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Getting Primed With Subtraction And Addition (Posted on 2007-12-26) Difficulty: 2 of 5
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    
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Solution: (Hide)
Only p=11 satisfies the conditions of the problem.


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.

Comments: ( You must be logged in to post comments.)
  Subject Author Date
re: solution slightly differentlyK Sengupta2008-03-06 04:32:20
Solutionsolution slightly differentlyPaul2007-12-27 21:05:57
SolutionSolutionPraneeth2007-12-27 00:49:41
solution/spoilerxdog2007-12-26 12:28:44
Some Thoughtscomputer exploration (spoiler)Charlie2007-12-26 11:59:39
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