Show that if $p , q\,$ are prime then the cyclotomic polynomial $\Phi_q(x)=1+x+..+x^{q-1}$
1) Has $q-1 $ roots in $Z_p$ if $p\equiv 1 \pmod q$
2) has 1 root if p=q
3) No root otherwise.
$(x-1)\Phi_q(x)=(x^q-1)$
By theorem , $\Phi(x)$ has at most $q-1$ roots.
I had proved there are $q-1$ roots in case of $p\equiv 1 \pmod q$
This proof I had convinced .
I had shown that 1 is root of $\Phi(x)$ for p=q .
I wanted to show that this is only possible solution? that I'm no't able to show.
Also I am not able to show last one.
Any help will be appreciated.
Suppose that $a$ is a root and $p\neq q$. Then $(a-1)\Phi(a)\equiv 0\pmod p$ and hence $a^q\equiv 1\pmod p$; therefore $q$ is a multiple of the order of $a$ in $\Bbb Z_p$. Since $q$ is prime and $a\neq 1$, this order is precisely $q$.
But the order of every element of a group divides the order of the group, that is, $q$ divides $p-1$.