If n is even, then 2 divides n^{4} + 4^{n}.

Clearly (n^{4} + 4^{n})/2 > 2, for n > 1.

This establishes the result for even n.

If n is odd, we will write n^{4} + 4^{n} as the difference of two squares of integers, and hence obtain a factorization. Setting n = 2m + 1, we have

<table><tbody><tr class="h"><td class="u">n

^{4} + 4

^{2m+1}</td><td> = (n

^{2})

^{2} + (2

^{2m+1})

^{2}</td></tr>
<tr><td> </td><td> = (n

^{2} + 2

^{2m+1})

^{2} − 2

^{2m+2}n

^{2}</td></tr><tr><td> </td><td> = (n

^{2} + 2

^{2m+1} + 2

^{m+1}n)(n

^{2} + 2

^{2m+1} − 2

^{m+1}n)</td></tr></tbody></table>

We must now show that both factors are greater than one.

Clearly n^{2} + 2^{2m+1} + 2^{m+1}n > 1, for n > 1.

<table><tbody><tr class="h"><td class="u">Next consider f(n,m)</td><td> = n

^{2} + 2

^{2m+1} − 2

^{m+1}n</td></tr><tr><td> </td><td> = (n − 2

^{m})

^{2} + 2

^{2m}</td></tr></tbody></table>

For n > 1 (m > 0), 2^{2m} > 1; hence f(n,m) > 1 for all odd n > 1.

Therefore n^{4} + 4^{n} is composite for all integers n > 1.