Show that if (a,m) = 1 and m has a prime factor p such that (p-1) | Q, then $\displaystyle (a^Q-1,m) > 1$
I am not sure how to do this, so any help would be great!
since p divides m, if we show that $\displaystyle p \text{ divides } a^Q-1$,then the statement is proved.
since p-1 divides Q, there exist integer k such that Q=(p-1)k.
by the fermat theorem: $\displaystyle p \text{ divides } a^{p-1}-1$;
and $\displaystyle a^Q-1=(a^{p-1}-1)(1+a^{p-1}+a^{2(p-1)}+ \cdot \cdot \cdot +a^{(k-1)(p-1)})$
thus p divides $\displaystyle a^Q-1$. QED