If n is an even number

Let n = 2k (k>1)

n^4+4^n

= (2k)^4 +4^{2k}

=2^4\cdot k^4+4\cdot 4^{2k-1}

=4(4k^4+4^{2k-1})

Therefore n^4+4^n is a composite number

If n is an odd number

Let n = 2k+1 (k>1)

n^4+4^n

= (n^2)^2 +(2^n)^2

= (n^2+2^n)^2 - 2n^22^n

= (n^2+2^n)^2 - n^22^{n+1}

= (n^2+2^n)^2 - n^22^{2k+2}

= (n^2+2^n)^2 - (n2^{k+})^2

=(n^2+2^n+n2^{k+})(n^2+2^n-n2^{k+})

Therefore

n^4+4^n is composite number when n is odd.

In summary, n^4+4^n is a composite number for both even and odd integers that is larger than 1