Tuesday, 7 January 2014

elementary number theory - A contradiction proof of "If $((n^q)-1)$ is divisible by $p,$ then show that $,q mid (p-1)"$.

Let $\,p, q\,$ be prime numbers and $\,n\,\in \mathbb N$ such that $(n-1)$ is not divisible by $p$. If $\,(n^q-1)\,$ is divisible by p then show that $q \mid (p-1)$.



How can I prove it by contradiction. Let us take $(p-1)$ is not divisible by $q$ then how can I achieve a contradiction to to show that $(n^q-1)$ is not divisible by $p$.



Please help me to solve it. Thanks in advance.

No comments:

Post a Comment

real analysis - How to find $lim_{hrightarrow 0}frac{sin(ha)}{h}$

How to find $\lim_{h\rightarrow 0}\frac{\sin(ha)}{h}$ without lhopital rule? I know when I use lhopital I easy get $$ \lim_{h\rightarrow 0}...