Mathematics problem (number theory)?

Question

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

Answer 1

try two diffrenet numbers like 2 or 3 and you can get that composite number

Answer 2

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

QED

Answer 3

One may think of the congruence modulo 5,

if n and 4 are prime together then 4^n = 4 (modulo 5)

and n^ 4 =1 (modulo 5) thus the sum is 0 (modulo 5).

if n and 4 are not prime together then 2 is a factor of the sum.

With some basic group theory, it would be easy to work it out.