
# 5.3: The Existence of Primitive Roots

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8847
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In this section, we demonstrate which integers have primitive roots. We start by showing that every power of an odd prime has a primitive root and to do this we start by showing that every square of an odd prime has a primitive root.

If $$p$$ is an odd prime with primitive root $$r$$, then one can have either $$r$$ or $$r+p$$ as a primitive root modulo $$p^2$$.

Notice that since $$r$$ is a primitive root modulo $$p$$, then $ord_pr=\phi(p)=p-1.$ Let $$m=ord_{p^2}r$$, then $r^m\equiv 1(mod \ p^2).$ Thus $r^m\equiv 1(mod \ p).$ By Theorem 54, we have $p-1\mid m.$ By Exercise 7 of section 6.1, we also have that $m\mid \phi(p^2).$ Also, $$\phi(p^2)=p(p-1)$$ and thus $$m$$ either divides $$p$$ or $$p-1$$. And since $$p-1\mid m$$ then we have $m=p-1 \ \ \mbox{or} \ \ m=p(p-1).$ If $$m=p(p-1)$$ and $$ord_{p^2}r=\phi(p^2)$$ then $$r$$ is a primitive root modulo $$p^2$$. Otherwise, we have $$m=p-1$$ and thus $r^{p-1}\equiv 1(mod \ p^2).$ Let $$s=r+p$$. Then $$s$$ is also a primitive root modulo $$p$$. Hence, $$ord_{p^2}s$$ equals either $$p-1$$ or $$p(p-1)$$. We will show that $$ord_{p^2}s\neq p-1$$ so that $$ord_{p^2}s=p(p-1)$$. Note that \begin{aligned} s^{p-1}=(r+p)^{p-1}&=&r^{p-1}+(p-1)r^{p-2}p+...+p^{p-1}\\&=& r^{p-1}+(p-1)p.r^{p-2}(mod \ p^2).\end{aligned} Hence $p^2\mid s^{p-1}-(1-pr^{p-2}.$ Note also that if $p^2 \mid (s^{p-1}-1),$ then $p^2\mid pr^{p-2}.$ Thus we have $p\mid r^{p-2}$ which is impossible because $$p\nmid r$$. Because $$ord_{p^2}s\neq p-1$$, we can conclude that $ord_{p^2}s=p(p-1)=\phi(p^2).$ Thus, $$s=r+p$$ is a primitive root of $$p^2$$.

Notice that 7 has 3 as a primitive root. Either $$ord_{49}3=6$$ or $$ord_{49}3=42$$. But since $$3^6\not\equiv 1(mod \ 49)$$. Hence $$ord_{49}3=42$$. Hence 3 is a primitive root of 49.

We now show that any power of an odd prime has a primitive root.

Let $$p$$ be an odd prime. Then any power of $$p$$ is a primitive root. Moreover, if $$r$$ is a primitive root modulo $$p^2$$, then $$r$$ is a primitive root modulo $$p^m$$ for all positive integers $$m$$.

By Theorem 62, we know that any prime $$p$$ has a primitive root $$r$$ which is also a primitive root modulo $$p^2$$, thus $\label{1} p^2\nmid (r^{p-1}-1).$ We will prove by induction that $\label{2} p^m\nmid (r^{p^{m-2}(p-1)}-1)$ for all integers $$m\geq 2$$. Once we prove the above congruence, we show that $$r$$ is also a primitive root modulo $$p^m$$. Let $$n=ord_{p^m}r$$. By Theorem 54, we know that $$n\mid \phi(p^m)$$. Also, we know that $$\phi(p^m)=p^m(p-1)$$. Hence $$n\mid p^m(p-1)$$. On the other hand, because $p^m\mid (r^n- 1),$ we also know that $p\mid (r^n-1).$ Since $$\phi(p)=p-1$$, we see that by Theorem 54, we have $$n=l(p-1)$$. also $$n\mid p^{m-1}(p-1)$$, we have that $$n=p^s(p-1)$$, where $$0 \leq s\leq m-1$$. If $$n=p^s(p-1)$$ with $$s\leq m-2$$, then $p^k\mid r^{p^{m-2}(p-1)}-1,$ which is a contradiction. Hence $ord_{p^m}r=\phi(p^m).$

We prove now ([2]) by induction. Assume that our assertion is true for all $$m\geq 2$$. Then $p^m\nmid (r^{p^{m-2}(p-1)}-1).$ Because $$(r,p)=1$$, we see that $$(r,p^{m-1})=1$$. We also know from Euler’s theorem that $p^{m-1}\mid (r^{p^{m-2}(p-1)}-1).$ Thus there exists an integer $$k$$ such that $r^{p^{m-2}(p-1)}=1+kp^{m-1}.$ where $$p\nmid k$$ because $$r^{p^{m-2}(p-1)}\not\equiv 1(mod \ p^m)$$. Thus we have now \begin{aligned} r^{p^{m-1}(p-1)}&=&(1+kp^{m-1})^p\\ &\equiv&1+kp^m(mod \ p^{m+1})\end{aligned} Because $$p\nmid k$$, we have $p^{m+1}\nmid (r^{p^{m-1}(p-1)}-1).$

Since 3 is a primitive root of 7, then 3 is a primitive root for $$7^k$$ for all positive integers $$k$$.

In the following theorem, we prove that no power of 2, other than 2 or 4, has a primitive root and that is because when $$m$$ is an odd integer, $$ord_2^km\neq \phi(2^k)$$ and this is because $$2^k\mid (a^{\phi(2^k)/2}-1)$$.

If $$m$$ is an odd integer, and if $$k\geq 3$$ is an integer, then $m^{2^{k-2}}\equiv 1(mod \ 2^k).$

We prove the result by induction. If $$m$$ is an odd integer, then $$m=2n+1$$ for some integer $$n$$. Hence, $m^2=4n^2+4n+1=4n(n+1)+1.$ It follows that $$8\mid (m^2-1)$$.

Assume now that $2^k\mid (m^{2^{k-2}}-1).$ Then there is an integer $$q$$ such that $m^{2^{k-2}}=1+q.2^{k}.$ Thus squaring both sides, we get $m^{2^{k-1}}=1+q.2^{k+1}+q^22^{2k}.$ Thus $2^{k+1}\mid (m^{2^{k-1}}-1).$

Note now that 2 and 4 have primitive roots 1 and 3 respectively.

We now list the set of integers that do not have primitive roots.

If $$m$$ is not $$p^a$$ or $$2p^a$$, then $$m$$ does not have a primitive root.

Let $$m=p_1^{s_1}p_2^{s_2}...p_i^{s_i}$$. If $$m$$ has a primitive root $$r$$ then $$r$$ and $$m$$ are relatively prime and $$ord_mr=\phi(m)$$. We also have, we have $$(r,p^s)=1$$ where $$p^s$$ is of the primes in the factorization of $$m$$. By Euler’s theorem, we have $p^s\mid (r^{\phi(p^s)}-1).$ Now let $L=[\phi(p_1^{s_1}), \phi(p_2^{s_2}),...,\phi(p_i^{s_i})].$ We know that $r^L\equiv 1(mod \ p_k^{s_k})$ for all $$1\leq k\leq m$$. Thus using the Chinese Remainder Theorem, we get $m\mid (r^L-1),$ which leads to $$ord_mr=\phi(m)\leq L$$. Now because $\phi(m)=\phi(p_1^{s_1})\phi(p_2^{s_2})...\phi(p_n^{s_n})\leq [\phi(p_1^{s_1}),\phi(p_2^{s_2}),...,\phi(p_n^{s_n})].$ Now the inequality above holds only if $\phi(p_1^{s_1}),\phi(p_2^{s_2}),...,\phi(p_n^{s_n})$ are relatively prime. Notice now that by Theorem 41, $\phi(p_1^{s_1}),\phi(p_2^{s_2}),...,\phi(p_n^{s_n})$ are not relatively prime unless $$m=p^s$$ or $$m=2p^s$$ where $$p$$ is an odd prime and $$t$$ is any positive integer.

We now show that all integers of the form $$m=2p^s$$ have primitive roots.

Consider a prime $$p\neq 2$$ and let $$s$$ is a positive integer, then $$2p^s$$ has a primitive root. In fact, if $$r$$ is an odd primitive root modulo $$p^s$$, then it is also a primitive root modulo $$2p^s$$ but if $$r$$ is even, $$r+p^s$$ is a primitive root modulo $$2p^s$$.

If $$r$$ is a primitive root modulo $$p^s$$, then $p^s\mid (r^{\phi(p^s)}-1)$ and no positive exponent smaller than $$\phi(p^s)$$ has this property. Note also that $\phi(2p^s)=\phi(p^s),$ so that $p^s\mid (r^{\phi(2p^s)}-1).$

If $$r$$ is odd, then $2\mid (r^{\phi(2p^s)}-1).$ Thus by Theorem 56, we get $2p^s\mid (r^{\phi(2p^s)}-1).$ It is important to note that no smaller power of $$r$$ is congruent to 1 modulo $$2p^s$$. This power as well would also be congruent to 1 modulo $$p^s$$ contradicting that $$r$$ is a primitive root of $$p^s$$. It follows that $$r$$ is a primitive root modulo $$2p^s$$.

While, if $$r$$ is even, then $$r+p^s$$ is odd. Hence $2\mid ((r+p^s)^{\phi(2p^s)}-1).$

Because $$p^s\mid (r+p^s-r)$$, we see that $p^s\mid ((r+p^s)^{\phi(2p^s)}-1).$ As a result, we see that $$2p^s\mid ((r+p^s)^{\phi(2p^s)}-1)$$ and since for no smaller power of $$r+p^s$$ is congruent to 1 modulo $$2p^s$$, we see that $$r+p^s$$ is a primitive root modulo $$2p^s$$.

As a result, by Theorem 63, Theorem 65 and Theorem 66, we see that

The positive integer $$m$$ has a primitive root if and only if $$n=2,4, p^s$$ or $$2p^s$$

for prime $$p\neq 2$$ and $$s$$ is a positive integer.
Exercises

1. Which of the following integers 4, 12, 28, 36, 125 have a primitive root.

2. Find a primitive root of 4, 25, 18.

3. Find all primitive roots modulo 22.

4. Show that there are the same number of primitive roots modulo $$2p ^s$$ as there are modulo $$p^s$$, where $$p$$ is an odd prime and $$s$$ is a positive integer.

5. Find all primitive roots modulo 25.

6. Show that the integer $$n$$ has a primitive root if and only if the only solutions of the congruence $$x^2\equiv 1(mod n)$$ are $$x\equiv \pm1 (mod \ n)$$.

### Contributors

• Dr. Wissam Raji, Ph.D., of the American University in Beirut. His work was selected by the Saylor Foundation’s Open Textbook Challenge for public release under a Creative Commons Attribution (CC BY) license.