Show that 3 is the only natural number p such that p, p + 2 and p + 4 are all pr

Question

Show that 3 is the only natural number p such that p, p + 2 and p + 4 are all prime.

Explanation / Answer

Suppose x, x+2, and x+4 are prime and x > 3. Well, x is not a multiple of three because if it were, then x would not be prime. So x is either one more than a multiple of three or two more than a multiple of three. In the first case (x is one more than a multiple of three), x+2 will be a multiple of three and hence won't be prime (contrary to our assumption). In the second case, x+4 will be a multiple of three - another contradiction. Thus we have a contradiction in all situations, which means that the assumption must be invalid. Thus 3, 5, 7 is the only such prime triplet.

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