For every $a>4$, $2^{-a-2}\cdot(3^{2^a(2b-1)}-1)$ is odd I'm trying to show that the expression:
$$\frac{3^{2^a(2b-1)}-1}{2^{a+2}},\{a,b\}\in\mathbb{Z^+}$$
is always odd for $2^a(2b-1)>4$. This expression came from working backwards from a pattern I noticed using a brute force method. I have already shown by induction that $\frac{3^{2c-1}-1}{2}$ is always odd, however I seem to be stuck with this case where the exponent of $3$ is even. I have tried induction on it without much success.
 A: First of all, note that $3^{2m}+1$ is not divisible by $4$ where $m$ is a non-negative integer since
$$3^{2m}+1\equiv (-1)^{2m}+1\equiv 1+1\equiv 2\pmod 4$$
Now, let us prove by induction on $b$ that $$3^{2^5(2b-1)}-1=2^7k\tag1$$ where $k$ is odd.
The base case : $$\frac{3^{2^5(2\cdot 1-1)}-1}{2^{5+2}}=\frac{(3^{16}+1)(3^8+1)(3^4+1)(3^2+1)\cdot 2^3}{2^7}$$
is odd.
Supposing that $3^{2^5(2b-1)}-1=2^7k$ where $k$ is odd gives
$$\begin{align}3^{2^5(2(b+1)-1)}-1&=3^{64}\cdot 3^{32(2b-1)}-1\\&=3^{64}(2^7k+1)-1\\&=3^{64}\cdot 2^7k+(3^{32}+1)(3^{16}+1)(3^8+1)(3^4+1)(3^2+1)\cdot 2^3\\&=2^7(3^{64}k+2k')\qquad (\text{where $k'$ is an integer})\\&=2^7\times(\text{odd integer})\qquad\blacksquare\end{align}$$
Next, let us prove by induction on $a$ that $$3^{2^a(2b-1)}-1=2^{a+2}k$$ where $k$ is odd.
The base case is $(1)$.
Suppose that $3^{2^a(2b-1)}-1=2^{a+2}k$ where $k$ is odd gives 
$$\begin{align}3^{2^{a+1}(2b-1)}-1&=(3^{2^a(2b-1)})^2-1\\&=(2^{a+2}k+1)^2-1\\&=2^{2a+4}k^2+2^{a+3}k\\&=2^{a+3}(2^{a+1}k^2+k)\\&=2^{a+3}\times (\text{odd integer})\qquad\blacksquare\end{align}$$
