Let F(n) = n^4+4^n

Suppose that n is even. If so, n^4 is divisible by 4 and consequently F(n) is always divisible by 4 and threfore composite....(i).

Suppose that n is odd. Then,

F(n) = n^4+4^n

= (n^2+2^n)^2 - ((2^(n+1)/2)*n)^2

= (n^2+2^n + 2^(n+1)/2)(n^2+2^n - n*2^((n+1)/2)) .....(ii)

Since n is odd, it follows that 2^(n+1)/2 is always an integer.

It can easily be shown that:

n^2+2^n - n*2^((n+1)/2 >= 5, whenever n>=3

Consequently,n^2+2^n +/- n*2^((n+1)/2 >= 5, whenever n>=3

Consequently, combining (i) and (ii), it follows that F(n) is always

composite whenever n>1.