3
$\begingroup$

$\newcommand{ind}{\operatorname{ind}}$ 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.

$\endgroup$
  • $\begingroup$ It should be $r\equiv-1\pmod p$ not $p\equiv-1$ $\endgroup$ – lab bhattacharjee May 8 '13 at 3:20
2
$\begingroup$

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 $s=\frac{p-1}{2}$.

$\endgroup$
2
$\begingroup$

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 $

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.