Math 412: Number TheoryLecture 16

Gexin Yugyu@wm.edu

College of William and Mary

Gexin Yu gyu@wm.edu Math 412: Number Theory Lecture 16

THM: there are φ(p − 1) primitive roots for each prime p.

Lagrange’s Theorem: let f (x) = anxn + an−1x

n−1 + . . .+ aax + a0 be apolynomial of degree n ≥ 1, with integer coefficients and with leadingcoefficient an not divisible by p. Then f (x) has at most n incongruent rootsmodulo p. (pf: induction on n)

Let p be a prime and let d be a divisor of p − 1. Then xd − 1 has exactly dincongruent roots mod p. (proof: consider xp−1 − 1, which has p − 1incongruent roots mod p, and factor that into xd − 1 and a polynomial ofdegree p − d − 1. )

Let p be a prime and d |p − 1. Then the number of integers of order d in areduced system of residues modulo p is 0 or φ(d). (proof: suppose there isone, say a, with order d , then the roots to xd ≡ 1 (mod p) are a, a2, . . . , ad .But ordp(ai ) = d iff (i , d) = 1, so there are exactly φ(d) such. )

Let p be a prime and d |p − 1. Then the number of integers of order d in areduced system of residues modulo p is φ(d). (pf: let F (d) be the number.Then F (d) ≤ φ(d). But p− 1 =

∑d |p − 1F (d) ≤

∑d |p − 1φ(d) = p− 1.

So F (d) = φ(d) for each d .)

For every prime p, there is a primitive root modulo p. In fact, there areφ(p − 1) primitive roots. (This means that U(p) is cyclic and has φ(p − 1)generators. )

Gexin Yu gyu@wm.edu Math 412: Number Theory Lecture 16

THM: there are φ(p − 1) primitive roots for each prime p.

Lagrange’s Theorem: let f (x) = anxn + an−1x

n−1 + . . .+ aax + a0 be apolynomial of degree n ≥ 1, with integer coefficients and with leadingcoefficient an not divisible by p. Then f (x) has at most n incongruent rootsmodulo p. (pf: induction on n)

Let p be a prime and let d be a divisor of p − 1. Then xd − 1 has exactly dincongruent roots mod p. (proof: consider xp−1 − 1, which has p − 1incongruent roots mod p, and factor that into xd − 1 and a polynomial ofdegree p − d − 1. )

Let p be a prime and d |p − 1. Then the number of integers of order d in areduced system of residues modulo p is 0 or φ(d). (proof: suppose there isone, say a, with order d , then the roots to xd ≡ 1 (mod p) are a, a2, . . . , ad .But ordp(ai ) = d iff (i , d) = 1, so there are exactly φ(d) such. )

Let p be a prime and d |p − 1. Then the number of integers of order d in areduced system of residues modulo p is φ(d). (pf: let F (d) be the number.Then F (d) ≤ φ(d). But p− 1 =

∑d |p − 1F (d) ≤

∑d |p − 1φ(d) = p− 1.

So F (d) = φ(d) for each d .)

For every prime p, there is a primitive root modulo p. In fact, there areφ(p − 1) primitive roots. (This means that U(p) is cyclic and has φ(p − 1)generators. )

Gexin Yu gyu@wm.edu Math 412: Number Theory Lecture 16

THM: there are φ(p − 1) primitive roots for each prime p.

Lagrange’s Theorem: let f (x) = anxn + an−1x

n−1 + . . .+ aax + a0 be apolynomial of degree n ≥ 1, with integer coefficients and with leadingcoefficient an not divisible by p. Then f (x) has at most n incongruent rootsmodulo p. (pf: induction on n)

Let p be a prime and let d be a divisor of p − 1. Then xd − 1 has exactly dincongruent roots mod p. (proof: consider xp−1 − 1, which has p − 1incongruent roots mod p, and factor that into xd − 1 and a polynomial ofdegree p − d − 1. )

Let p be a prime and d |p − 1. Then the number of integers of order d in areduced system of residues modulo p is 0 or φ(d). (proof: suppose there isone, say a, with order d , then the roots to xd ≡ 1 (mod p) are a, a2, . . . , ad .But ordp(ai ) = d iff (i , d) = 1, so there are exactly φ(d) such. )

Let p be a prime and d |p − 1. Then the number of integers of order d in areduced system of residues modulo p is φ(d). (pf: let F (d) be the number.Then F (d) ≤ φ(d). But p− 1 =

∑d |p − 1F (d) ≤

∑d |p − 1φ(d) = p− 1.

So F (d) = φ(d) for each d .)

For every prime p, there is a primitive root modulo p. In fact, there areφ(p − 1) primitive roots. (This means that U(p) is cyclic and has φ(p − 1)generators. )

Gexin Yu gyu@wm.edu Math 412: Number Theory Lecture 16

THM: there are φ(p − 1) primitive roots for each prime p.

Lagrange’s Theorem: let f (x) = anxn + an−1x

n−1 + . . .+ aax + a0 be apolynomial of degree n ≥ 1, with integer coefficients and with leadingcoefficient an not divisible by p. Then f (x) has at most n incongruent rootsmodulo p. (pf: induction on n)

Let p be a prime and let d be a divisor of p − 1. Then xd − 1 has exactly dincongruent roots mod p. (proof: consider xp−1 − 1, which has p − 1incongruent roots mod p, and factor that into xd − 1 and a polynomial ofdegree p − d − 1. )

Let p be a prime and d |p − 1. Then the number of integers of order d in areduced system of residues modulo p is 0 or φ(d). (proof: suppose there isone, say a, with order d , then the roots to xd ≡ 1 (mod p) are a, a2, . . . , ad .But ordp(ai ) = d iff (i , d) = 1, so there are exactly φ(d) such. )

Let p be a prime and d |p − 1. Then the number of integers of order d in areduced system of residues modulo p is φ(d). (pf: let F (d) be the number.Then F (d) ≤ φ(d). But p− 1 =

∑d |p − 1F (d) ≤

∑d |p − 1φ(d) = p− 1.

So F (d) = φ(d) for each d .)

For every prime p, there is a primitive root modulo p. In fact, there areφ(p − 1) primitive roots. (This means that U(p) is cyclic and has φ(p − 1)generators. )

Gexin Yu gyu@wm.edu Math 412: Number Theory Lecture 16

THM: there are φ(p − 1) primitive roots for each prime p.

Lagrange’s Theorem: let f (x) = anxn + an−1x

n−1 + . . .+ aax + a0 be apolynomial of degree n ≥ 1, with integer coefficients and with leadingcoefficient an not divisible by p. Then f (x) has at most n incongruent rootsmodulo p. (pf: induction on n)

Let p be a prime and let d be a divisor of p − 1. Then xd − 1 has exactly dincongruent roots mod p. (proof: consider xp−1 − 1, which has p − 1incongruent roots mod p, and factor that into xd − 1 and a polynomial ofdegree p − d − 1. )

Let p be a prime and d |p − 1. Then the number of integers of order d in areduced system of residues modulo p is 0 or φ(d). (proof: suppose there isone, say a, with order d , then the roots to xd ≡ 1 (mod p) are a, a2, . . . , ad .But ordp(ai ) = d iff (i , d) = 1, so there are exactly φ(d) such. )

Let p be a prime and d |p − 1. Then the number of integers of order d in areduced system of residues modulo p is φ(d). (pf: let F (d) be the number.Then F (d) ≤ φ(d). But p− 1 =

∑d |p − 1F (d) ≤

∑d |p − 1φ(d) = p− 1.

So F (d) = φ(d) for each d .)

For every prime p, there is a primitive root modulo p. In fact, there areφ(p − 1) primitive roots. (This means that U(p) is cyclic and has φ(p − 1)generators. )

Gexin Yu gyu@wm.edu Math 412: Number Theory Lecture 16

THM: there are φ(p − 1) primitive roots for each prime p.

Lagrange’s Theorem: let f (x) = anxn + an−1x

n−1 + . . .+ aax + a0 be apolynomial of degree n ≥ 1, with integer coefficients and with leadingcoefficient an not divisible by p. Then f (x) has at most n incongruent rootsmodulo p. (pf: induction on n)

Let p be a prime and let d be a divisor of p − 1. Then xd − 1 has exactly dincongruent roots mod p. (proof: consider xp−1 − 1, which has p − 1incongruent roots mod p, and factor that into xd − 1 and a polynomial ofdegree p − d − 1. )

Let p be a prime and d |p − 1. Then the number of integers of order d in areduced system of residues modulo p is 0 or φ(d). (proof: suppose there isone, say a, with order d , then the roots to xd ≡ 1 (mod p) are a, a2, . . . , ad .But ordp(ai ) = d iff (i , d) = 1, so there are exactly φ(d) such. )

Let p be a prime and d |p − 1. Then the number of integers of order d in areduced system of residues modulo p is φ(d). (pf: let F (d) be the number.Then F (d) ≤ φ(d). But p− 1 =

∑d |p − 1F (d) ≤

∑d |p − 1φ(d) = p− 1.

So F (d) = φ(d) for each d .)

For every prime p, there is a primitive root modulo p. In fact, there areφ(p − 1) primitive roots. (This means that U(p) is cyclic and has φ(p − 1)generators. )

Gexin Yu gyu@wm.edu Math 412: Number Theory Lecture 16

People observed that 2 is a primitive root for many primes.

Is 2 a primitive root for infinitely many primes?

Artin’s conjecture states that the integer a is a primitive root ofinfinitely many primes if a 6= 1,−1 and a is not a perfect square.

Not a single value has been confirmed. But people have shown thatthere are at most two possible exceptions, if any, to the conjecture. Itfollows that 2,3, or 5 must be a primitive root for infinitely manyprimes.

Gexin Yu gyu@wm.edu Math 412: Number Theory Lecture 16

People observed that 2 is a primitive root for many primes.

Is 2 a primitive root for infinitely many primes?

Artin’s conjecture states that the integer a is a primitive root ofinfinitely many primes if a 6= 1,−1 and a is not a perfect square.

Not a single value has been confirmed. But people have shown thatthere are at most two possible exceptions, if any, to the conjecture. Itfollows that 2,3, or 5 must be a primitive root for infinitely manyprimes.

Gexin Yu gyu@wm.edu Math 412: Number Theory Lecture 16

People observed that 2 is a primitive root for many primes.

Is 2 a primitive root for infinitely many primes?

Artin’s conjecture states that the integer a is a primitive root ofinfinitely many primes if a 6= 1,−1 and a is not a perfect square.

Not a single value has been confirmed. But people have shown thatthere are at most two possible exceptions, if any, to the conjecture. Itfollows that 2,3, or 5 must be a primitive root for infinitely manyprimes.

Gexin Yu gyu@wm.edu Math 412: Number Theory Lecture 16

People observed that 2 is a primitive root for many primes.

Is 2 a primitive root for infinitely many primes?

Artin’s conjecture states that the integer a is a primitive root ofinfinitely many primes if a 6= 1,−1 and a is not a perfect square.

Not a single value has been confirmed. But people have shown thatthere are at most two possible exceptions, if any, to the conjecture. Itfollows that 2,3, or 5 must be a primitive root for infinitely manyprimes.

Gexin Yu gyu@wm.edu Math 412: Number Theory Lecture 16

Thm: an integer n has a primitive root if and only if n = 1, 2, 4, pa, 2pa,where p is odd prime.

if r is a primitive root modulo p, then r or r + p is a primitive rootmodulo p2.

If r is a primitive root for p2, then r is a primitive root for pk withk ≥ 2.

If r is a primitive root for pk , then r (if r odd) or r + pk (if r even) isalso a primitive root for 2pk .

if a is odd and k ≥ 3, then aφ(2k )/2 ≡ 1 (mod 2k)

ord2k (5) = φ(2k)/2.

Gexin Yu gyu@wm.edu Math 412: Number Theory Lecture 16

Thm: an integer n has a primitive root if and only if n = 1, 2, 4, pa, 2pa,where p is odd prime.

if r is a primitive root modulo p, then r or r + p is a primitive rootmodulo p2.

If r is a primitive root for p2, then r is a primitive root for pk withk ≥ 2.

If r is a primitive root for pk , then r (if r odd) or r + pk (if r even) isalso a primitive root for 2pk .

if a is odd and k ≥ 3, then aφ(2k )/2 ≡ 1 (mod 2k)

ord2k (5) = φ(2k)/2.

Gexin Yu gyu@wm.edu Math 412: Number Theory Lecture 16

Thm: an integer n has a primitive root if and only if n = 1, 2, 4, pa, 2pa,where p is odd prime.

if r is a primitive root modulo p, then r or r + p is a primitive rootmodulo p2.

If r is a primitive root for p2, then r is a primitive root for pk withk ≥ 2.

If r is a primitive root for pk , then r (if r odd) or r + pk (if r even) isalso a primitive root for 2pk .

if a is odd and k ≥ 3, then aφ(2k )/2 ≡ 1 (mod 2k)

ord2k (5) = φ(2k)/2.

Gexin Yu gyu@wm.edu Math 412: Number Theory Lecture 16

Thm: an integer n has a primitive root if and only if n = 1, 2, 4, pa, 2pa,where p is odd prime.

if r is a primitive root modulo p, then r or r + p is a primitive rootmodulo p2.

If r is a primitive root for p2, then r is a primitive root for pk withk ≥ 2.

If r is a primitive root for pk , then r (if r odd) or r + pk (if r even) isalso a primitive root for 2pk .

if a is odd and k ≥ 3, then aφ(2k )/2 ≡ 1 (mod 2k)

ord2k (5) = φ(2k)/2.

Gexin Yu gyu@wm.edu Math 412: Number Theory Lecture 16

Thm: an integer n has a primitive root if and only if n = 1, 2, 4, pa, 2pa,where p is odd prime.

if r is a primitive root modulo p, then r or r + p is a primitive rootmodulo p2.

If r is a primitive root for p2, then r is a primitive root for pk withk ≥ 2.

If r is a primitive root for pk , then r (if r odd) or r + pk (if r even) isalso a primitive root for 2pk .

if a is odd and k ≥ 3, then aφ(2k )/2 ≡ 1 (mod 2k)

ord2k (5) = φ(2k)/2.

Gexin Yu gyu@wm.edu Math 412: Number Theory Lecture 16

Thm: an integer n has a primitive root if and only if n = 1, 2, 4, pa, 2pa,where p is odd prime.

if r is a primitive root modulo p, then r or r + p is a primitive rootmodulo p2.

If r is a primitive root for p2, then r is a primitive root for pk withk ≥ 2.

If r is a primitive root for pk , then r (if r odd) or r + pk (if r even) isalso a primitive root for 2pk .

if a is odd and k ≥ 3, then aφ(2k )/2 ≡ 1 (mod 2k)

ord2k (5) = φ(2k)/2.

Gexin Yu gyu@wm.edu Math 412: Number Theory Lecture 16

Lemma: if r is a primitive root modulo p, then r or r + p is a primitiveroot modulo p2.

Gexin Yu gyu@wm.edu Math 412: Number Theory Lecture 16

Lemma: If r is a primitive root for p2, then r is a primitive root for pk

with k ≥ 2.

Idea of proof: use induction on k to show that rpk−2(p−1) 6≡ 1 (mod pk).

Then deduce that ordpk r = φ(pk) = pk−1(p − 1).

Gexin Yu gyu@wm.edu Math 412: Number Theory Lecture 16

Lemma: If r is a primitive root for p2, then r is a primitive root for pk

with k ≥ 2.

Idea of proof: use induction on k to show that rpk−2(p−1) 6≡ 1 (mod pk).

Then deduce that ordpk r = φ(pk) = pk−1(p − 1).

Gexin Yu gyu@wm.edu Math 412: Number Theory Lecture 16

Lemma: If r is a primitive root for pk , then r (if r odd) or r + pk (if reven) is also a primitive root for 2pk .

Gexin Yu gyu@wm.edu Math 412: Number Theory Lecture 16

Lemma: if a is odd and k ≥ 3, then aφ(2k )/2 ≡ 1 (mod 2k).

Idea of proof: induction on k .

Gexin Yu gyu@wm.edu Math 412: Number Theory Lecture 16

Lemma: if a is odd and k ≥ 3, then aφ(2k )/2 ≡ 1 (mod 2k).

Idea of proof: induction on k .

Gexin Yu gyu@wm.edu Math 412: Number Theory Lecture 16

Lemma: ord2k (5) = φ(2k)/2.

pf: indcution on k to show that 52k−3 ≡ 1 + 2k−1 6≡ 1 (mod 2k).

Gexin Yu gyu@wm.edu Math 412: Number Theory Lecture 16

Lemma: ord2k (5) = φ(2k)/2.

pf: indcution on k to show that 52k−3 ≡ 1 + 2k−1 6≡ 1 (mod 2k).

Gexin Yu gyu@wm.edu Math 412: Number Theory Lecture 16

Discrete Logarithm and index arithmetic

Def: Let m be a integer with primitive root r , and (a,m) = 1. Theunique integer x with 1 ≤ x ≤ φ(m) and r x ≡ a (mod m) is calledthe index (discrete logarithm) of a to the base r modulo m is denotedas indra.

Two simply (but important) properties:

I r indra = a

I a ≡ b (mod m) if and only if indra = indrb.

Ex: Let m = 7. Then 10 is a primitive root of 7. Let r = 10. Thenind101 = 6, ind102 = 2, ind103 = 1, ind104 = 4, ind105 = 5, ind106 = 3.

Gexin Yu gyu@wm.edu Math 412: Number Theory Lecture 16

Discrete Logarithm and index arithmetic

Def: Let m be a integer with primitive root r , and (a,m) = 1. Theunique integer x with 1 ≤ x ≤ φ(m) and r x ≡ a (mod m) is calledthe index (discrete logarithm) of a to the base r modulo m is denotedas indra.

Two simply (but important) properties:

I r indra = a

I a ≡ b (mod m) if and only if indra = indrb.

Ex: Let m = 7. Then 10 is a primitive root of 7. Let r = 10. Thenind101 = 6, ind102 = 2, ind103 = 1, ind104 = 4, ind105 = 5, ind106 = 3.

Gexin Yu gyu@wm.edu Math 412: Number Theory Lecture 16

Discrete Logarithm and index arithmetic

Def: Let m be a integer with primitive root r , and (a,m) = 1. Theunique integer x with 1 ≤ x ≤ φ(m) and r x ≡ a (mod m) is calledthe index (discrete logarithm) of a to the base r modulo m is denotedas indra.

Two simply (but important) properties:

I r indra = a

I a ≡ b (mod m) if and only if indra = indrb.

Ex: Let m = 7. Then 10 is a primitive root of 7. Let r = 10. Thenind101 = 6, ind102 = 2, ind103 = 1, ind104 = 4, ind105 = 5, ind106 = 3.

Gexin Yu gyu@wm.edu Math 412: Number Theory Lecture 16

Discrete Logarithm and index arithmetic

Def: Let m be a integer with primitive root r , and (a,m) = 1. Theunique integer x with 1 ≤ x ≤ φ(m) and r x ≡ a (mod m) is calledthe index (discrete logarithm) of a to the base r modulo m is denotedas indra.

Two simply (but important) properties:

I r indra = a

I a ≡ b (mod m) if and only if indra = indrb.

Ex: Let m = 7. Then 10 is a primitive root of 7. Let r = 10. Thenind101 = 6, ind102 = 2, ind103 = 1, ind104 = 4, ind105 = 5, ind106 = 3.

Gexin Yu gyu@wm.edu Math 412: Number Theory Lecture 16

Discrete Logarithm and index arithmetic

Def: Let m be a integer with primitive root r , and (a,m) = 1. Theunique integer x with 1 ≤ x ≤ φ(m) and r x ≡ a (mod m) is calledthe index (discrete logarithm) of a to the base r modulo m is denotedas indra.

Two simply (but important) properties:

I r indra = a

I a ≡ b (mod m) if and only if indra = indrb.

Ex: Let m = 7. Then 10 is a primitive root of 7. Let r = 10. Thenind101 = 6, ind102 = 2, ind103 = 1, ind104 = 4, ind105 = 5, ind106 = 3.

Gexin Yu gyu@wm.edu Math 412: Number Theory Lecture 16

Thm: let r be a primitive root for m, and a, b are cop rimes to m. Then

indr1 ≡ 0 (mod φ(m)),

indr (ab) ≡ indra + indrb (mod φ(m)).

indr (ak) ≡ k · indr (a) (mod φ(m)) if k > 0.

Gexin Yu gyu@wm.edu Math 412: Number Theory Lecture 16

Thm: let r be a primitive root for m, and a, b are cop rimes to m. Then

indr1 ≡ 0 (mod φ(m)),

indr (ab) ≡ indra + indrb (mod φ(m)).

indr (ak) ≡ k · indr (a) (mod φ(m)) if k > 0.

Gexin Yu gyu@wm.edu Math 412: Number Theory Lecture 16

Thm: let r be a primitive root for m, and a, b are cop rimes to m. Then

indr1 ≡ 0 (mod φ(m)),

indr (ab) ≡ indra + indrb (mod φ(m)).

indr (ak) ≡ k · indr (a) (mod φ(m)) if k > 0.

Gexin Yu gyu@wm.edu Math 412: Number Theory Lecture 16

Ex: Solve equation 6x12 ≡ 11 (mod 17). (note that 3 is a primitive root)

Gexin Yu gyu@wm.edu Math 412: Number Theory Lecture 16

Ex: Solve 7x ≡ 6 (mod 17). (again, 3 is a primitive root).

Gexin Yu gyu@wm.edu Math 412: Number Theory Lecture 16