math 412: number theory lecture 16 · djp 1f(d) p djp 1˚(d) = p 1. so f(d) = ˚(d) for each d.)...
TRANSCRIPT
![Page 1: Math 412: Number Theory Lecture 16 · djp 1F(d) P djp 1˚(d) = p 1. So F(d) = ˚(d) for each d.) For every prime p, there is a primitive root modulo p. In fact, there are ˚(p 1)](https://reader034.vdocuments.site/reader034/viewer/2022050506/5f981bf5cb5cdf61ff493fbf/html5/thumbnails/1.jpg)
Math 412: Number TheoryLecture 16
Gexin [email protected]
College of William and Mary
Gexin Yu [email protected] Math 412: Number Theory Lecture 16
![Page 2: Math 412: Number Theory Lecture 16 · djp 1F(d) P djp 1˚(d) = p 1. So F(d) = ˚(d) for each d.) For every prime p, there is a primitive root modulo p. In fact, there are ˚(p 1)](https://reader034.vdocuments.site/reader034/viewer/2022050506/5f981bf5cb5cdf61ff493fbf/html5/thumbnails/2.jpg)
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 [email protected] Math 412: Number Theory Lecture 16
![Page 3: Math 412: Number Theory Lecture 16 · djp 1F(d) P djp 1˚(d) = p 1. So F(d) = ˚(d) for each d.) For every prime p, there is a primitive root modulo p. In fact, there are ˚(p 1)](https://reader034.vdocuments.site/reader034/viewer/2022050506/5f981bf5cb5cdf61ff493fbf/html5/thumbnails/3.jpg)
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 [email protected] Math 412: Number Theory Lecture 16
![Page 4: Math 412: Number Theory Lecture 16 · djp 1F(d) P djp 1˚(d) = p 1. So F(d) = ˚(d) for each d.) For every prime p, there is a primitive root modulo p. In fact, there are ˚(p 1)](https://reader034.vdocuments.site/reader034/viewer/2022050506/5f981bf5cb5cdf61ff493fbf/html5/thumbnails/4.jpg)
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 [email protected] Math 412: Number Theory Lecture 16
![Page 5: Math 412: Number Theory Lecture 16 · djp 1F(d) P djp 1˚(d) = p 1. So F(d) = ˚(d) for each d.) For every prime p, there is a primitive root modulo p. In fact, there are ˚(p 1)](https://reader034.vdocuments.site/reader034/viewer/2022050506/5f981bf5cb5cdf61ff493fbf/html5/thumbnails/5.jpg)
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 [email protected] Math 412: Number Theory Lecture 16
![Page 6: Math 412: Number Theory Lecture 16 · djp 1F(d) P djp 1˚(d) = p 1. So F(d) = ˚(d) for each d.) For every prime p, there is a primitive root modulo p. In fact, there are ˚(p 1)](https://reader034.vdocuments.site/reader034/viewer/2022050506/5f981bf5cb5cdf61ff493fbf/html5/thumbnails/6.jpg)
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 [email protected] Math 412: Number Theory Lecture 16
![Page 7: Math 412: Number Theory Lecture 16 · djp 1F(d) P djp 1˚(d) = p 1. So F(d) = ˚(d) for each d.) For every prime p, there is a primitive root modulo p. In fact, there are ˚(p 1)](https://reader034.vdocuments.site/reader034/viewer/2022050506/5f981bf5cb5cdf61ff493fbf/html5/thumbnails/7.jpg)
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 [email protected] Math 412: Number Theory Lecture 16
![Page 8: Math 412: Number Theory Lecture 16 · djp 1F(d) P djp 1˚(d) = p 1. So F(d) = ˚(d) for each d.) For every prime p, there is a primitive root modulo p. In fact, there are ˚(p 1)](https://reader034.vdocuments.site/reader034/viewer/2022050506/5f981bf5cb5cdf61ff493fbf/html5/thumbnails/8.jpg)
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 [email protected] Math 412: Number Theory Lecture 16
![Page 9: Math 412: Number Theory Lecture 16 · djp 1F(d) P djp 1˚(d) = p 1. So F(d) = ˚(d) for each d.) For every prime p, there is a primitive root modulo p. In fact, there are ˚(p 1)](https://reader034.vdocuments.site/reader034/viewer/2022050506/5f981bf5cb5cdf61ff493fbf/html5/thumbnails/9.jpg)
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 [email protected] Math 412: Number Theory Lecture 16
![Page 10: Math 412: Number Theory Lecture 16 · djp 1F(d) P djp 1˚(d) = p 1. So F(d) = ˚(d) for each d.) For every prime p, there is a primitive root modulo p. In fact, there are ˚(p 1)](https://reader034.vdocuments.site/reader034/viewer/2022050506/5f981bf5cb5cdf61ff493fbf/html5/thumbnails/10.jpg)
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 [email protected] Math 412: Number Theory Lecture 16
![Page 11: Math 412: Number Theory Lecture 16 · djp 1F(d) P djp 1˚(d) = p 1. So F(d) = ˚(d) for each d.) For every prime p, there is a primitive root modulo p. In fact, there are ˚(p 1)](https://reader034.vdocuments.site/reader034/viewer/2022050506/5f981bf5cb5cdf61ff493fbf/html5/thumbnails/11.jpg)
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 [email protected] Math 412: Number Theory Lecture 16
![Page 12: Math 412: Number Theory Lecture 16 · djp 1F(d) P djp 1˚(d) = p 1. So F(d) = ˚(d) for each d.) For every prime p, there is a primitive root modulo p. In fact, there are ˚(p 1)](https://reader034.vdocuments.site/reader034/viewer/2022050506/5f981bf5cb5cdf61ff493fbf/html5/thumbnails/12.jpg)
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 [email protected] Math 412: Number Theory Lecture 16
![Page 13: Math 412: Number Theory Lecture 16 · djp 1F(d) P djp 1˚(d) = p 1. So F(d) = ˚(d) for each d.) For every prime p, there is a primitive root modulo p. In fact, there are ˚(p 1)](https://reader034.vdocuments.site/reader034/viewer/2022050506/5f981bf5cb5cdf61ff493fbf/html5/thumbnails/13.jpg)
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 [email protected] Math 412: Number Theory Lecture 16
![Page 14: Math 412: Number Theory Lecture 16 · djp 1F(d) P djp 1˚(d) = p 1. So F(d) = ˚(d) for each d.) For every prime p, there is a primitive root modulo p. In fact, there are ˚(p 1)](https://reader034.vdocuments.site/reader034/viewer/2022050506/5f981bf5cb5cdf61ff493fbf/html5/thumbnails/14.jpg)
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 [email protected] Math 412: Number Theory Lecture 16
![Page 15: Math 412: Number Theory Lecture 16 · djp 1F(d) P djp 1˚(d) = p 1. So F(d) = ˚(d) for each d.) For every prime p, there is a primitive root modulo p. In fact, there are ˚(p 1)](https://reader034.vdocuments.site/reader034/viewer/2022050506/5f981bf5cb5cdf61ff493fbf/html5/thumbnails/15.jpg)
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 [email protected] Math 412: Number Theory Lecture 16
![Page 16: Math 412: Number Theory Lecture 16 · djp 1F(d) P djp 1˚(d) = p 1. So F(d) = ˚(d) for each d.) For every prime p, there is a primitive root modulo p. In fact, there are ˚(p 1)](https://reader034.vdocuments.site/reader034/viewer/2022050506/5f981bf5cb5cdf61ff493fbf/html5/thumbnails/16.jpg)
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 [email protected] Math 412: Number Theory Lecture 16
![Page 17: Math 412: Number Theory Lecture 16 · djp 1F(d) P djp 1˚(d) = p 1. So F(d) = ˚(d) for each d.) For every prime p, there is a primitive root modulo p. In fact, there are ˚(p 1)](https://reader034.vdocuments.site/reader034/viewer/2022050506/5f981bf5cb5cdf61ff493fbf/html5/thumbnails/17.jpg)
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 [email protected] Math 412: Number Theory Lecture 16
![Page 18: Math 412: Number Theory Lecture 16 · djp 1F(d) P djp 1˚(d) = p 1. So F(d) = ˚(d) for each d.) For every prime p, there is a primitive root modulo p. In fact, there are ˚(p 1)](https://reader034.vdocuments.site/reader034/viewer/2022050506/5f981bf5cb5cdf61ff493fbf/html5/thumbnails/18.jpg)
Lemma: if r is a primitive root modulo p, then r or r + p is a primitiveroot modulo p2.
Gexin Yu [email protected] Math 412: Number Theory Lecture 16
![Page 19: Math 412: Number Theory Lecture 16 · djp 1F(d) P djp 1˚(d) = p 1. So F(d) = ˚(d) for each d.) For every prime p, there is a primitive root modulo p. In fact, there are ˚(p 1)](https://reader034.vdocuments.site/reader034/viewer/2022050506/5f981bf5cb5cdf61ff493fbf/html5/thumbnails/19.jpg)
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 [email protected] Math 412: Number Theory Lecture 16
![Page 20: Math 412: Number Theory Lecture 16 · djp 1F(d) P djp 1˚(d) = p 1. So F(d) = ˚(d) for each d.) For every prime p, there is a primitive root modulo p. In fact, there are ˚(p 1)](https://reader034.vdocuments.site/reader034/viewer/2022050506/5f981bf5cb5cdf61ff493fbf/html5/thumbnails/20.jpg)
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 [email protected] Math 412: Number Theory Lecture 16
![Page 21: Math 412: Number Theory Lecture 16 · djp 1F(d) P djp 1˚(d) = p 1. So F(d) = ˚(d) for each d.) For every prime p, there is a primitive root modulo p. In fact, there are ˚(p 1)](https://reader034.vdocuments.site/reader034/viewer/2022050506/5f981bf5cb5cdf61ff493fbf/html5/thumbnails/21.jpg)
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 [email protected] Math 412: Number Theory Lecture 16
![Page 22: Math 412: Number Theory Lecture 16 · djp 1F(d) P djp 1˚(d) = p 1. So F(d) = ˚(d) for each d.) For every prime p, there is a primitive root modulo p. In fact, there are ˚(p 1)](https://reader034.vdocuments.site/reader034/viewer/2022050506/5f981bf5cb5cdf61ff493fbf/html5/thumbnails/22.jpg)
Lemma: if a is odd and k ≥ 3, then aφ(2k )/2 ≡ 1 (mod 2k).
Idea of proof: induction on k .
Gexin Yu [email protected] Math 412: Number Theory Lecture 16
![Page 23: Math 412: Number Theory Lecture 16 · djp 1F(d) P djp 1˚(d) = p 1. So F(d) = ˚(d) for each d.) For every prime p, there is a primitive root modulo p. In fact, there are ˚(p 1)](https://reader034.vdocuments.site/reader034/viewer/2022050506/5f981bf5cb5cdf61ff493fbf/html5/thumbnails/23.jpg)
Lemma: if a is odd and k ≥ 3, then aφ(2k )/2 ≡ 1 (mod 2k).
Idea of proof: induction on k .
Gexin Yu [email protected] Math 412: Number Theory Lecture 16
![Page 24: Math 412: Number Theory Lecture 16 · djp 1F(d) P djp 1˚(d) = p 1. So F(d) = ˚(d) for each d.) For every prime p, there is a primitive root modulo p. In fact, there are ˚(p 1)](https://reader034.vdocuments.site/reader034/viewer/2022050506/5f981bf5cb5cdf61ff493fbf/html5/thumbnails/24.jpg)
Lemma: ord2k (5) = φ(2k)/2.
pf: indcution on k to show that 52k−3 ≡ 1 + 2k−1 6≡ 1 (mod 2k).
Gexin Yu [email protected] Math 412: Number Theory Lecture 16
![Page 25: Math 412: Number Theory Lecture 16 · djp 1F(d) P djp 1˚(d) = p 1. So F(d) = ˚(d) for each d.) For every prime p, there is a primitive root modulo p. In fact, there are ˚(p 1)](https://reader034.vdocuments.site/reader034/viewer/2022050506/5f981bf5cb5cdf61ff493fbf/html5/thumbnails/25.jpg)
Lemma: ord2k (5) = φ(2k)/2.
pf: indcution on k to show that 52k−3 ≡ 1 + 2k−1 6≡ 1 (mod 2k).
Gexin Yu [email protected] Math 412: Number Theory Lecture 16
![Page 26: Math 412: Number Theory Lecture 16 · djp 1F(d) P djp 1˚(d) = p 1. So F(d) = ˚(d) for each d.) For every prime p, there is a primitive root modulo p. In fact, there are ˚(p 1)](https://reader034.vdocuments.site/reader034/viewer/2022050506/5f981bf5cb5cdf61ff493fbf/html5/thumbnails/26.jpg)
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 [email protected] Math 412: Number Theory Lecture 16
![Page 27: Math 412: Number Theory Lecture 16 · djp 1F(d) P djp 1˚(d) = p 1. So F(d) = ˚(d) for each d.) For every prime p, there is a primitive root modulo p. In fact, there are ˚(p 1)](https://reader034.vdocuments.site/reader034/viewer/2022050506/5f981bf5cb5cdf61ff493fbf/html5/thumbnails/27.jpg)
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 [email protected] Math 412: Number Theory Lecture 16
![Page 28: Math 412: Number Theory Lecture 16 · djp 1F(d) P djp 1˚(d) = p 1. So F(d) = ˚(d) for each d.) For every prime p, there is a primitive root modulo p. In fact, there are ˚(p 1)](https://reader034.vdocuments.site/reader034/viewer/2022050506/5f981bf5cb5cdf61ff493fbf/html5/thumbnails/28.jpg)
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 [email protected] Math 412: Number Theory Lecture 16
![Page 29: Math 412: Number Theory Lecture 16 · djp 1F(d) P djp 1˚(d) = p 1. So F(d) = ˚(d) for each d.) For every prime p, there is a primitive root modulo p. In fact, there are ˚(p 1)](https://reader034.vdocuments.site/reader034/viewer/2022050506/5f981bf5cb5cdf61ff493fbf/html5/thumbnails/29.jpg)
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 [email protected] Math 412: Number Theory Lecture 16
![Page 30: Math 412: Number Theory Lecture 16 · djp 1F(d) P djp 1˚(d) = p 1. So F(d) = ˚(d) for each d.) For every prime p, there is a primitive root modulo p. In fact, there are ˚(p 1)](https://reader034.vdocuments.site/reader034/viewer/2022050506/5f981bf5cb5cdf61ff493fbf/html5/thumbnails/30.jpg)
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 [email protected] Math 412: Number Theory Lecture 16
![Page 31: Math 412: Number Theory Lecture 16 · djp 1F(d) P djp 1˚(d) = p 1. So F(d) = ˚(d) for each d.) For every prime p, there is a primitive root modulo p. In fact, there are ˚(p 1)](https://reader034.vdocuments.site/reader034/viewer/2022050506/5f981bf5cb5cdf61ff493fbf/html5/thumbnails/31.jpg)
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 [email protected] Math 412: Number Theory Lecture 16
![Page 32: Math 412: Number Theory Lecture 16 · djp 1F(d) P djp 1˚(d) = p 1. So F(d) = ˚(d) for each d.) For every prime p, there is a primitive root modulo p. In fact, there are ˚(p 1)](https://reader034.vdocuments.site/reader034/viewer/2022050506/5f981bf5cb5cdf61ff493fbf/html5/thumbnails/32.jpg)
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 [email protected] Math 412: Number Theory Lecture 16
![Page 33: Math 412: Number Theory Lecture 16 · djp 1F(d) P djp 1˚(d) = p 1. So F(d) = ˚(d) for each d.) For every prime p, there is a primitive root modulo p. In fact, there are ˚(p 1)](https://reader034.vdocuments.site/reader034/viewer/2022050506/5f981bf5cb5cdf61ff493fbf/html5/thumbnails/33.jpg)
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 [email protected] Math 412: Number Theory Lecture 16
![Page 34: Math 412: Number Theory Lecture 16 · djp 1F(d) P djp 1˚(d) = p 1. So F(d) = ˚(d) for each d.) For every prime p, there is a primitive root modulo p. In fact, there are ˚(p 1)](https://reader034.vdocuments.site/reader034/viewer/2022050506/5f981bf5cb5cdf61ff493fbf/html5/thumbnails/34.jpg)
Ex: Solve equation 6x12 ≡ 11 (mod 17). (note that 3 is a primitive root)
Gexin Yu [email protected] Math 412: Number Theory Lecture 16
![Page 35: Math 412: Number Theory Lecture 16 · djp 1F(d) P djp 1˚(d) = p 1. So F(d) = ˚(d) for each d.) For every prime p, there is a primitive root modulo p. In fact, there are ˚(p 1)](https://reader034.vdocuments.site/reader034/viewer/2022050506/5f981bf5cb5cdf61ff493fbf/html5/thumbnails/35.jpg)
Ex: Solve 7x ≡ 6 (mod 17). (again, 3 is a primitive root).
Gexin Yu [email protected] Math 412: Number Theory Lecture 16