<s>[INST] Show that $2^{y-1}|(\alpha_{0}+\alpha_{1}+...+\alpha_{2^{y-1}-1})$ 
Problem show that
Let $\alpha_{0},\alpha_{1},\alpha_{2},...,\alpha_{2^{y-1}-1}$ are smallest non negative integers such that
$$2^y|(3^{(2^{y-1}+i)}-\alpha_{i})$$
For $0\le i \le 2^{y-1}-1$
Then $$2^{y-1}|(\alpha_{0}+\alpha_{1}+...+\alpha_{2^{y-1}-1})$$
Where $y$ is given positive integer.

Example
Let $y=3 \implies \alpha_{0}=1,\alpha_{1}=3,\alpha_{2}=1,\alpha_{3}=3$
and $2^{y-1}=4|(\alpha_{0}+\alpha_{1}+\alpha_{2}+\alpha_{3})=8$
 [/INST] Let $m=2^{y-1}$.
Then $\alpha_i\equiv 3^{m+i}\pmod{2m}$.
Since $\varphi(2m)=m$, we have $3^m\equiv 1\pmod{2m}$.
Consequently:
\begin{align}
s
&=\sum_{i=0}^{m-1}\alpha_i\\
&\equiv\sum_{i=0}^{m-1}3^{m+i}\\
&\equiv\sum_{i=0}^{m-1}3^{i}
\pmod{2m}
\end{align}
Then
\begin{align}
3s
&\equiv\sum_{i=0}^{m-1}3^{i+1}\\
&\equiv s\pmod{2m}
\end{align}
hence $2s\equiv 0\pmod{2m}$ from which $s\equiv 0\pmod m$.
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