If x and (x² + 8) are both primes, then prove that (x³ +16) is also a prime.
(In reply to
re(2): Second thought by SilverKnight)
I think I've got it!
When x is a prime number, it is (by definition) not divisible by 3 (except when x = 3). Now, note that in every sequence of 3 consecutive integers ONE of them IS divisible by 3. Therefore, for any given prime number x (other than 3), either (x-1) is divisible by 3 or (x+1) is divisible by 3.
so... x²+8 = (x²-1) + 9
= (x-1)(x+1) + 9
Since either (x-1) or (x+1) is divisible by 3 when x is prime, (x-1)(x+1) is ALSO divisible by 3. And if we then add nine to it, it remains divisible by 3.
Therefore, if x is prime then x²+8 cannot be prime unless x=3.
Since x must be 3, and 3³+16 (which happens to be 43) is prime, the initial hypothesis is true.
--- SK :-) (okay... now my head hurts more)
re(3): Second thought
