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Prove that for every pair of twin primes except for the pair (3,5) that the number between them is divisible by 6.”

User Amolv
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Answer:

(3,5)

Explanation:

Twin primes are a pair of prime numbers that differ by 2. Let's consider a pair of twin primes, p and p + 2 (where p > 3). We want to prove that the number between them, p + 1, is divisible by 6.

We know that every prime number greater than 3 can be written in the form 6k ± 1 for some integer k. This is because any integer can be written in one of six possible forms: 6k, 6k + 1, 6k + 2, 6k + 3, 6k + 4, or 6k + 5. However, we can eliminate the forms that are divisible by 2 or 3 (except for 2 and 3 themselves), leaving only 6k ± 1 and 6k ± 5. Since twin primes differ by 2, they must both be of the form 6k ± 1.

Let's consider p, the smaller of the twin primes. We know that p is of the form 6k ± 1 for some integer k. If p is of the form 6k + 1, then p + 2 is of the form 6k + 3, which is not prime (since it is divisible by 3). Therefore, p must be of the form 6k - 1. Then, p + 1 is of the form 6k, which means it is divisible by 2 and by 3.

Similarly, if p + 2 is of the form 6k - 1, then p is of the form 6k - 3, which is not prime (since it is divisible by 3). Therefore, p + 2 must be of the form 6k + 1. Then, p + 1 is of the form 6k, which means it is divisible by 2 and by 3.

Therefore, in either case, the number between the twin primes, p + 1, is divisible by 6. We have shown that this is true for any pair of twin primes except for the pair (3, 5).

User Fielding
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