Divisibility by $2$ and powers of $3$ While dabbling in a similar but different problem, I sort of noticed this.

Is it true that $3^n-1$, for all odd $n\in\mathrm{N}$, is no more than once divisible by $2$? If so, is there a conclusive proof for this statement? In other words, if:
$$3^n-1=2a$$
where $a\in\mathrm{N}$, is it true that $a$ must be odd too? 

I attempted to disprove this in the following way-

Consider that $n=2k+1$, and $j$ and $a$ where $k,j,a\in\mathrm{N}$ and:
$$\begin{split}
3^n-1&amp;=3\cdot3^{2k}-1\\
&amp;=2^ja
\end{split}$$
Using induction on $k$ here led to the following:
$$ 3\cdot3^2\cdot3^{2k}-1=2^{j}\cdot3^2\cdot a+8$$
which doesn't really put a bound on $j$. I haven't been able to find a counterexample yet which makes me think that the above isn't correct.

So is the statement correct? Please provide an answer with a proof, if possible.
Well, $3^n-1=2(1+3+3^2+...+3^{n-1})$

When $n$ is odd there are odd number of odd numbers plus together in the bracket.
$3^n-1\equiv 0 \pmod 4$ if $n$ is a power of $2,n\gt1$.