Answer to Question #100902 in Algebra for Salman

Question

Prove that n^4+4^n is not a prime number for any integer n greater than 1

Answer 1

Consider 4ⁿ+n⁴;

$\forall$n, n$\in$ N /{1} it is true that

n=2m+1 or n=2m.

If n=2m+1,

then 4^(2m+1)+(2m+1)⁴ = 2^2(2m+1)+ ((2m+1)²)² = 2^2(2m+1)+ ((2m+1)²)² + 2 ×2^2m+1 ×(2m+1)² - 2 ×2^2m+1 ×(2m+1)²= (2^2m+1 + (2m+1)²)² - 2^2m+2 × (2m+1)² = (2^2m+1 + (2m+1)²)² - (2^m+1×(2m+1))²=(2^2m+1+(2m+1) - 2^m+1×(2m+1))(2^2m+1+(2m+1)+2^m+1×(2m+1)) is a product of two multipliers, they are not 1 simultaneously, hence the product is not prime.

If n=2m,

then

4^2m+(2m)⁴ is an even number because

4^2m is an even number $\forall$ m, m$\in$ N,

(2m)⁴ is an even number $\forall$ m, m$\isin$ N.

Thus, 4^2m+(2m)⁴ is an even number.