For any odd prime number p there exists a positive integer k where `1 < k < p`, such that the remainder of `(k^2)/p` is 1. Then the number of positive integers k is :

Options

A. a. `0`
B. b. b. `1`
C. c. `p - 1`
D. d. can't be determined

Answer

To solve the problem, we need to find the number of positive integers \( k \) such that \( 1 < k < p \) and the remainder of \( k^2 \) when divided by \( p \) is 1.

### Step-by-Step Solution:

1. **Understanding the Condition**: We need to find integers \( k \) such that \( k^2 \equiv 1 \mod p \). This means that when \( k^2 \) is divided by \( p \), the remainder is 1.

2. **Rearranging the Equation**: The equation \( k^2 \equiv 1 \mod p \) can be rewritten as:
\[ k^2 - 1 \equiv 0 \mod p \]
This implies that \( (k - 1)(k + 1) \equiv 0 \mod p \).

3. **Finding the Roots**: For the product \( (k - 1)(k + 1) \equiv 0 \mod p \) to hold, at least one of the factors must be congruent to 0 modulo \( p \). This gives us two cases:
- \( k - 1 \equiv 0 \mod p \) which simplifies to \( k \equiv 1 \mod p \)
- \( k + 1 \equiv 0 \mod p \) which simplifies to \( k \equiv -1 \mod p \) or \( k \equiv p - 1 \mod p \)

4. **Identifying Valid Values of \( k \)**:
- From \( k \equiv 1 \mod p \), we find \( k = 1 \), which does not satisfy the condition \( 1 < k < p \).
- From \( k \equiv p - 1 \mod p \), we find \( k = p - 1 \), which satisfies \( 1 < k < p \).

5. **Conclusion**: The only valid value of \( k \) that satisfies the condition \( 1 < k < p \) is \( k = p - 1 \). Therefore, there is exactly one positive integer \( k \) for any odd prime \( p \).

### Final Answer:
The number of positive integers \( k \) such that \( 1 < k < p \) and \( k^2 \equiv 1 \mod p \) is **1**.