Conditions for Integer to have Primitive Root

Theorem
Let $$n \in \Z: n > 1$$.

Then $$n$$ has a primitive root iff: where $$p > 2$$ is prime and $$k \ge 1$$.
 * $$n = 2$$;
 * $$n = 4$$;
 * $$n = p^k$$, or
 * $$n = 2 p^k$$

Proof of Necessity
This proof comes in several sections, so as to be able to cover all cases.

The cases where n = 2 and n = 4

 * $$1$$ is trivially a primitive root of $$2$$.


 * $$3$$ is a primitive root of $$4$$, as $$\phi \left({4}\right) = 2$$ and $$3^2 = 9 \equiv 1\left({\bmod\, 4}\right)$$.

Power of 2 greater than 4: No primitive root
Let $$n = 2^k$$ where $$k \ge 3$$.

From Euler Phi Function of a Prime, $$\phi \left({2^k}\right) = 2^{k-1}$$.

The $$2^{k-1}$$ least positive residues prime to $2^k$ are the odd integers up to $$2^k - 1$$.

What we need to show is that for any odd integer $$a$$ there exists a positive integer $$r < 2^{k-1}$$ such that $$a \equiv 1 \left({\bmod\, 2^k}\right)$$.

We assert that $$r = 2^{k-2}$$ has exactly this property, which we prove by induction.

For all $$k \in \N^*$$, let $$P \left({k}\right)$$ be the proposition $$a^{\left({2^{k-2}}\right)} \equiv 1 \left({\bmod\, 2^k}\right)$$.

Basis for the Induction

 * $$P(3)$$ is true, as follows:

$$k = 3 \Longrightarrow 2^k = 8, 2^{k-2} = 2^1 = 2$$.

The odd integers less than $$8$$ are $$1, 3, 5, 7$$. So:

$$ $$ $$ $$

Thus $$P(3)$$ holds.

This is our basis for the induction.

Induction Hypothesis

 * Now we need to show that, if $$P \left({h}\right)$$ is true, where $$h \ge 3$$, then it logically follows that $$P \left({h+1}\right)$$ is true.

So this is our induction hypothesis:


 * $$a^{\left({2^{h-2}}\right)} \equiv 1 \left({\bmod\, 2^h}\right)$$.

Then we need to show:


 * $$a^{\left({2^{h-1}}\right)} \equiv 1 \left({\bmod\, 2^{h+1}}\right)$$.

Induction Step
We can write our induction hypothesis more conveniently as $$a^{\left({2^{h-2}}\right)} = 1 + m 2^h$$ for some $$m \in \Z$$.

Hence:

$$ $$ $$ $$

But $$\left({a^{\left({2^{h-2}}\right)}}\right)^2 = a^{\left({2^{h-2}}\right)} a^{\left({2^{h-2}}\right)} = a^{\left({2^{h-2} + 2^{h-2}}\right)} = a^{2 \left({2^{h-2}}\right)} = a^{\left({2^{h-1}}\right)}$$.

So $$P \left({h}\right) \Longrightarrow P \left({h+1}\right)$$ and the result follows by the Principle of Mathematical Induction.

Therefore $$\forall k \ge 3: a^{\left({2^{k-2}}\right)} \equiv 1 \left({\bmod\, 2^k}\right)$$.

Two Coprime Divisors Greater than 2: No Primitive Root
Let us take $$n = r s$$ where $$r > 2, s > 2, r \perp s$$.

Let $$a \perp r s$$.

From Euler Phi Function Even for Argument Greater than 2, $$\phi \left({r}\right)$$ and $$\phi \left({s}\right)$$ are both even.

So $$\frac {\phi \left({r}\right)} 2$$ and $$\frac {\phi \left({s}\right)} 2$$ are both integers, and hence so is $$\frac {\phi \left({r s}\right)} 2$$.

As $$a \perp r s$$ we have that $$a \perp r$$ which gives $$a^{\phi \left({r}\right)} \equiv 1 \left({\bmod\, r}\right)$$ from Euler's Theorem.

So putting $$k = \frac {\phi \left({r s}\right)} 2$$, we have:

$$ $$ $$ $$

Interchanging the roles of $$r$$ and $$s$$ shows that $$a^k \equiv 1 \left({\bmod\, s}\right)$$ as well.

So we see that $$a^k \equiv 1 \left({\bmod\, r s}\right)$$.

Thus for any $$a$$, we have $$a^k \equiv 1 \left({\bmod\, n}\right)$$ where $$k = \frac {\phi \left({n}\right)} 2 < \phi \left({n}\right)$$.

Hence $$n$$ has no primitive root.

Hence we see that in order for $$n$$ to have a primitive root, it is necessary that: where $$p > 2$$ is prime and $$k \ge 1$$.
 * $$n = 2$$;
 * $$n = 4$$;
 * $$n = p^k$$, or
 * $$n = 2 p^k$$

Proof of Sufficiency
We have covered the cases where $$n = 2$$ and $$n = 4$$.

Now we need to show that if: where $$p > 2$$ is prime and $$k \ge 1$$, then $$n$$ has a primitive root.
 * $$n = p^k$$, or
 * $$n = 2 p^k$$