Every prime number of the form 3k+1 can be represented in th form 6m+1 (where k, m are integers), when

Options

A. k is odd
B. k is even
C. k can be both odd and even
D. No such form is possible

Answer

To solve the problem, we need to show that every prime number of the form \(3k + 1\) can be represented in the form \(6m + 1\), where \(k\) and \(m\) are integers.

### Step-by-Step Solution:

1. **Understanding the Forms**:
- We start with the prime number expressed as \(3k + 1\).
- We want to see if this can be expressed in the form \(6m + 1\).

2. **Assuming Values for \(k\)**:
- Let's first consider the case when \(k\) is even.
- We can express \(k\) as \(k = 2n\) for some integer \(n\).

3. **Calculating the Prime Number**:
- Substituting \(k = 2n\) into \(3k + 1\):
\[ 3k + 1 = 3(2n) + 1 = 6n + 1 \]
- This is of the form \(6m + 1\) where \(m = n\). Thus, when \(k\) is even, the prime number can be expressed in the desired form.

4. **Considering Odd Values of \(k\)**:
- Now, let’s consider \(k\) as odd. We can express \(k\) as \(k = 2n + 1\) for some integer \(n\).
- Substituting this into \(3k + 1\):
\[ 3k + 1 = 3(2n + 1) + 1 = 6n + 3 + 1 = 6n + 4 \]
- This can be rewritten as \(6m + 4\), which is not of the form \(6m + 1\).

5. **Conclusion**:
- From our analysis, we find that when \(k\) is even, \(3k + 1\) can be expressed as \(6m + 1\).
- However, when \(k\) is odd, \(3k + 1\) cannot be expressed in the form \(6m + 1\).
- Therefore, every prime number of the form \(3k + 1\) can be represented in the form \(6m + 1\) only when \(k\) is even.

### Final Answer:
Every prime number of the form \(3k + 1\) can be represented in the form \(6m + 1\) when \(k\) is even.