2011/046) let n be any positive integer >1. then show that n^4+4^n is a composite number

there are 2 cases

1) n is even.

n^4+4^n > 2 and even so composite

2) n is odd

n^4 + 4^n
= n^4 + 2n^2*2^n + 4^n - 2n^2*2^n
= (n^2 + 2^n)^2 - 2n^2*2^n
= (n^2 + 2^n)^2 - n^2*2^(n+1)
= (n^2 + 2^n - n * 2^((n+1)/2)) * (n^2 + 2^n + n * 2^((n+1)/2)). Therefore,

if n is odd it has 2 factors so composite if 1st term is not 1 as second term > 1

for n > 1 1st term is > 1 and hence the given expression is composite