Prove this is Composite?

We can patch this : For even n, the number n^4+4^n

is even and >2 making it composite . For odd n

we have n^4+4^n = n^4+2^(2n)+n^2* 2^(n+1)-n^2*2^(n+1)

=n^4+2^(n+1)*n^2+2^(2n)-n^2*2^(n+1)

={n^2 + 2^n}^2 - {n*2^((n+1)/2)}^2

where (n+1)/2 is an integer with n odd and we have the

factors,

{n^2+n*2^((n+1)/2)+2^n}{n^2 -n*2^((n+1)/2)+2^n}

and the factor on the far right exceeds 1 if n>1.

Therefore n^4+4^n is composite for all natural n>1.

The statement is obvious true when n is even since n^4 + 4^n will be an even number greater than 2.

Also, if n > 1 and n = 10k + 1, 10k +3, 10k + 7 or 10k +9, let t = 1, 3, 7, or 9, we will have n^4 = (10k + t)^4 = 1 mod 10 and 4^(10k + t) = 4 mod 10, so n^4 + 4^n = 5 mod 10 i.e., it's divisible by 5, so it's composite.

Finally, for n = 10k + 5, observe that n^4 + 4^n = (n^2)^2 + (2^n)^2 and (n^2 + 2^n)^2 = n^4 + 4^n + 2(n^2)(2^n) i.e., n^4 + 4^n = (n^2 + 2^n)^2 - 2(n^2)(2^n) = (n^2 + 2^n)^2 - n^2*2^(n+1). But n^2*2^(n+1) is a perfect square since n^2 is a square and 2^(n+1) is also a square since n=10k +5 is odd, so it makes (n^2 + 2^n)^2 - (n^2)(2^(n+1)) a difference of two squares, which is factorable. So n^4 + 4^n is composite when n=10k +5. [Note: I actually can include all n = 2k + 1 = odd in this paragraph and omit the 2nd paragraph I wrote, but I didn't realize the difference of two squares until I single out the case when n = 10k +5, but anyway the proof is complete. ]

The statement is false. Let x = 3, n = 1. Then N = 3^1 - 1 = 2, a prime.

n=1

1^4 + 4^1

= 1 + 4

= 5

5 is prime.

Something is wrong with your statement.