \) divides \( n^{128} - 1 \)
- \( n^{16} + 1 \) divides \( n^{128} - 1 \)
- \( n^{32} + 1 \) divides \( n^{128} - 1 \)
- \( n^{64} + 1 \) divides \( n^{128} - 1 \)

The largest \( m \) for which \( n^m + 1 \) divides \( n^{128} - 1 \) is therefore \( m = 64 \).

### Conclusion:
The largest integer \( m \) such that \( n^m + 1 \) divides \( 1 + n + n^2 + n^3 + \ldots + n^{127} \) is:
\[ \boxed{64} \]