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I would like to propose generalization of this question :

Let $p$ be a prime number such that : $p\equiv 1 \pmod 4$

Show that $~k\cdot p \pm a~$ is a primitive root modulo $p~$ iff

$a$ is a primitive root modulo $p$ , where $k \in \mathbb{Z}$ .

So we want to show that :

If $(kp\pm a) ^m \equiv 1 \pmod p ~$ then $(p-1) \mid m$

According to Freshman Dream Theorem it follows that :

$(kp+(\pm a)) ^p \equiv (kp)^p +(\pm a)^p \pmod p$

And from Fermat Little Theorem we know that :

$a^{p-1} \equiv 1 \pmod p$

How could we proceed from this point ?

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    Does$n$'t this follow directly from [this one](http://math.stackexchange.com/questions/105225/basic-question-on-primitive-roots)?2012-02-03

2 Answers 2

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Assume $p\equiv1\pmod4$ and $-a$ is a not a primitive root modulo $p$. Then $(-a)^{(p-1)/q}\equiv1\pmod p$ for some prime $q$ dividing $p-1$. But from $p\equiv1\pmod4$ we deduce that $(p-1)/q$ is even (the critical case is $q=2$), so $(-a)^{(p-1)/q}=a^{(p-1)/q}$, so $a$ is not a primitive root. Thus we have proved that if $a$ is a primitive root then so is $-a$.

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What you have is not a generalization of the linked question, but is instead something very close to the definition of primitivity. By definition, if $g\mod p$ is a primitive root, then $\operatorname{ord}g=\varphi(p)=p-1$, and it's a basic fact that $u^k=1\mod v$ implies $(\operatorname{ord}u)|k$, so there you have it.

If you're not familiar with this basic fact, say $b=\operatorname{ord}u$ and write $k=tb+s$ with $0\le s. Then we have $1=u^k=(u^t)^b u^s=u^s$ which implies $\operatorname{ord}u\le s$, a contradiction, unless $s=0$.

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    Yes, primitive of $a$ implying primitivity of $-a$ requires $p\equiv1$, which is the hypothesis here, and what I've given applies in this case because $p$ is prime and $\pm a$ primitive.2012-02-03