How can I prove that the sum $4^n+n^4$ is a composite number for all $n>1$.

Question

How can I prove that $4^n+n^4$ is always a composite number? For all $n>1$ and $n$ is an integer.

Answer 1

Hint:

Use Sophie Germain's identity: $$x^4+4y^4=(x^2-2xy+2y^2)(x^2+2xy+2y^2).$$

Answer 2

Consider the question in two cases:

$1$.If $n$ is even, $n^4+4^n$ is divisible by $4$ and hence it is composite.

$2$.If n is odd, suppose $n=2p+1$, where $p$ is positive integer.

Then $n^4+4^n$ = $n^4+4*4^{2p}$= ${n^4+4(2^p)^4}$ which is of the form $n^4+4b^4 $ where $b$ is positive integer ($=2^p$).

$n^4+4b^4 =(n^4+4b^2+4b^4)-4b^2=(n^2+2b+2b^2)(n^2-2b+2b^2)$ .

Now $(n^2-2b+2b^2)$ can not be equal to $1$ because $n>1$ and $2b(b-1)>1$ {since $b=2^p$}.

Also by the same reason $(n^2+2b+2b^2)$ can't be equal to $1$. So we get that $(n^2+2b+2b^2)(n^2-2b+2b^2)$ is not same as $(n^4+4b^4)(1)$ and hence we find that $n^4+4b^4$ is composite number consequently $n^4+4^n$ is composite number when $n$ is odd.

Hence, $n^4+4^n$ is composite for $n>1$. We are done.

Answer 3

HINT:

There are two cases to consider, $n$ even and $n$ odd. Clearly 2 divides $n^4+4^n$ for $n$ even. For the odd case: $$n^4 + 4^{2m+1}= (n^2)^2 + (2^{2m+1})^2 = (n^2 + 2^{2m+1})^2 − 2^{2m+2}n^2 = (n^2 + 2^{2m+1} + 2^{m+1}n)(n^2 + 2^{2m+1} − 2^{m+1}n)$$