## Never Prime

#### Problem

Prove that 14^{n} + 11 is never prime.

#### Solution

In problems of this nature it is often helpful to substitute values for `n` to see if anything useful can be pertained. Thus the first seven terms are 25, 207, 2755, 38427, 537835, 7529547, respectively. Although we cannot be certain, it seems when `n` is odd, 14^{n} + 11 is divisible by 5, and, although not entirely obvious, when `n` is even it is divisible by 3.

Let us consider `n` being even: 14^{2k} = 196^{k}. As 196 1 mod 3, it follows that 196^{k} 1 mod 3. Therefore 14^{2k} + 11 is divisible by 3.

When `n` is odd: 14^{2k+1} = 1414^{2k} = 14196^{k}. As 196 1 mod 5, it follows that 196^{k} 1 mod 5, and 14196^{k} 14 4 mod 5. Therefore 14^{2k+1} + 11 is divisible by 5.

Hence 14^{n} + 11 is divisible by 5 and 3 alternately, and can never be prime.

Of course, with this insight we can approach it far more efficiently by noting that 14 -1 mod 15, hence 14^{n} will be alternately -1/1.

When `n` is odd: 14^{n} + 11 -1 + 11 = 10 mod 15 divisible by 5

When `n` is even: 14^{n} + 11 1 + 11 = 12 mod 15 divisible by 3

Prove that (14^{n} + 4)/2 is never prime.