Q:

Questions regarding congruences of multiples and relatively prime numbers

If we say $n=p_1^{\alpha_1}\times p_2^{\alpha2}\times \cdots \times p_k^{\alpha_k}$, where $p_i$ are prime numbers, $\alpha_i$ are natural numbers, can or can we not say that:
Choose a $p_i$ such that it minimizes the quantity of $v_2(p_i-1)$.
1) Write $p_i=1+2^{\beta_i}\gamma_i$, where $\gamma_i$ is an odd integer, then $n\equiv 1 (\mbox{mod }2^{\beta_i})$ (I actually copied this out from a book. Just curious why is it true.)
Indeed, if the above is true, $n-1=2^{\beta_i}t$, for some integer $t$.
2) Is this true and why: $2^{2^{\beta_i}t}\equiv -1 (\mbox{mod }p_i)$

A:

As I have commented on the questions and answer provided, assertion #2 is indeed invalid due to my misunderstanding. Apologies to everyone.
As for assertion #1, let us rearrange the $p_i$ such that $v_2(p_1-1)\le v_2(p_2-1)\le \cdots \le v_2(p_k-1)$
Now we can write $p_i=1+2^{\beta_i}\gamma_1$ for all $i\in\{1, 2, \cdots, k\}$ and we can see that $v_2(p_i-1)=\beta_i$. Hence, $\beta_1\le \beta_2\le\cdots\le \beta_k$. From this assumption, $p_i$ is the problem is actually $p_1$ here as it has the smallest value of $v_2(p_i-1)$
Notice that $p_i\equiv 1 (\mbox{mod }2^{\beta_1})$ for all $i\in\{1, 2, \cdots, k\}$. Take note that here the $p_i$ are congruent to $\beta_1$ not $\beta_i$ (although the congruence establish as well.) And hence, multiply up all the congruences, we get $n\equiv 1(\mbox{mod }2^{\beta_1})$.

