(A) Determine all possible prime numbers f such that each of (f+1)/2 and (f-1)/4 are prime numbers.

(B) What are the possible prime numbers g such that each of (g+1)/4 and (g-1)/2 are prime numbers?

By inspection, f=2 isn't a solution since (2-1)/4 is not an integer. So f is odd. Also, if f = 4k-1 for some k then f fails because (f+1)/2 = (4k-1+1)/2 =2k is non prime. (Technically, k=1 might work but that makes f=3 and (3-1)/4 is not an integer.) So f = 4k+1 for some k

Now, f (4k+1) is prime, and (f+1)/2 = 2k+1 is prime and (f-1)/4 = k is prime. Consider these three modulo three. At least one must = 0 mod 3 (if k=0 mod 3, it's k, if k=1 mod 3 then 2k+1 = 0 mod three and if k=2 mod three then 4k+1 = 0 mod three.) Since tall are assumed to be prime, at least one must = 3 itself, and only k=3 results in all integer values for the rest, making f=13 the complete set of primes meeting these conditions.

g can be handled in the same way, except now g must = 4k-1 where k and 2k-1 are both prime. The same mod 3 logic forces k=3 and the result is that g=11 is the only solution.

Posted by Paul on 2007-08-21 15:56:52