Let $a$ be an odd integer such that $a^{2^{n-2}}\equiv 1\; \mod {2^{n}}$. I want to show that $a^{2^{n-1}}\equiv 1\; \mod {2^{n+1}}$.
My try:
The integer $a^{2^{n-1}}$ is obtained from $a^{2^{n-2}}$ by the relation $a^{2^{n-1}}=a^{2^{n-2}}*a^{2^{n-2}}$. Hence since $a^{2^{n-2}}=1+ \alpha {2^{n}}$ for some $\alpha \in \mathbb Z$ then $a^{2^{n-1}}=a^{2^{n-2}}+ \alpha {2^{n}} a^{2^{n-2}}$ but I can't go any further. Thank you for your help!!
You are almost finished. The square of $1+\alpha2^n$ is $1+(2)\alpha 2^n +\alpha^22^{2n}$. The second term is $\alpha 2^{n+1}$ and the third is divisible by $2^{n+1}$ if $n\ge 1$, since $2n\ge n+1$.