3. Recall that if p and p + 2 are both prime we call them twin primes, and it is

Question

3. Recall that if p and p + 2 are both prime we call them twin primes, and it is not known if there are infinitely many twin primes. Now suppose we say that if p, p +2, and p+ 4 are all prime, we'll call them triplet primes. The primes 3, 5, and 7 are triplet primes. We can determine whether there are infinitely many triplet primes as follows. (a) When any positive integer is divided by 6 the remainder must be 0, 1, 2, 3, 4, or 5. So any positive integer can be written in one of the following forms: 6k, 6k +1, 6k + 2, 6k +3, 6k+ 4, or 6k +5 Use this to show that any prime number greater than 3 can only be of the form 6k +1 or 6k +5. (b) Now, using part (a), detcrmine whether or not there are infinitely many triplet primes.

Explanation / Answer

a. Clearly 6k is not a prime since it has 2 factors 2 and 3

Let us spare the case 6k+1

And 6k+2 =2(3k+1) , hence it is even, hence not a prime

And, 6k+3 = 3(2k+1) , hence it is a multiple of 3 and greater than 3, hence not a prime

And, 6k+4 = 2(3k+2) which is a multiple of 2, hence not a prime

So we are left with the case 6k+1 and 6k+5

So, any prime is of the form either 6k+1 or 6k+5

But certainly all numbers of the 2 forms are not prime (take, 25 and 35)

b. If we are to find 3 prime triplets (other than 3,5,7) such that they differ by 2 each, then, the first prime is either 6k+1 or 6k+5 for some k

If, the first prime p=6k+1, then, p+2 = 6k+3 which is not a prime.

Hence the first prime may be p=6k+5

Then, p+4= 6k+9, which is a multiple of 3, hence not a prime.

Thus we have exhausted all the cases annd conclude that there are no more prime triplets other than the given one.

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