If n is even, n^4+4^n is even, and greater than 2, so it's not a prime, so we need consider only odd numbers. We write n^4+4^n= (n²)²+(2n)²= (n²+2n)²-n²2^(n+1)= (n²+2n+n2^(n+1)/2)(n²+2n-n2^(n+1)/2). As n is odd, (n+1)/2 is an integer, both factors are integers. Also, it's easy to prove that for n>1, both factors are also greater than 1, so n^4+4^n is composite. |