Divisible problem

If $n$ is even then $2^{n+1}+(-1)^n \equiv 2+1 \equiv 0 \bmod 3$ . If $n$ is odd then $2^{n+1}+(-1)^n \equiv 1+2 \equiv 0 \bmod 3$ .

Let $f(n) = 2^{n+1}+(-1)^n$ . When $n$ is odd, let $n=2m-1$ , where $m$ is a positive integer. Then $f(n) = 2^{2m} - 1$ . Since $\gcd(2,3)=1$ , we can apply Euler's theorem and

$\begin{aligned} f(n) & \equiv 2^{2m\ \bmod \color{#3D99F6}\ \phi(3)} - 1 & \small \color{#3D99F6} \text{where }\phi (\cdot) \text{ denotes the Euler's totient function,} \\ & \equiv 2^{2m\ \bmod \color{#3D99F6}\ 2} - 1 & \small \color{#3D99F6} \text{and since 3 is a prime, }\phi (3) = 3-1 = 2. \\ & \equiv 0 \text{ (mod 3)} \end{aligned}$

Therefore, $f(n)$ is divisible by 3 when $n$ is odd.

When $n$ is even, let $n=2m$ . Then $f(n) \equiv 2^{2m+1} + 1 \equiv 2^{(2m+1) \bmod 2} + 1 \equiv 2+1 \equiv 0 \text{ (mod 3)}$ . Again $f(n)$ is divisible by 3 when $n$ is even.

Yes , $f(n) 2^{n+1}+(-1)^n$ is divisible by 3 for every positive integer $n$ .

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Reference: Euler's totient function