Numbers

I. p and p² + 8 both are prime
Since p is prime, we can say that for p ≥ 5, p = 6k ± 1

=> p² + 8 = 36k² ± 12k + 9
(a multiple of 3, so can not be prime)

=> p can only be 3

For p = 3, p³ + 8p + 2 = 53 which is prime

So, correct

II. a⁴ + 4 = (a² + 2)² – 4a² = (a² + 2)² = (2a)²
=> a⁴ + 4 = (a² – 2a + 2)(a² + 2a + 2)

Now, (a² + 2a + 2) = (a + 1)² + 1, so always greater than 5 for a > 1
(a² – 2a + 2) = (a – 1)² + 1, so always greater than or equal to 2 for a > 1

Hence always composite for a > 1