# Summary

Today we proved the (majority of) the primitive root theorem, and so we are now able to determine exactly which moduli *m* have primitive roots. Indeed, this means we can count precisely how many primitive roots a number *m* has just by looking at a factorization of *m*.

# Integers with Primitive Roots

Last class period we were able to show that there are a handful of integers which do not have primitive roots. Our two big classes of such moduli were

- powers of 2 that are at least 8
- products
*mn*so that $(m,n) = 1$ and so that $m,n>2$.

Today we'll aim to show that the remaining integers — i.e., the integers $1,24$ together with integers of the form $p^e$ or $2p^e$, where *p* is an odd prime — all *do* have primitive roots. We already know that primitive roots exist modulo *p*, so we'll start our investigation by verifying that primitive roots exist modulo $p^2$.

Theorem: A modulus

mhas primitive roots if and only if $m = 1,2,4,p^e \mbox{ or } 2p^e$, wherepis an odd prime.

#### Example: When do primitive roots exist?

From the previous theorem we know that there are no primitive roots modulo 65, since $65 = 3\cdot 13$ and hence doesn't match any of the "right" forms from our theorem. On the other hand, we do know that $242 = 2 \cdot 11^2$ does have primitive roots. $\square$

Proposition: If

pis an odd prime, then there exists a primitive root mod $p^2$. In fact, if $r$ is a primitive root modulop, thenris a primitive root modulo $p^2$ if and only if$r^{p-1} \nequiv 1 \mod{p^2}$.

In the case that $r^{p-1} \equiv 1 \mod{p^2}$, then the number $r+p$ is a primitive root modulo $p^2$.

Notice that this proposition doesn't just say there are primitive roots mod $p^2$, but gives us a test we can run to actually product primitive roots modulo $p^2$ quickly.

Proof: Suppose that *r* is a primitive root mod *p*; we already know such an element exists. Now we know that if $n = \mbox{ord}_{p^2}(r)$, then we have $n \mid \phi(p^2) = p(p-1)$. We also know that $r^n \equiv 1 \mod{p^2}$ implies $r^n \equiv 1 \mod{p}$, and so we get $p-1 = \mbox{ord}_p(r) \mid n$. Together, this means that we have either $n = p-1$ or $n = p(p-1)$. If the latter case is true, then we're finished: such an *r* is a primitive root mod $p^2$.

Otherwise, consider the element $r+p$. Since it is congruence to *r* mod *p*, it is still a primitive root mod *p*. Hence its order is either $p-1$ or $p(p-1)$. We'll show that the former is impossible, and so we will be able to conclude that $r+p$ is a primitive root mod $p^2$. To see that $r+p$ can't have order $p-1$, we compute:

Now the "tail" of this expression is clearly divisible by $p^2$, and so modulo $p^2$ we have

(2)We are already assuming that the order of *r* mod $p^2$ is $p-1$, and hence we get $r^{p-1} \equiv 1 \mod{p^2}$. Also, we know that $pr^{p-2} \not\equiv 0 \mod{p^2}$ since $(r,p) = 1$. Therefore the previous equation says

Hence we get that $\mbox{ord}_{p^2}(r+p) \neq p-1$, and so we must have that $r+p$ is a primitive root mod $p^2$. $\square$

#### Example: Finding a primitive root mod 121

From the book we know that *2* is a primitive root modulo 11 (note that we could prove that 2 is a primitive root by calculating $2^1,2^2$ and $2^5$ mod 11 and showing that none of these quantities is $1 \mod{11}$). Hence to find a primitive root modulo 121, we need to determine the value of $2^{10} \mod{121}$. From the result above, if this quantity isn't 1, then we'll know 2 is a primitive root; on the other hand, if $2^{10} \equiv 1 \mod{121}$, then we'll know that *2* isn't a primitive root mod 121, but that $2+11 = 13$ is. Using successive squaring we were able to see that

and so we know that 2 is a primitive root mod 121. $\square$

This might seem like we've only made slight progress, moving from primitive roots mod *p* to primitive roots mod $p^2$. As it turns out, though, this firs step is all one needs to do: primitive roots mod $p^2$ are always primitive roots mod $p^m$.

Proposition: Any primitive root mod $p^2$ is a primitive root mod $p^e$, where

pis an odd prime.

Proof: Let *r* be a primitive root mod $p^2$, and write *n* for $\mbox{ord}_{p^e}(r)$. As before, we know that

Moreover we have that

(6)Since we know that $\mbox{ord}_{p^2}(r) = p(p-1)$ (since *r* is a primitive root mod $p^2$), Equations (5) and (6) together imply

Now we would like to show that $n = p^{e-1}(p-1)$, so we need to show that no choice of $k<e-1$ is possible in the above equation. To do this, we'll show that

(8)in the next lemma. In particular, this equation will force $\mbox{ord}_{p^e}(r) \neq p^k(p-1)$ for $k<e-1$, and hence we will be forced into the condition $\mbox{ord}_{p^e}(r) = p^{e-1}(p-1)$; i.e., *r* will be a primitive root. $\square$

Lemma: Suppose that

ris a primitive root mod $p^2$. Then$r^{p^{e-2}(p-1)} \not\equiv 1 \mod{p^e}$.

Proof: We'll prove this result by induction. The case $e=2$ is already taken care of since we know that

(9)(since *r* is a primitive root, the smallest exponent which sends *r* to 1 is $\phi(p^2) = p(p-1)$). So assume that we know the result holds for $e-1$.

Now we know that

(10)and hence we have

(11)notice that $p \nmid c$ since otherwise we'd have $r^{p^{e-3}(p-1)} \equiv 1 \mod{p^e-1}$, contrary to the induction hypothesis. Now we'll raise both sides of Equation (11) to the p^{th} power. We get

Notice that all the first two terms are the only which survive when we consider this equation mod $p^e$. For instance, we know that $\left(\begin{array}{cc}p\\2\end{array}\right)(cp^{e-2})^2 = \frac{p(p-1)}{2}c^2p^{2e-4}$. Since *p* is odd, the term $\frac{p(p-1)}{2}$ is a multiple of *p*, and so this term is divisible by $p^{2e-3}$. Since we know $e \geq 3$, this means $2e-3 \geq e$. Hence the third term of this expansion is 0 modulo $p^e$. Any other term has a factor of the form $(p^{e-2})^k$ for $k \geq 3$, and since $ke-2k \geq e$ for such *k*, we know these terms are also congruent to 0 modulo $p^e$.

Hence we have

(13)Now since $p \nmid c$ we get $cp^{e-1} \not\equiv 0 \mod{p^e}$, and so we have

(14)as desired. $\square$

Having proven this result, we then stated the following

Proposition: If

ris an odd primitive root modulo $p^2$, then it is also a primitive root modulo $2p^e$ for everye.