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