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If $r$ is a primitive root of odd prime $p$, prove that $\text{ind}_r (-1) = \frac{p-1}{2}$


I know $r^{p-1}\equiv 1 \pmod {p} \implies r^{(p-1)/2}\equiv -1 \pmod{p}$ But some how I feel the question wants me prove the result using indices properties (w/o factoring.. ) So I am posting it here to see if there are any other ways to prove this XD thanks in advance!

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Note that $\left(r^{(p-1)/2}\right)^2\equiv 1\pmod{p}$, so $r^{(p-1)/2}$ is a solution of the congruence $x^2\equiv 1\pmod{p}$.

The congruence $x^2\equiv 1\pmod{p}$ has precisely two solutions, $x\equiv 1\pmod{p}$ and $x\equiv -1\pmod{p}$.

Since $r^{(p-1)/2}\not\equiv 1\pmod{p}$, we must have $r^{(p-1)/2}\equiv -1\pmod p$, so $-1$ has index $(p-1)/2$.

Another way: If you want to just use "algebraic" properties of indices, let $d$ be the index of $-1$. Then the index of $(-1)^2$ is congruent to $2d$ modulo $p-1$. (Index of a product is, sort of, the sum of the indices.) But the index of $(-1)^2$ is $p-1$. so $2d\equiv 0\pmod{p-1}$ and $1\le d\lt p-1$. This forces $d=\frac{p-1}{2}$.

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  • $\begingroup$ I see.. $r$ is a primitive root so $r^{(p-1)/2}\not\equiv 1\pmod{p}$. That makes the index of -1 equal to (p-1)/2.I feel a bit more confident now xD Thanks to you! $\endgroup$ – pooja Jan 29 '15 at 5:40
  • $\begingroup$ I added an index only proof. $\endgroup$ – André Nicolas Jan 29 '15 at 5:43

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