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Let m be a positive integer greater than 1.

Prove that if r is a primitive root of m, then $r^{φ(m)/2} ≡ -1$ (mod m).

Gustavo
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2 Answers2

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To work without touching group theory too much, the identity $r^{\varphi(n)} \equiv 1 \mod n$ holds because:

  • The order of an integer $a \mod n$ is defined to be the least positive $c$ such that $a^c \equiv 1 \mod{n}$.
  • A primitive root $r \mod n$ is said to be an integer $r$ of order $\varphi(n)$.

Put the two together and you get the identity $r^{\varphi(n)} \equiv 1 \mod{n}$.

So now re-arrange and solve $$\frac{r^{\varphi(n)}}{r^2} \equiv 1 \mod{n}$$

There's still a bonus difficulty - why cannot $r\equiv 1 \mod{n}$?

There are some group theory ideas that enlighten the solution here. These are:

  • The set of integers $1, \ldots, n-1$ coprime to $n$ form a group $\mod {n}$. You can check this yourself with the group axioms.
  • A generator of a group is said to be an element $a$ in the group whose successive powers uniquely produce elements of $G$.
  • $|G|$, the order of $G$ is said to be the number of elements in $G$.
  • $|g|$, the order of an element $g \in G$ is said to be the least positive $c$ such that $g^c = e$ where $e$ is the identity element of $G$, in the case of groups $\mathbb{Z}_m^*$ this is usually 1.

So here's some ideas to explore:

  1. The order of $\mathbb{Z}_m^*$ is $\varphi(m)$.
  2. $a$ is a generator of a group iff $a^{\varphi(m)} \equiv 1 \mod{n}$. Particularly, why can't I pick a number less than $\varphi(m)$ and generate the group.
  3. Is there anything special about $g \in G$ where $g$ has an order less than $G$? What can you say about the orders of these $g$? In exploring this I recommend picking $m$ not prime, then prime, and see what happens.
намерение
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If $r$ is a primitive root of $m$, then we know that $r$ is a generator of $\mathbb{Z}_m^\times$. Since $|\mathbb{Z}_m^\times| = \varphi(m)$, then $r^{\varphi(m)} \equiv 1 \pmod{m}$ where $\varphi(m)$ is the smallest natural number to satisfy this congruence.

Can you take it from here?

Kaj Hansen
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  • Hi, thank you! We have not done generators. Do you have a simpler way of approaching this? – Gustavo Apr 29 '14 at 07:57
  • @Gustavo What definition of "primitive root" are you using? How can you say both that "We have not done generators" and "I'm studying this on my own"? You need to add a large amount of detail to your question before it can be answered. – Erick Wong Apr 29 '14 at 08:40