Another way to prove that $81^{3^{n}}+4^{2n+1}$ is not prime for any positive integer $n$ The question is to prove that $81^{3^{n}}+4^{2n+1}$ is not prime for any positive integer $n$.
The easy way to solve this is by taking the last digit of each number:

*

*$81^{3^{n}}$ ends in $1$ because $81$ also ends in $1$.

*$4^{2n+1}$ ends in $4$ as $2n + 1$ is odd.

So the sum will end in $4+1=5$ and because $81^{3^{n}}+4^{2n+1}>5$, it is $5|81^{3^{n}}+4^{2n+1}$, thus the number is not prime.

Are/is there any other proof(s) which do not reuse the same idea?

\begin{align}
81^{3n} + 4^{2n+1} &= (3^4)^{3n} + (2^2)^{2n+1} \\
&= (3^{6n})^2 + (2^{2n+1})^2 \\
&= (3^{6n})^2 + (2^{2n+1})^2 + 2 \times 3^{6n} 2^{2n+1}  - 2 \times 3^{6n} 2^{2n+1} \\
&= (3^{6n} + 2^{2n+1})^2 - 3^{6n} 2^{2n+2} \\
&= (3^{6n} + 2^{2n+1})^2 - (3^{3n} 2^{n+1})^2 \\
&= (3^{6n} + 2^{2n+1} - 3^{3n} 2^{n+1}) (3^{6n} + 2^{2n+1} + 3^{3n} 2^{n+1})
\end{align}
As you see $81^{3n} + 4^{2n+1}$ can be written as a product of two integers and none of them are $1$ for positive integers1, so it is not prime.

 1. Because both of them are increasing functions and above $1$ for $n=1$. 
