1. Kim Correll Holly Bochsler
Primitive Roots
Kim Correll
Holly Bochsler

2. Order mod n
Definition:
If (a,n)=1 then the order of
a(mod n) , denoted ordn (a) , is the smallest integer k, such that
ak = 1 (mod n).

3. Example 1 Order of 6 mod 7 ord7 (6) 61 mod 7 = 6 mod 7
The order of 6 mod 7 is obviously 2, since 2 is the smallest integer to result in 1 mod 7.

4. Example 2
ord7 (3)
31 mod 7 = 3 mod 7
32 mod 7 = 2 mod 7
33 mod 7 = 6 mod 7
34 mod 7 = 4 mod 7
35 mod 7 = 5 mod 7
36 mod 7 = 1 mod 7
In this case, ord7 (3) = 6. Notice also 3k mod 7 includes all the residue classes of mod 7.
3 is a primitive root of 7

5. Primitive Root
Let p be prime, then b is a primitive root for p if the powers of b include all the residue classes mod p (not including 0).
There will obviously be p-1 residues
i.e. the first p-1 powers of b have a different mod p, as with the example of 3(mod7).

6. Example 2 31 mod 7 = 3 mod 7 32 mod 7 = 2 mod 7 33 mod 7 = 6 mod 7
As you can see, all of these have a different residue class, p-1 of them.
All of the residue classes of 7 are included. (1,2,3,4,5,6)

7. Example 3 2 mod 7 21 mod 7 = 2 mod 7 22 mod 7 = 4 mod 7
Obviously, 2 cannot be a primitive root for 7, since it does not contain all the residues of mod 7.

8. Euler’s Phi Function Definition:
Euler’s phi function, denoted Φ(n), is defined as the amount of integers greater than or equal to 1 and less than n, that is relatively prime to n.
Example: Φ(4) = 2 because 1 and 3 are the only integers between 1 and 4 relatively prime to 4.
The phi function is both an arithmetic function and multiplicative

9. Euler’s Phi Function
Definition: An arithmetic function is a function defined for all positive integers.
Example: sine and cosine functions are arithmetic but the tangent function is not

10. Euler’s Phi Function
Definition: An arithmetic function is called multiplicative if F(mn)=F(m)*F(n) whenever m and n are relatively prime. A function is called completely multiplicative if F(mn)=F(m)*F(n) for all positive integers m and n.
Example: f(x)=x is multiplicative. Let x=40. We can rewrite this as f(8*5)=f(8)*f(5) ((8,5)=1).
This function is also completely multiplicative because we can also rewrite this as f(2*20)=f(2)*f(20) ((2,20)=2).

11. Euler Phi Function
The Euler phi function is multiplicative but not completely.
Proof: Assume the phi function is arithmetic. Suppose the greatest common divisor of m and n is one. If n=(p1)x1…(pr)xr and m=(q1)x1…(qs)xs are the prime factorizations of n and m then no pr can occur among qs because m and n are relatively prime. Hence Φ(mn)= (p1)x1…(pr)xr(q1)x1…(qs)xs =Φ(n)Φ(m).

12. Euler Phi Function
When n is a prime number p, Φ(p)=p-1. This makes sense because the only divisors of p are 1 and itself, and no number less than p equals p.
Example :
Φ(19)=18

13. Theorem
Theorem 1: Let p be a prime and a a positive integer. Then Φ(pja)= (pj)a-(pj)a-1
Proof: The positive integers less than pa that are not relatively prime to p are those integers not exceeding pa , that are divisible by p. These are the integers kp, where 1≤k≤ pa-1. Since there are exactly pa-1 such integers, there are pa -pa-1 integers less than pa. Hence Φ(pa)= (pj)a-(pj)a-1 .

14. Example
What is Φ(100)?
Using the multiplicative principle we can rewrite this as Φ(4)*Φ(25).
We know from a previous example that Φ(4)=2.
According to theorem 2 we know Φ(25)=52-5=20.
Therefore Φ(100)=2*20=40.

15. For all other numbers we have the following theorem:
Theorem 2: If n=(p1)a1…(pr)ai is the prime factorization then Φ(n)=n(1-1/p1) …(1-1/pi) .
Proof: Since Φ is multiplicative we can rewrite Φ as Φ(n)=Φ(p1a1)…Φ(pkak). We can rewrite the previous theorem as Φ(pjaj)= pjaj (1-1/pj) for j=1,2,…k. Hence Φ(n)= p1a1(1-1/p1) p2a2 (1-1/p2) …pkak (1-1/pk). This can then be rewritten as n(1-1/p1) …(1-1/pi) .

16. Example
Φ(100)= 22 * 52
= 100(1-1/2)(1-1/5)
= 100(1/2)(1/5)
= 40
Notice this is the same answer we arrived at using the multiplicative property and the prime power properties.

17. Theorem
Theorem : If r and n are relatively prime integers with n>0 and if r is a primitive root modulo n, then the integers r1,r2,…,rΦ(n) form a reduced residue set modulo n.
Theorem : The integer a is a primitive root
(mod n) if and only if ordn (a) = Φ(n), where Φ(n) is the largest possible order mod n.

18. Proof
If ordn (a) < Φ(n), then the residue class of 1 appears more than once, since Φ(n) is the largest possible order mod n. Most importantly
aordn(a) ≡ aΦ(n) ≡ 1 mod n.
Meaning the residue classes are not all different, so there are not p-1 residues, and therefore cannot be a primitive root.

19. Proof If ordn (a) = Φ(n), then ai ≡ aj mod n.
If ai ≡ aj mod n then i ≡ j (mod Φ(n).
for 1 ≤ i ≤ j ≤ Φ(n), the congruence i ≡ j (mod Φ(n), implies that i = j.
They are a reduced congruency class.
Corollary: Let r be a primitive root modulo n, where n is an integer, n>1. Then ru is a primitive root modulo n if and only if (u,Φ(n))=1.

20. Theorem
Theorem: If the positive integer n has a primitive root, then it has a total of Φ(Φ(n)) incongruent primitive roots.
Proof: Let r be a primitive root modulo n. Then theorem 3 tells us that the integers r1,r2,…,rΦ(n) form a reduced residue system modulo n. By the corollary we know that ru is a primitive root modulo n if and only if (u,Φ(n))=1. Because there are exactly Φ(Φ(n)) such integers u, there are exactly Φ(Φ(n)) primitive roots modulo n.