Prove that $2^{r}m+\frac{2^{r-1}(3-2(-1)^{r})-2}{3} \in \mathbb{N}$

Question

I am trying to prove that $2^{r}m+\frac{2^{r-1}(3-2(-1)^{r})-2}{3} \in \mathbb{N}$ for $r \in \mathbb{Z^{+}}$ and $m \in \mathbb{N}$. Note that here $\mathbb{N} = \{0,1,2,...\}$

It's obvious that $2^{r}m \in \mathbb{N}$. But I'm having trouble proving the fraction is an element of $\mathbb{N}$. I tried proof by induction but got stuck on the last step. Is there a quick divisibility proof I can use to prove this?

Answer 1

First, note that $2^a\equiv 1 \bmod 3$ when $a$ is even and $2^a\equiv -1 \bmod 3$ when $a$ is odd.

Now consider the two cases that $r$ is odd or even.

Case 1: $r$ is even. Then the numerator becomes $(-1)(3-2)-2=-3\equiv 0 \bmod 3$, i.e. the numerator is evenly divisible by $3$. Also, the smallest possible even value $r=2$ renders the entire fraction equal to $0$, with larger even values of $r$ yielding larger values for the fraction. So the fraction is a natural number when $r$ is even.

Case 2: $r$ is odd. Then the numerator becomes $(1)(3+2)-2=3\equiv 0 \bmod 3$, i.e. the numerator is evenly divisible by $3$. Also, the smallest possible odd value $r=1$ renders the entire fraction equal to $1$, with larger odd values of $r$ yielding larger values for the fraction. So the fraction is a natural number when $r$ is odd.

Since the fraction is always a natural number, the expression as a whole is a natural number.