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Assume $g$ is a primitive root modulo a prime $p$. Show that $p-g$ is a primitive root if and only if $p \equiv 1 \pmod 4$.

I am studying for a number theory exam that is why I am posting a lot of questions related to primitive roots. I would really appreciate your help because they seem like really nice questions.

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We solve the first problem. The case $p=2$ is simple, so let $p$ be an odd prime.

If $p\equiv 3\pmod{4}$, then $p-g$ is not a primitive root of $p$. For $p-g\equiv -g\pmod{p}$. But $-1$ is a quadratic non-residue of $p$, and therefore $-g$ is a QR of $p$, so cannot be a primitive root of $p$.

Conversely, let $p\equiv 1\pmod{4}$. We show that $-g$ is a primitive root of $p$.

By Fermat's Theorem, $g\equiv g^p\equiv -(-g)^p\pmod{p}$. But $-1$ is a QR of $p$, so $-1\equiv g^{2k}\equiv (-g)^{2k}\pmod{p}$ for some integer $k$.

t follows that $g\equiv (-g)^{2k}(-g)^p\pmod{p}$. Thus $g$ is congruent to a power of $-g$. Since the powers of $g$ travel, modulo $p$, through $1,2,\dots, p-1$, so do the powers of $-g$, and therefore $-g$ is a primitive root of $p$.

Remark: For other problems that you may post (and this one also) please indicate what you have tried, any progress you may have made, and where difficulties remain.

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  • $\begingroup$ Thank you for solving the first question, but I believe I need some help for the 2nd one as well which seems more complicated.So since g is a primitive root mod p that we know that $g^{p-1}=1$ (mod p). We also know that if g is a primitive root then g or g+p is a primitive root of $p^2$.But How do I relate this facts to the 2nd question? @AndreNicolas $\endgroup$ – user2214 Apr 11 '15 at 20:13
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    $\begingroup$ Your are welcome. Multipart questions in which the parts are not closely related often get closed quickly. I would prefer two questions, the second about primitive roots mod $p^2$. Recall that $g+kp$ is a primitive root mod $p^2$ for all but one $k$ in the interval $0$ to $p-1$. $\endgroup$ – André Nicolas Apr 11 '15 at 22:34
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$$-g=g(-1)\equiv g^{1+(p-1)/2}\equiv g^{(p+1)/2}$$

We know, ord$_ma=d, $ord$_m(a^k)=\dfrac d{(d,k)}$ (Proof @Page$\#95$)

ord$_p g^{(p+1)/2}=\dfrac{p-1}{\left(p-1,\dfrac{p+1}2\right)}$

Now, if integer $d(>0)$ divides both, $d$ must divide $-(p-1)+2\cdot\dfrac{p+1}2=2$

As for odd prime $p,p-1$ is even

$\left(p-1,\dfrac{p+1}2\right)=2$ if $\dfrac{p+1}2$ is even $=2k$(say) $\iff p=4k-1\equiv-1\pmod4$

$\left(p-1,\dfrac{p+1}2\right)=1$ if $\dfrac{p+1}2$ is odd $=2k+1$(say) $\iff p=4k+1\equiv1\pmod4$

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