Mathematical Reflections 2011, Issue 6 - Problem O211

Problem:
Prove that for any positive integer $n$ the number $(2^n+4^n)^2+4(6^n+9^n+12^n)$ has at least $9$ positive divisors.

Proposed by Titu Andreescu.

Solution:
Rewriting,
$$(2^n+4^n)^2+4\cdot3^n(2^n+4^n)+4\cdot9^n = (2^n+4^n+2\cdot3^n)^2 = 2^2(2^{n-1}+2^{2n-1}+3^n)^2.$$ Since $4$ has three positive divisors and $2^{n-1}+2^{2n-1}+3^n \neq 2^k, k \in \mathbb{N}$ for every positive integer $n$, then $2^{n-1}+2^{2n-1}+3^n$ has one prime divisor $p > 2$. So, there are at least $3 \cdot 3 = 9$ positive divisors for the given number, namely $$\{1,2,4,p,2p,4p,p^2,2p^2,4p^2\}.$$