Implication in congruential equation 
  
*
  
*I dont' understand why $ a \equiv_{} 1  (2) \implies a^2 \equiv_{} 1 \space(8)  $ . I thought that a possible reason could be the following: $a^2 \equiv_{} 1 \space (8) \implies (a-1)\cdot(a+1) \equiv_{} 0 \space (8) \implies (a-1) \equiv_{} 0 \space (8) \implies (a-1) \equiv_{} 0 \space (2) \implies a \equiv_{} 1 \space (2)$
  
*Why $ a \equiv_{} 1 \space(3) \implies a^3 \equiv_{} 1 \space (9) $ ?  I thought that a possible reason could be the following: $a^3 \equiv_{} 1 \space (9) \implies (a-1)\cdot(a^2 +a+1) \equiv_{} 0 \space (9) \implies (a-1) \equiv_{} 0 \space (9) \implies (a-1) \equiv_{} 0 \space (3) \implies a \equiv_{} 1 \space (3)$ 
  

However I dont' know if there is an easier way to prove that. Furthermore I've made these deductions from the "conclusion": there is a way of starting the proof from the fact that $ a \equiv_{} 1 \space (2)$ (first) and $a \equiv 1 \space (3)$ ?
 A: Because we have
$$
\eqalign{
  & a \equiv 1\quad \left( {\bmod 2} \right)\quad  \Rightarrow \quad a = 1 + 2n\quad  \Rightarrow   \cr 
  &  \Rightarrow \quad \quad a^{\,2}  = 1 + 4n + 4n^{\,2}  = 1 + 4\,n\left( {n + 1} \right) = 1 + 8\,\left( \matrix{
  n + 1 \cr 
  2 \cr}  \right) = 1 + 8\,m \cr} 
$$
and
$$
a \equiv 1\quad \left( {\bmod q} \right)\quad  \Rightarrow \quad a = 1 + q\,n\quad  \Rightarrow 
$$
$$
 \Rightarrow \quad \quad a^{\,q}  = 1 + \sum\limits_{1\, \le \,k\, \le \,q} {\left( \matrix{
  q \cr 
  k \cr}  \right)q^{\,k} n^{\,k} }  = 1 + \sum\limits_{0\, \le \,k\, \le \,q - 1} {\left( \matrix{
  q \cr 
  k + 1 \cr}  \right)q^{\,k + 1} n^{\,k + 1} }  = 
$$
$$
 = 1 + \left( \matrix{
  q \cr 
  1 \cr}  \right)q\,n + \sum\limits_{0\, \le \,k\, \le \,q - 2} {\left( \matrix{
  q \cr 
  k + 2 \cr}  \right)q^{\,k + 2} n^{\,k + 2} }  = 
$$
$$
 = 1 + q^{\,2} \left( {n + \sum\limits_{0\, \le \,k\, \le \,q - 2} {\left( \matrix{
  q \cr 
  k + 2 \cr}  \right)q^{\,k} n^{\,k + 2} } } \right)\quad \left| {\,2 \le q} \right.
$$
