Primitive Root Question

**Janu42** · October 27th 2010, 08:55 PM

Let r be a primitive root of the prime p with p congruent to 1 (mod 4). Show that -r is also a primitive root.

**PaulRS** · October 28th 2010, 01:54 AM

Note that $x=r^{\tfrac{p-1}{2}} \equiv -1 (\bmod. p)$ since $x^2 \equiv 1 (\bmod. p)$ and so -since p is prime- $x\equiv 1 (\bmod. p)$ or $x \equiv -1 (\bmod. p)$ , but the former is not possible since r is a primitive root.

Thus $-r \equiv r^{\frac{p+1}{2}}(\bmod. p)$ so your problem is equivalent to showing that $\displaystyle\frac{p+1}{2}$ is coprime to $p-1$ .

But $2\cdot \left(\displaystyle\frac{p+1}{2}\right) - (p-1) = 2$ and so $\text{gcd}\left(\tfrac{p+1}{2}, p-1\right)$ is either 1 or 2.

Now note that 4 doesn't divide p+1, since p-1 is divisible by 4, and so (p+1)/2 must be odd.

**userit8** · March 22nd 2013, 03:33 AM

Originally Posted by PaulRS

Note that $x=r^{\tfrac{p-1}{2}} \equiv -1 (\bmod. p)$ since $x^2 \equiv 1 (\bmod. p)$ and so -since p is prime- $x\equiv 1 (\bmod. p)$ or $x \equiv -1 (\bmod. p)$ , but the former is not possible since r is a primitive root.

Thus $-r \equiv r^{\frac{p+1}{2}}(\bmod. p)$

why $-r \equiv r^{\frac{p+1}{2}}(\bmod. p)$ is true?

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