Form analysis — Every prime of the form 3k + 1 can be written as 6m + 1 when which condition on k holds?

Introduction / Context:
Primes greater than 3 are of the form 6m ± 1, because any other residue class modulo 6 makes the number divisible by 2 or 3. The question ties this fact to primes represented as 3k + 1 and asks what must be true about k to rewrite the prime as 6m + 1.

Given Data / Assumptions:

• Let p be a prime number with p = 3k + 1.
• p must be odd (since p > 2 for primes in this set).
• We seek when p can be written as 6m + 1.

Concept / Approach:
Analyze parity: if p = 3k + 1 is to be odd, then 3k must be even, which forces k to be even. Writing k = 2t leads to p = 3(2t) + 1 = 6t + 1. Hence the required condition is that k be even.

Step-by-Step Solution:
Let p = 3k + 1 be prime and > 3, so p is odd.For p to be odd, 3k must be even → k is even.Set k = 2t → p = 3(2t) + 1 = 6t + 1.Therefore, any prime of the form 3k + 1 corresponds to 6m + 1 if and only if k is even.

Verification / Alternative check:
Try examples: p = 7 → 3k + 1 = 7 → k = 2 (even), and 7 = 6*1 + 1; p = 13 → k = 4 (even), and 13 = 6*2 + 1.

Why Other Options Are Wrong:

• k odd / both odd and even: Violates parity; 3k + 1 would be even when k is odd (and > 2, therefore composite).
• No such form: Contradicted by numerous examples (7, 13, 19, 31, …).

Common Pitfalls:
Overlooking parity constraints; assuming all 3k + 1 primes can take both 6m ± 1 forms; forgetting that primes > 3 must be 6m ± 1.

Final Answer:
k is even