If 1+2^n+4^n is prime then prove that n is 3^k.Both variables are natural numbers.
Help would be appreciated.
What have you tried so far? My gut is telling me to use the division algorithm $\displaystyle n = 3^k q + r$. Then since the order of 2 in the multiplicative group of $\displaystyle \mathbb{Z} / 3^{k+1}\mathbb{Z}$ is known to be $\displaystyle (3-1)\cdot 3^{k+1-1} = 2\cdot 3^k$, you can further use the division algorithm on $\displaystyle 4^{3^kq+r} = 4^r\left(4^{3^k}\right)^q = 4^r\left(1+3^{k+1} x\right)^q$ for some $\displaystyle x \in \mathbb{Z}$. Maybe an induction argument will arise that will yield that if $\displaystyle n \neq 3^k$ then 3 divides $\displaystyle 1+2^n + 4^n$.