Determine if a number is a primitive root

Question

Let $p$ be an odd prime, let $g$ be a primitive root of $p$. Prove $-g$ is a primitive root of $p$ if and only if $$p\equiv1\pmod{4}$$ Hint: express $-g$ as $g^{k}$, then use property 6 ... verify $$gcd((p+1)/2,p-1) = gcd((p-1)/2,((p-1)/2)-1)$$ Unfortunately I am not entirely sure what property 6 is, but I believe it is: "If $g$ is a primitive root of $p$, then $g^k$ is a primitive root of $p$ if and only if $gcd(ϕ(p),k)=1$
Can anyone confirm that this is the property I need for this proof?
So far I have:
Let $g$ be a primitive root of an odd prime $p$. Because $g$ is a primitive root all other equivalence classes $mod(p)$ can be expressed as a power of $g$. Therefore there is some integer $k$ such that $-g\equiv g^k$.

From here I'm not sure how to continue or even what the rest of the hint is suggesting.

Who gave that hint? How come you're not sure what "property 6" is?? — DonAntonio, Mar 07 '18 at 09:39
My professor gave the hint. I'm not sure what property 6 is because he doesn't specify what properties he's referring to, and I can't find a list of 6+ properties in my notes or on any of the handouts i have. I messaged a classmate, but he never responded so that leaves me in a bit of a pickle — JakeRyan34, Mar 07 '18 at 09:46
What about asking your prof. ...?? Could it be that $;-1;$ is a quadratic residue modulo a prime $;p;$ iff $;p=-1\pmod 4;$ ?\ — DonAntonio, Mar 07 '18 at 09:51
Well it's 4am here so I don't imagine he would be very responsive lol If I can't figure this out I plan to ask him in the morning — JakeRyan34, Mar 07 '18 at 09:53

score 2 · Answer 1 · answered Mar 07 '18 at 09:44

2

More generally ord$_pa=d\implies$ ord $_p(a^k)=\dfrac d{(k,d)}$ (Proof @Page#95)

Using Prove that $a^{(p-1)/2} \equiv 1$ (mod p) and $a^{(p-1)/2} \equiv -1$ (mod p),

$$\displaystyle -g\equiv g^{\frac{p-1}2+1}\pmod p$$

Now if $d$ divides both $\dfrac{p+1}2, p-1$

$d$ must divide $2\cdot\dfrac{p+1}2-(p-1)=2$

$\implies\left(\dfrac{p+1}2, p-1\right)=1$ iff $\dfrac{p+1}2$ is odd as $p-1$ is even

$\dfrac{p+1}2=2n+1$(say)$\implies p=4n+1\equiv1\pmod4$

answered Mar 07 '18 at 09:44

lab bhattacharjee

274,582

do you mean g^(((p-1)/2))+1) or g^((p-1)/2)+1? The formatting on this site is hard to deal with lol – JakeRyan34 Mar 07 '18 at 10:09
@JakeRyan34, Have you understood uptil then? – lab bhattacharjee Mar 07 '18 at 10:12
I think so. I'm working my way through that other thread and comparing it to this problem. – JakeRyan34 Mar 07 '18 at 10:17
@JakeRyan34, The exponent is $$1+\dfrac{(p-1)}2$$ – lab bhattacharjee Mar 07 '18 at 10:17
In that case I don't understand haha. Whose answer should I be looking at on that thread? – JakeRyan34 Mar 07 '18 at 10:20
From the way I'm understanding this, by Fermat's Little Theorem, we know that $g^{p-1} \equiv 1 mod(p)$. So, $0 \equiv g^{p-1}-1 \equiv (g^{(p-1)/2}-1)(g^{(p-1)/2}+1) mod(p)$. And since p is prime, $g^{(p-1)/2} \equiv 1$ or $-1 mod(p)$ – JakeRyan34 Mar 07 '18 at 10:27
But I'm not seeing how you represented $-g$ as $g^{\frac{p-1}2+1}$ – JakeRyan34 Mar 07 '18 at 10:30
@JakeRyan34, What is the order of $g\pmod p?$ – lab bhattacharjee Mar 07 '18 at 10:31
It's $p-1$ right? – JakeRyan34 Mar 07 '18 at 10:33
@JakeRyan34, Then $$g^{(p-1)/2}\not\equiv1\pmod p$$ right? – lab bhattacharjee Mar 07 '18 at 10:36
That I'm not convinced about. To consider a specific example, take $p=5$ with $g=2$ as a primitive root. Then $2^{\frac{5-1}2} \equiv 1 mod(5)$ => $4 \equiv 1 mod(5)$ doesn't hold right? – JakeRyan34 Mar 07 '18 at 10:40
Oh never mind you edited that. I see your point now – JakeRyan34 Mar 07 '18 at 10:40
I'm convinced that you're right but how did you derive that exponent? – JakeRyan34 Mar 07 '18 at 10:54

score 1 · Accepted Answer · answered Mar 07 '18 at 13:26

1

Suppose $g$ is a primitive root, then $$g^{p-1}\equiv_p1\iff (g^{\frac{p-1}{2}}-1)(g^{\frac{p-1}{2}}+1)\equiv_p 0$$ Since $g$ is a primitive root $g^{\frac{p-1}{2}}\not\equiv_p 1$, we must have $$g^{\frac{p-1}{2}}\equiv_p -1$$

Therefore $(-1)^{\frac{p-1}{2}}g^{\frac{p-1}{2}}\equiv_p (-g)^{\frac{p-1}{2}}\equiv_p (-1)^{\frac{p+1}{2}}$, so $-g$ is a primitive root $\iff p\equiv_4 1$.

answered Mar 07 '18 at 13:26

cansomeonehelpmeout

12,782
3
22
49

$p \equiv 1 mod(4)$ implies that $\frac{p-1}2$ is even right? – JakeRyan34 Mar 07 '18 at 14:18
1

That's right. But it also implies that $(-g)^{\frac{p-1}{2}}\equiv_p -1$, which means that $-g$ is a primitive root. – cansomeonehelpmeout Mar 07 '18 at 14:20
Wouldn't that mean that $(-1)^\frac{p-1}2 = 1$ since $\frac{p-1}2$ is even? – JakeRyan34 Mar 07 '18 at 15:09
Yes, that's right. – cansomeonehelpmeout Mar 07 '18 at 15:23
Hm, well then why is $(−1)^\frac{p−1}2 g^\frac{p−1}2 ≡(−g)^\frac{p−1}2 mod(p)$ valid? – JakeRyan34 Mar 07 '18 at 15:35
Sorry if that's a dumb question I just have a hard time seeing the relationships in congruences – JakeRyan34 Mar 07 '18 at 15:36
1

Do you agree that $(-1)^2 3^2=(-3)^2$? That's the same idea. – cansomeonehelpmeout Mar 07 '18 at 15:39
1

ohhh i see what you mean, thanks – JakeRyan34 Mar 07 '18 at 15:51

Determine if a number is a primitive root

2 Answers2