Show that two $(p-1)$ th roots of unity cannot be congruent modulo $p$.
I was thinking in this way: That suppose two $(p-1)$th roots of unity $a_i, a_j$ are congruent to mod p. Then $a_i \equiv a_j$ mod $p$ Then we are end up with $1 \equiv 1$ mod $p$, which is true. Not able to find any contradiction.
Thanks in advance!
On
The $p-1$th roots of unity are $$\displaystyle e^{\dfrac{2\pi i r}{p-1}}$$ where $0\le r\le p-2$
$$\text{If }\displaystyle e^{\dfrac{2\pi i r_1}{p-1}}\equiv e^{\dfrac{2\pi i r_2}{p-1}}\pmod p$$
$$e^{\dfrac{2\pi(r_2-r_1)i}{p-1}}\equiv1\pmod p$$
$$e^{\dfrac{2\pi(r_2-r_1)i}{p-1}}\equiv e^{2\pi i s}\pmod p$$ for some integer $s$
$$\displaystyle\implies \frac{r_2-r_1}{p-1}\equiv s\pmod{p-1}$$ as $\phi(p)=p-1$ which is impossible unless $s=0\iff r_2=r_1$ as $\displaystyle 0\le r\le p-2, |r_2-r_1|<p-1$
Hint If $a_i \equiv a_j \pmod{p}$, then $p|(a_i-a_j)$.