If x and (x² + 8) are both primes, then prove that (x³ +16) is also a prime.
(Explained more clearly than in my earlier comment).
First, note that in every sequence of 3 consecutive integers, exactly ONE of them is evenly divisible by 3.
Also note that every prime (other than 3) is NOT evenly divisible by 3.
Therefore for EVERY prime other than 3, EITHER the integer BEFORE, or the integer AFTER *is* divisible by 3.
Now, given that x is a prime, not equal to 3:
x²+8 = (x²-1) + 9
= (x-1)(x+1) + 9
Since either (x-1) or (x+1) is divisible by 3, (x-1)(x+1) is ALSO divisible by 3. And if we then add nine to it, the result remains divisible by 3.
Restated: If x is a prime other than 3, then x²+8 is divisible by 3.
Therefore, the only occasion that x and (x² + 8) are BOTH prime is when x=3.
Since (3³ + 16) = 43 which is prime, we've proved the initial supposition!
Edited on September 10, 2003, 1:41 am
Solution (spoiler)
