# The index of $-1$ modulo an odd prime.

$\newcommand{ind}{\operatorname{ind}}$ Let $p$ be an odd prime and $r$ a primitive root of $p$. Find $\ind_r(-1)$, where $\ind$ stands for $\;\text{index}$.

My solution: $p \equiv -1\pmod p$. By squaring both sides we get $r^2 \equiv 1\pmod p$. so $r=1$. Not sure if this is right.

• It should be $r\equiv-1\pmod p$ not $p\equiv-1$ – lab bhattacharjee May 8 '13 at 3:20

Set $s$ to be the index of $-1$, i.e. $r^s\equiv -1$. Squaring, we get $r^{2s}\equiv 1\pmod{p}$. Since $r$ is a primitive root, $r^{p-1}\equiv 1\pmod{p}$ and hence $(p-1)|2s$ or $\frac{p-1}{2}|s$. Hence $s=\frac{p-1}{2}$ or $s=2\frac{p-1}{2}$ or $s=3\frac{p-1}{2}$ etc. But $s\le p-1$ since $\{r,r^2,\ldots,r^{p-1}\}$ is a reduced residue system. Further, $s\neq p-1$ since $r^{p-1}\equiv 1\not\equiv -1$. Thus $s=\frac{p-1}{2}$.

Let $$a\equiv-1\pmod p \implies a^2\equiv1\pmod p$$

Taking Discrete Logarithm, $$2ind_ra\equiv0\pmod {p-1}\implies 2ind_ra=k\cdot(p-1)$$ where $k$ is any integer

$$\implies ind_ra=\frac{k\cdot(p-1)}2$$

As $0\le ind_ra< p-1, 0\le k\le 1$ as $3\cdot\frac{p-1}2\ge p-1\iff p\ge 1$

If $k=0, ind_ra=0\implies a\equiv r^0\pmod p\equiv1$

But $a\equiv-1\pmod p\implies 1\equiv-1\pmod p\implies p$ divides $2$ which impossible

So, $k=1\implies ind_ra=\frac{(p-1)}2$