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 Getting Primed With Subtraction And Addition (Posted on 2007-12-26)
Determine all possible primes p such that each of p-8, p-4, p+8 and p+12 are also primes.

 Submitted by K Sengupta No Rating Solution: (Hide) Only p=11 satisfies the conditions of the problem. EXPLANATION: Let us substitute p-8 = q, so that: (p-8, p-4, p+8, p+12) = (q, q+4, q+16, q+20) If q (Mod 3) = 0, then q is non prime unless q=3 If q (Mod 3) = 1, then q+20 is divisible by 3, an hence composite. This is a contradiction. If q(Mod 3) = 2, then each of q+4 and q+16 is divisible by 3, and hence both are composites. This is a contradiction. Accordingly, q=3 and consequently, p-8=3, so that p=11 is the only possible solution to the given problem. *** Also refer to the solution submitted by Paul in this location.

 Subject Author Date re: solution slightly differently K Sengupta 2008-03-06 04:32:20 solution slightly differently Paul 2007-12-27 21:05:57 Solution Praneeth 2007-12-27 00:49:41 solution/spoiler xdog 2007-12-26 12:28:44 computer exploration (spoiler) Charlie 2007-12-26 11:59:39

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