Show that $2^n - (-1)^n = 3m$

Question

Show that $2^n - (-1)^n$ is divisible by 3 for all positive integers n

I tried:

let $n = 2k$ (Even numbers)

$n = 2k + 1$ (odd numbers)

but also, I didn't find the solution.

Answer 1

Using Modular_arithmetic,

For $n = 2k$, $$2^n - (-1)^n=4^k-1$$

Since $4^k \equiv 1 \quad\text{(mod 3)}$, $$2^n - (-1)^n\equiv 1-1\equiv 0 \quad\text{(mod 3)}$$

So $2^n - (-1)^n$ is divisible by 3 for $n = 2k$.

Leave another case for you.

Answer 2

In general, we have $$a^n - b^n = (a - b) (a^{n-1} + a^{n-2}b + a^{n-3}b^2 + \dots + b^{n-1}).$$ Hence, $$2^n - (-1)^n=3\times \text{something}.$$

Answer 3

Use $(x^n-a^n) =(x-a) f(x, a)$, then $(2^n-(-1)^n) =(2--1)N=3N.$

Answer 4

The case is clear for $n=1$.

Suppose $2^n - (-1)^n$ is a multiple of $3$, say $3m$.
Then \begin{align} 2^{n+1} - (-1)^{n+1} &= 2\cdot 2^n + (-1)^n \\ &= 2(2^n - (-1)^n) + 2(-1)^n +(-1)^n\\ &= 2(3m) + 3(-1)^n. \end{align} I suppose you can reach a conclusion now.

Answer 5

$$2^n-(-1)^n=3m$$

Let $n=2k$

$2^n-(-1)^n=2^{2k}-(-1)^{2k}=2^{2k}-1=(2^k-1)(2^k+1)$

Since an integer is one of these forms $3t-1, 3t$ or $3t+1$ and a positive integer power of $2$ is not a multiple of $3$ then it must be one of the forms $3k-1$ or $3k+1$ and $(2^k-1)(2^k+1)$ is a multiple of $3$

Let $n=2k+1$

$2^n-(-1)^n=2^{2k+1}-(-1)^{2k+1}=2^{2k+1}+1=$

$=(2+1)(2^{2k} - a^{2k-1} + \ldots + 1)=$

$=3\times (2^{2k} - a^{2k-1} + \ldots - 2 + 1)$

In both cases $2^n-(-1)^n$ is a multiple of $3$.