- 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 $a \equiv 1 \space (3)$ ?
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 $$ a \equiv 1\quad \left( {\bmod q} \right)\quad \Rightarrow \quad a = 1 + q\,n\quad \Rightarrow $$ $$ \Rightarrow \quad \quad a^{\,q} = 1 + \sum\limits_{1\, \le \,k\, \le \,q} {\left( \matrix{ q \cr k \cr} \right)q^{\,k} n^{\,k} } = 1 + \sum\limits_{0\, \le \,k\, \le \,q - 1} {\left( \matrix{ q \cr k + 1 \cr} \right)q^{\,k + 1} n^{\,k + 1} } = $$ $$ = 1 + \left( \matrix{ q \cr 1 \cr} \right)q\,n + \sum\limits_{0\, \le \,k\, \le \,q - 2} {\left( \matrix{ q \cr k + 2 \cr} \right)q^{\,k + 2} n^{\,k + 2} } = $$ $$ = 1 + q^{\,2} \left( {n + \sum\limits_{0\, \le \,k\, \le \,q - 2} {\left( \matrix{ q \cr k + 2 \cr} \right)q^{\,k} n^{\,k + 2} } } \right)\quad \left| {\,2 \le q} \right. $$