Show $\gcd(8^n, 4^n +2^n +1)=1$

Question

Show that for all $n\in\mathbb{N},\,\gcd(8^n, 4^n +2^n +1)=1$. I understand one way is to use Extended Euclidean algorithm but the process seems infinitely long. How may I use any easy method to show this identity?

Answer 1

$8^n-(2^n-1)(4^n+2^n+1)=1$ Hence $gcd(8^n,4^n+2^n+1)=1$.

Answer 2

Hint: The only prime that divides $8^n$ is $2$. Does $2$ divide $4^n+2^n+1$?

Answer 3

Observe that:

• $8^n=2^{3n}$
• $4^n+2^n+1=$ even $+$ even $+$ odd $=$ odd

In other words:

• $2$ is the only prime factor of $8^n$
• $2$ is not a prime factor of $4^n+2^n+1$

Therefore $\gcd(8^n,4^n+2^n+1)=1$.