Class #19 - Composite Numbers having Primitive
Roots: n = 2pr case
1. Recall: It can be shown that n > 1 has a primitive root exactly in the following cases:
(1) n = pr for some odd prime p
(2) n = 2pr for some odd prime p
(3) n = 2 or n = 4
2. Theorem: Suppose α is a primitive root of pr where p is an odd prime and r ≥ 1. Then
(a) α is a primitive root of 2pr if α is odd.
(b) α + pr is a primitive root of 2pr if α is even.
Proof: Fix r ≥ 1. Then Ordpr α =
.
r
What is φ(2p )?
If α is odd, then gcd(α, 2pr ) =
Then Ordpr β =
. In this case, let β = α.
.
If α is even, then gcd(α + pr , 2pr ) =
Then Ordpr β =
. In this case, let β = α + pr .
because
.
So in either case, we have β such that gcd(β, 2pr ) = 1 and
Ordpr β =
= φ(
).
3. This will not be asked in an exam:
It is left to show that if n 6= 2, 4, pk , 2pk , where p is an odd prime, then n has no
primitive root.
This can be done by means of so-called universal exponent:
If n = 2k0 p1 k1 p2 k2 · · · pr kr is the prime factorization on n > 1, define the universal
exponent λ(n) of n by
λ(n) = lcm(λ(2k0 ), φ(p1 k1 ), · · · , φ(pr kr ))
where λ(2) = 1, λ(22 ) = 2, and λ(2k ) = 2k−2 for k ≥ 3. The following are true:
(a) For n = 2, 4, pk , 2pk , where p is an odd prime, λ(n) = φ(n).
(b) Except for the cases 2, 4, pk , 2pk , we have λ(n)| 21 φ(n).
(c) If gcd(a, n) = 1, then aλ(n) ≡ 1 (mod n).
4. Recall: One can show that 2 is a primitive root of 13. Then, in the set {21 , 22 , · · · , 212 },
2j such that
1
2 ≡ 2,
5
2 ≡ 6,
will have the order 12.
7
2 ≡ 11,
211 ≡ 7 (mod
1
).
5. Def: If α is a primitive root (mod n) and gcd(r, n) = 1, by the index of r relative to
α, we mean the number j, 0 ≤ j < φ(n), such that αj ≡ r (mod n).
Notation: j = Indα r
Example: 2 is a primitive root of 13. r ∈ {
} satisfies
gcd(r, n) = 1.
,
Ind2 1 =
Ind2 2 =
,
Ind2 6 =
,
Ind2 7 =
,
Ind2 3 =
6. Rules for Indices: Let α be a primitive root of an odd prime p. Then
(1) Indα (ab) ≡ Indα a + Indα b (mod (p − 1))
(2) Indα (ak ) ≡ kIndα a (mod (p − 1))
Note: p = 2 case is not interesting because p − 1 =
= gcd(b, p) to have
gcd(a, p) =
.
defined.
Proof: Since α is a primitive root of p, Ordp α =
.
x
Take a with gcd(a, p) = 1. Then a ≡ r ≡ α (mod p) where r ∈ {
Because α is a primitive root of p,
}.
. Since αp−1 ≡ α0 (mod p),
x
WLOG, we can pick x such that 0 ≤ x < p − 1. By def., such x =
.
Proof of (1):
. So, ab ≡ r ≡ αx (mod
LHS: Note gcd(ab, p) =
r∈{
} and
x
RHS: gcd(a, p) = 1, so a ≡ r1 ≡ αx1 (mod
} and
r1 ∈ {
Similarly, b ≡ r2 ≡ α
x2
) where
. So, x =
.
) where
x1
(mod p) and x2 =
. So, x1 =
.
.
≡(modp) αx1 +x2 . On the other hand, from above discussion,
Then ab ≡(modp)
we have ab ≡(modp) αx .
By D.A., x1 + x2 = (p − 1)q + l where
Since α
x1 +x2
x
≡(modp) α , l =
.
. Therefore,
(Proof of (2) will be assigned as a homework.)
2
.