If x and (x² + 8) are both primes, then prove that (x³ +16) is also a prime.
At the outset, we note that 2 is the only even prime.
But for x=2, we have x^3+8=16, which is not a prime. Contradiction.
If x is a odd prime then either x=3k+/-1 or x=3
Now, x=3k+/-1=> x =+/-1 (mod 3)
=> x^2 =1(mod 3)
=> x^2+8 = 9 (mod 3) = 0 (mod 3)
This is a contradiction, and accordingly:
x=3k, which is composite for all k>=2, and accordingly:
k=1, giving x=3
Thus, if x and (x^2+8) are both primes, then we must have x=3
And 3^3+16= 43, which is a prime number.
Consequently, when x and x^2+8 are both primes then x^3+16 must be a prime, and this is satisfied for a lone x value, viz, 3.
As a check, we can easily verify that for x^2+8=17, which is indeed a prime number.
Edited on December 14, 2021, 2:14 am
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