How to prove by induction that $a^{2^{k-2}} \equiv 1\pmod {2^k}$ for odd $a$? Let $a \in $ Z be odd. How to prove by induction that $a^{2^{k-2}} \equiv 1\pmod {2^k}$ for all $k \geq 3$.

My attempt: let $k=3$
  $$\begin{align*}
a^{2^{k-2}}-1&= a^{2^{3-2}}-1=a^{2}-1\\
& \stackrel{\text{a is odd}}{=} (2l+1)^2-1=4l^2+4l=4l(l+1)\\ &\stackrel{\text{choose } l=2}{=} 8 \cdot 3 = 8q
\end{align*}$$
  or $2^3|a^2-1 \Longleftrightarrow a^2 \equiv 1\pmod {2^3}$.
Induction assumption: 
  Let us assume that claim holds when $k=n$ or 
  $2^n|a^{2^{n-2}}-1$ or $a^{2^{n+2}} =q'2^n+1 $. 
  Look at the statement $a^{2^{k-2}}-1$ with the value $k=n+1$. 
  Then 
  $$\begin{align*}
a^{2^{n+1-2}}-1 &= a^{2^{n-2}\cdot 2}-1=(a^{2^{n-2}})^2-1\\
&\stackrel{\text{assumption}}{=} (q')^2(2^n)^2+2q'\cdot 2^n +1 -1\\
&= 2^n((q')^22^n+2q')=2^nq''\\
&\Longrightarrow 2^n|a^{2^{n+1-2}}-1.
\end{align*}$$ 
  Claim holds, i.e. $a^{2^{k-2}} \equiv 1\pmod {2^k}$.


For the case $k=3$, you cannot choose $l$. You must argue that $4l(l+1)$ is a multiple of $8$, which can be done by arguing that $l(l+1)$ must be even.
Your induction step seems ok, but it could be written more simply as
$$
a^{2^{n+1-2}}-1 = a^{2^{n-2}\cdot 2}-1=(a^{2^{n-2}}-1)(a^{2^{n-2}}+1)
$$
and use that $a^{2^{n-2}}+1$ is even.
