Another way to prove that $81^{3^{n}}+4^{2n+1}$ is not prime for any positive integer $n$

Question

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.

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

Answer 1

\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.

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

Answer 2

Yes, there are other proofs. Instead of showing $81^{3^{n}}+4^{2n+1}\equiv 0\bmod 5$ you can also show that $81^{3^{n}}+4^{2n+1}\equiv 0 \bmod 17$ for all $n\in \Bbb N$.

Answer 3

As has already been noted, the problem is much easier modulo $5$:

$81^{3^{n}}+4^{2n+1}\equiv1^{3n}+(-1)^{2n+1} \equiv 0\pmod 5$.

Since $81^{3^{n}}+4^{2n+1} > 5$ for all $n$ we are done.