Question

Prove that $n^4+4^n$ is a composite number for all integer value of $n>1.$

Answer 1

If n is even, $n^4+4^n$ is divisible by 4
$\therefore$ It is composite number
If n is odd, suppose $n=2p+1$, where p is a positive integer
Then $n^4+4^n=n^4+{4.4}^{2p}=n^4+4(2^p{)}^4$
which is of the form $n^4+4b^4,$ where b is a positive integer $(=2^p)$
$n^4+4b^4=(n^4+4b^2+4b^4)-4b^2$
$=(n^2-2b^2{)}^2-(2b{)}^2$
$=(n^2+2b+2b^2)(n^2-2b+2b^2)$
We find that $n^4+4b^4$ is a composite number consequently $n^4+4^n$ is composite when n is odd.
Hence $n^4+4^n$ is composite for all integer values of n> 1.