Prove $4^{n} -1$ is divisible by $3$ for all $n\geq1$ [duplicate]

For any $n\ge1$ it will be the case that exactly one of $2^n-1,2^n, 2^n+1$ will be divisible by $3$.

Since $2^n$ is not divisible by $3$ then the product $(2^n-1)(2^n+1)=4^n-1$ must be divisible by $3$.

You could do it this way: You have $4^k - 1 = 3m$, so $4^k=3m+1$. Now multiply both sides by $4$:

$$4^{k+1}=4(3m+1) = 12m+4 = 12m+3+1=3(4m+1)+1=3n+1$$

Thus, $4^{k+1}-1 = 3n$, as desired.

$$ 4^{k+1} - 1 = 4\cdot4^k-1 = 3\cdot4^k + 4^k-1. $$

Therefore via induction we know $4^k-1$ is divisible by three, and the $3\cdot 4^{k}$ is clearly divisible by $3$.

You can use the method of induction to prove the exercise.

For $n=1$, $4^n-1=4^1-1=3$ is divisible by $3$.

For $n=k$, assume $4^k-1$ is divisible by $3$, so $4^k-1=3m$ for some integer $m$.

For $n=k+1$, $4^n-1=4^{k+1}-1=4^k.4-4+3=4.(4^k-1)+3=4.3m+3=3.(4m+1)$ which is definitely divisible by $3$.

Thus, by the method of induction, our problem is solved for all $n\ge 1$