Quadratic Reciprocity

Home , Cubic reciprocity, Quadratic reciprocity

Jennifer Li

Department of Mathematics Louisiana State University Baton Rouge Theorem. Let n be a nonzero integer. Then there is a prime factorization Y n ( 1)²(n) pa(p) = − p 0 > with the exponents uniquely determined by n. Note: ²(n) = 0 or 1 (depends on sign of n).

More generally: k[x]: Ring of polynomials with coefﬁcients in ﬁeld k • Even more generally: Principal Ideal Domains •

Unique Factorization

Jennifer Li (Louisiana State University) Quadratic Reciprocity May 8, 2015 2 / 79 Note: ²(n) = 0 or 1 (depends on sign of n).

More generally: k[x]: Ring of polynomials with coefﬁcients in ﬁeld k • Even more generally: Principal Ideal Domains •

Unique Factorization

Theorem. Let n be a nonzero integer. Then there is a prime factorization Y n ( 1)²(n) pa(p) = − p 0 > with the exponents uniquely determined by n.

Jennifer Li (Louisiana State University) Quadratic Reciprocity May 8, 2015 2 / 79 More generally: k[x]: Ring of polynomials with coefﬁcients in ﬁeld k • Even more generally: Principal Ideal Domains •

Unique Factorization

Theorem. Let n be a nonzero integer. Then there is a prime factorization Y n ( 1)²(n) pa(p) = − p 0 > with the exponents uniquely determined by n. Note: ²(n) = 0 or 1 (depends on sign of n).

Jennifer Li (Louisiana State University) Quadratic Reciprocity May 8, 2015 2 / 79 Even more generally: Principal Ideal Domains •

Unique Factorization

Theorem. Let n be a nonzero integer. Then there is a prime factorization Y n ( 1)²(n) pa(p) = − p 0 > with the exponents uniquely determined by n. Note: ²(n) = 0 or 1 (depends on sign of n).

More generally: k[x]: Ring of polynomials with coefﬁcients in ﬁeld k •

Jennifer Li (Louisiana State University) Quadratic Reciprocity May 8, 2015 2 / 79 Unique Factorization

Theorem. Let n be a nonzero integer. Then there is a prime factorization Y n ( 1)²(n) pa(p) = − p 0 > with the exponents uniquely determined by n. Note: ²(n) = 0 or 1 (depends on sign of n).

More generally: k[x]: Ring of polynomials with coefﬁcients in ﬁeld k • Even more generally: Principal Ideal Domains •

Jennifer Li (Louisiana State University) Quadratic Reciprocity May 8, 2015 2 / 79 such that for all a,b R, with b 0, there exists c,d R where ∈ 6= ∈ a cb d and = + N(d) N(b) or d 0 < = Examples: Z and k[x]

An integral domain R is a Principal Ideal Domain if every ideal of R is principal.

Every Euclidean Domain is a Principal Ideal Domain

Euclidean and Principal Ideal Domains

An integral domain R is a Euclidean Domain if there is a normN

N : R\{0} N −→

Jennifer Li (Louisiana State University) Quadratic Reciprocity May 8, 2015 3 / 79 Examples: Z and k[x]

An integral domain R is a Principal Ideal Domain if every ideal of R is principal.

Every Euclidean Domain is a Principal Ideal Domain

Euclidean and Principal Ideal Domains

An integral domain R is a Euclidean Domain if there is a normN

N : R\{0} N −→ such that for all a,b R, with b 0, there exists c,d R where ∈ 6= ∈ a cb d and = + N(d) N(b) or d 0 < =

Jennifer Li (Louisiana State University) Quadratic Reciprocity May 8, 2015 3 / 79 An integral domain R is a Principal Ideal Domain if every ideal of R is principal.

Every Euclidean Domain is a Principal Ideal Domain

Euclidean and Principal Ideal Domains

An integral domain R is a Euclidean Domain if there is a normN

N : R\{0} N −→ such that for all a,b R, with b 0, there exists c,d R where ∈ 6= ∈ a cb d and = + N(d) N(b) or d 0 < = Examples: Z and k[x]

Jennifer Li (Louisiana State University) Quadratic Reciprocity May 8, 2015 3 / 79 Every Euclidean Domain is a Principal Ideal Domain

Euclidean and Principal Ideal Domains

An integral domain R is a Euclidean Domain if there is a normN

N : R\{0} N −→ such that for all a,b R, with b 0, there exists c,d R where ∈ 6= ∈ a cb d and = + N(d) N(b) or d 0 < = Examples: Z and k[x]

An integral domain R is a Principal Ideal Domain if every ideal of R is principal.

Jennifer Li (Louisiana State University) Quadratic Reciprocity May 8, 2015 3 / 79 Euclidean and Principal Ideal Domains

An integral domain R is a Euclidean Domain if there is a normN

N : R\{0} N −→ such that for all a,b R, with b 0, there exists c,d R where ∈ 6= ∈ a cb d and = + N(d) N(b) or d 0 < = Examples: Z and k[x]

An integral domain R is a Principal Ideal Domain if every ideal of R is principal.

Every Euclidean Domain is a Principal Ideal Domain

Jennifer Li (Louisiana State University) Quadratic Reciprocity May 8, 2015 3 / 79 For α a bω Z[ω], deﬁne = + ∈ N(α) a2 ab b2 = − + Z[ω] is a Euclidean Domain. ⇒ Z[ω] is also a Principal Ideal Domain. ⇒

Example

The ring Z[ω] {a bω a,b Z} where ω ( 1 p 3)/2 = + | ∈ = − + −

Jennifer Li (Louisiana State University) Quadratic Reciprocity May 8, 2015 4 / 79 Z[ω] is a Euclidean Domain. ⇒ Z[ω] is also a Principal Ideal Domain. ⇒

Example

The ring Z[ω] {a bω a,b Z} where ω ( 1 p 3)/2 = + | ∈ = − + − For α a bω Z[ω], deﬁne = + ∈ N(α) a2 ab b2 = − +

Jennifer Li (Louisiana State University) Quadratic Reciprocity May 8, 2015 4 / 79 Z[ω] is also a Principal Ideal Domain. ⇒

Example

The ring Z[ω] {a bω a,b Z} where ω ( 1 p 3)/2 = + | ∈ = − + − For α a bω Z[ω], deﬁne = + ∈ N(α) a2 ab b2 = − + Z[ω] is a Euclidean Domain. ⇒

Jennifer Li (Louisiana State University) Quadratic Reciprocity May 8, 2015 4 / 79 Example

The ring Z[ω] {a bω a,b Z} where ω ( 1 p 3)/2 = + | ∈ = − + − For α a bω Z[ω], deﬁne = + ∈ N(α) a2 ab b2 = − + Z[ω] is a Euclidean Domain. ⇒ Z[ω] is also a Principal Ideal Domain. ⇒

Jennifer Li (Louisiana State University) Quadratic Reciprocity May 8, 2015 4 / 79 Caution B A prime in Z may not be a prime in Z[ω]

Example: p 7 is prime in Z. = But in Z[ω]:

7 (3 ω)(2 ω) = + −

In Z: “rational primes p" In Z[ω]: “primes π"

Primes in Z[ω]

Jennifer Li (Louisiana State University) Quadratic Reciprocity May 8, 2015 5 / 79 A prime in Z may not be a prime in Z[ω]

Example: p 7 is prime in Z. = But in Z[ω]:

7 (3 ω)(2 ω) = + −

In Z: “rational primes p" In Z[ω]: “primes π"

Primes in Z[ω]

Caution B

Jennifer Li (Louisiana State University) Quadratic Reciprocity May 8, 2015 5 / 79 Example: p 7 is prime in Z. = But in Z[ω]:

7 (3 ω)(2 ω) = + −

In Z: “rational primes p" In Z[ω]: “primes π"

Primes in Z[ω]

Caution B A prime in Z may not be a prime in Z[ω]

Jennifer Li (Louisiana State University) Quadratic Reciprocity May 8, 2015 5 / 79 In Z: “rational primes p" In Z[ω]: “primes π"

Primes in Z[ω]

Caution B A prime in Z may not be a prime in Z[ω]

Example: p 7 is prime in Z. = But in Z[ω]:

7 (3 ω)(2 ω) = + −

Jennifer Li (Louisiana State University) Quadratic Reciprocity May 8, 2015 5 / 79 In Z[ω]: “primes π"

Primes in Z[ω]

Caution B A prime in Z may not be a prime in Z[ω]

Example: p 7 is prime in Z. = But in Z[ω]:

7 (3 ω)(2 ω) = + −

In Z: “rational primes p"

Jennifer Li (Louisiana State University) Quadratic Reciprocity May 8, 2015 5 / 79 Primes in Z[ω]

Caution B A prime in Z may not be a prime in Z[ω]

Example: p 7 is prime in Z. = But in Z[ω]:

7 (3 ω)(2 ω) = + −

In Z: “rational primes p" In Z[ω]: “primes π"

Jennifer Li (Louisiana State University) Quadratic Reciprocity May 8, 2015 5 / 79 Then a b(mod m) means that m divides b a. ≡ −

Congruence

Suppose a,b,m Z and m 0 ∈ 6=

Jennifer Li (Louisiana State University) Quadratic Reciprocity May 8, 2015 6 / 79 Congruence

Suppose a,b,m Z and m 0 ∈ 6= Then a b(mod m) means that m divides b a. ≡ −

Jennifer Li (Louisiana State University) Quadratic Reciprocity May 8, 2015 6 / 79 xn a(mod p) ≡

Fp: Finite ﬁeld with p elements.

Finite Fields

Let a,n Z. ∈ Suppose p is prime and and gcd (a,p) 1. =

Jennifer Li (Louisiana State University) Quadratic Reciprocity May 8, 2015 7 / 79 Fp: Finite ﬁeld with p elements.

Finite Fields

Let a,n Z. ∈ Suppose p is prime and and gcd (a,p) 1. = xn a(mod p) ≡

Jennifer Li (Louisiana State University) Quadratic Reciprocity May 8, 2015 7 / 79 Finite Fields

Let a,n Z. ∈ Suppose p is prime and and gcd (a,p) 1. = xn a(mod p) ≡

Fp: Finite ﬁeld with p elements.

Jennifer Li (Louisiana State University) Quadratic Reciprocity May 8, 2015 7 / 79 Elements: roots of xq x. − Multiplicative group of F: denoted Fq∗.

F∗ q 1 | q | = − Fq∗ is cyclic with generator b.

2 q 2 {1, b, b , ..., b − }

q 1 Note: b − 1. =

Finite Fields

Fq: ﬁnite ﬁeld with q elements.

Jennifer Li (Louisiana State University) Quadratic Reciprocity May 8, 2015 8 / 79 Multiplicative group of F: denoted Fq∗.

F∗ q 1 | q | = − Fq∗ is cyclic with generator b.

2 q 2 {1, b, b , ..., b − }

q 1 Note: b − 1. =

Finite Fields

Fq: ﬁnite ﬁeld with q elements. Elements: roots of xq x. −

Jennifer Li (Louisiana State University) Quadratic Reciprocity May 8, 2015 8 / 79 F∗ q 1 | q | = − Fq∗ is cyclic with generator b.

2 q 2 {1, b, b , ..., b − }

q 1 Note: b − 1. =

Finite Fields

Fq: ﬁnite ﬁeld with q elements. Elements: roots of xq x. − Multiplicative group of F: denoted Fq∗.

Jennifer Li (Louisiana State University) Quadratic Reciprocity May 8, 2015 8 / 79 Fq∗ is cyclic with generator b.

2 q 2 {1, b, b , ..., b − }

q 1 Note: b − 1. =

Finite Fields

Fq: ﬁnite ﬁeld with q elements. Elements: roots of xq x. − Multiplicative group of F: denoted Fq∗.

F∗ q 1 | q | = −

Jennifer Li (Louisiana State University) Quadratic Reciprocity May 8, 2015 8 / 79 Finite Fields

Fq: ﬁnite ﬁeld with q elements. Elements: roots of xq x. − Multiplicative group of F: denoted Fq∗.

F∗ q 1 | q | = − Fq∗ is cyclic with generator b.

2 q 2 {1, b, b , ..., b − }

q 1 Note: b − 1. =

Jennifer Li (Louisiana State University) Quadratic Reciprocity May 8, 2015 8 / 79 Primitive Roots

q 1 If q 1 is the smallest positive integer such that b − 1(mod q), − ≡ then b is a primitive root modulo q.

Jennifer Li (Louisiana State University) Quadratic Reciprocity May 8, 2015 9 / 79 Primitive Roots

Theorem. If h is the smallest positive integer such that

ah 1(mod q), ≡ and if r Z such that ∈ ar 1(mod q), ≡ then h r. |

Jennifer Li (Louisiana State University) Quadratic Reciprocity May 8, 2015 10 / 79 For which primes p is x2 a(mod p) solvable? ≡

Quadratic Reciprocity

Let p be prime and suppose p - a.

Jennifer Li (Louisiana State University) Quadratic Reciprocity May 8, 2015 11 / 79 Quadratic Reciprocity

Quadratic Reciprocity

Let p be prime and suppose p - a. For which primes p is x2 a(mod p) solvable? ≡

Jennifer Li (Louisiana State University) Quadratic Reciprocity May 8, 2015 11 / 79 Quadratic Reciprocity

Let p be prime and suppose p - a. For which primes p is x2 a(mod p) solvable? ≡

Quadratic Reciprocity

Jennifer Li (Louisiana State University) Quadratic Reciprocity May 8, 2015 11 / 79 YES a is a quadratic residue modulo p. ⇒ NO a is a quadratic nonresidue modulo p. ⇒

Quadratic Residues

Is there a solution to x2 a(mod p)? ≡

Jennifer Li (Louisiana State University) Quadratic Reciprocity May 8, 2015 12 / 79 NO a is a quadratic nonresidue modulo p. ⇒

Quadratic Residues

Is there a solution to x2 a(mod p)? ≡ YES a is a quadratic residue modulo p. ⇒

Jennifer Li (Louisiana State University) Quadratic Reciprocity May 8, 2015 12 / 79 Quadratic Residues

Is there a solution to x2 a(mod p)? ≡ YES a is a quadratic residue modulo p. ⇒ NO a is a quadratic nonresidue modulo p. ⇒

Jennifer Li (Louisiana State University) Quadratic Reciprocity May 8, 2015 12 / 79 3 is a quadratic nonresidue modulo 7, since there is no solution to • x2 3(mod 7) ≡

Quadratic Residues: Example

2 is a quadratic residue modulo 7, since • 32 9 2(mod 7) = ≡ x 3 is a solution to x2 2(mod 7)! = ≡

Jennifer Li (Louisiana State University) Quadratic Reciprocity May 8, 2015 13 / 79 Quadratic Residues: Example

2 is a quadratic residue modulo 7, since • 32 9 2(mod 7) = ≡ x 3 is a solution to x2 2(mod 7)! = ≡ 3 is a quadratic nonresidue modulo 7, since there is no solution to • x2 3(mod 7) ≡

Jennifer Li (Louisiana State University) Quadratic Reciprocity May 8, 2015 13 / 79 {1,2,4}: quadratic residues modulo 7

Quadratic Residues: Example

12 1 1(mod 7) = ≡ 22 4 4(mod 7) = ≡ 32 9 2(mod 7) = ≡ 42 16 2(mod 7) = ≡ 52 25 4(mod 7) = ≡ 62 36 1(mod 7) = ≡. .

Jennifer Li (Louisiana State University) Quadratic Reciprocity May 8, 2015 14 / 79 Quadratic Residues: Example

12 1 1(mod 7) = ≡ 22 4 4(mod 7) = ≡ 32 9 2(mod 7) = ≡ 42 16 2(mod 7) = ≡ 52 25 4(mod 7) = ≡ 62 36 1(mod 7) = ≡. .

{1,2,4}: quadratic residues modulo 7

Jennifer Li (Louisiana State University) Quadratic Reciprocity May 8, 2015 14 / 79  1 if a is a quadratic residue modulo p  (a/p) 1 if a is a quadratic nonresidue modulo p = −  0 if p divides a

µ a ¶ Sometimes denoted p

Legendre Symbol

p : an odd prime

Jennifer Li (Louisiana State University) Quadratic Reciprocity May 8, 2015 15 / 79 µ a ¶ Sometimes denoted p

Legendre Symbol

p : an odd prime

 1 if a is a quadratic residue modulo p  (a/p) 1 if a is a quadratic nonresidue modulo p = −  0 if p divides a

Jennifer Li (Louisiana State University) Quadratic Reciprocity May 8, 2015 15 / 79 Legendre Symbol

p : an odd prime

 1 if a is a quadratic residue modulo p  (a/p) 1 if a is a quadratic nonresidue modulo p = −  0 if p divides a

µ a ¶ Sometimes denoted p

Jennifer Li (Louisiana State University) Quadratic Reciprocity May 8, 2015 15 / 79 3 is a quadratic nonresidule modulo 7: • (3/7) 1 = −

Legendre Symbol: Example

2 is a quadratic residule modulo 7: • (2/7) 1 =

Jennifer Li (Louisiana State University) Quadratic Reciprocity May 8, 2015 16 / 79 Legendre Symbol: Example

2 is a quadratic residule modulo 7: • (2/7) 1 = 3 is a quadratic nonresidule modulo 7: • (3/7) 1 = −

Jennifer Li (Louisiana State University) Quadratic Reciprocity May 8, 2015 16 / 79 (p 1)/2 (1) a − (a/p)(mod p) ≡ 1 − 1

Legendre Symbol: 3 Properties

Suppose p - a.

Jennifer Li (Louisiana State University) Quadratic Reciprocity May 8, 2015 17 / 79 Legendre Symbol: 3 Properties

Suppose p - a.

(p 1)/2 (1) a − (a/p)(mod p) ≡ 1 − 1

Jennifer Li (Louisiana State University) Quadratic Reciprocity May 8, 2015 17 / 79 Importance: Residue x Residue = Residue Residue x Nonresidue = Nonresidue

Legendre Symbol: 3 Properties

(2) (ab/p) (a/p) (b/p) = ·

Jennifer Li (Louisiana State University) Quadratic Reciprocity May 8, 2015 18 / 79 Legendre Symbol: 3 Properties

(2) (ab/p) (a/p) (b/p) = · Importance: Residue x Residue = Residue Residue x Nonresidue = Nonresidue

Jennifer Li (Louisiana State University) Quadratic Reciprocity May 8, 2015 18 / 79 Legendre Symbol: 3 Properties

(3) If a b(mod p), then (a/p) (b/p) ≡ =

Jennifer Li (Louisiana State University) Quadratic Reciprocity May 8, 2015 19 / 79 The set of least residues modulo p :

n (p 1) (p 3) (p 1)o S − , − , , 1, 1, 2, , − = − 2 − 2 ··· − ··· 2

Let n Z. ∈ The least residue of n modulo p is an integer x S s.t. ∈ n x(mod p) ≡

Denoted by LRp(n)

Least Residues Modulo p

Reminder: p is an odd prime!

Jennifer Li (Louisiana State University) Quadratic Reciprocity May 8, 2015 20 / 79 Let n Z. ∈ The least residue of n modulo p is an integer x S s.t. ∈ n x(mod p) ≡

Denoted by LRp(n)

Least Residues Modulo p

Reminder: p is an odd prime!

The set of least residues modulo p :

n (p 1) (p 3) (p 1)o S − , − , , 1, 1, 2, , − = − 2 − 2 ··· − ··· 2

Jennifer Li (Louisiana State University) Quadratic Reciprocity May 8, 2015 20 / 79 Denoted by LRp(n)

Least Residues Modulo p

Reminder: p is an odd prime!

The set of least residues modulo p :

n (p 1) (p 3) (p 1)o S − , − , , 1, 1, 2, , − = − 2 − 2 ··· − ··· 2

Let n Z. ∈ The least residue of n modulo p is an integer x S s.t. ∈ n x(mod p) ≡

Jennifer Li (Louisiana State University) Quadratic Reciprocity May 8, 2015 20 / 79 Least Residues Modulo p

Reminder: p is an odd prime!

The set of least residues modulo p :

n (p 1) (p 3) (p 1)o S − , − , , 1, 1, 2, , − = − 2 − 2 ··· − ··· 2

Let n Z. ∈ The least residue of n modulo p is an integer x S s.t. ∈ n x(mod p) ≡

Denoted by LRp(n)

Jennifer Li (Louisiana State University) Quadratic Reciprocity May 8, 2015 20 / 79 Let p 11. = Then the set of least residues modulo 11 is:

S { 5, 4, 3, 2, 1, 1, 2, 3, 4, 5}. = − − − − − Then, LR (21) 1 since 21 1(mod 11) 11 =− ≡−

Least Residues Modulo p : Example

n (p 1) (p 3) (p 1)o S − , − , , 1, 1, 2, , − = − 2 − 2 ··· − ··· 2

Jennifer Li (Louisiana State University) Quadratic Reciprocity May 8, 2015 21 / 79 Then the set of least residues modulo 11 is:

S { 5, 4, 3, 2, 1, 1, 2, 3, 4, 5}. = − − − − − Then, LR (21) 1 since 21 1(mod 11) 11 =− ≡−

Least Residues Modulo p : Example

n (p 1) (p 3) (p 1)o S − , − , , 1, 1, 2, , − = − 2 − 2 ··· − ··· 2

Let p 11. =

Jennifer Li (Louisiana State University) Quadratic Reciprocity May 8, 2015 21 / 79 Least Residues Modulo p : Example

n (p 1) (p 3) (p 1)o S − , − , , 1, 1, 2, , − = − 2 − 2 ··· − ··· 2

Let p 11. = Then the set of least residues modulo 11 is:

S { 5, 4, 3, 2, 1, 1, 2, 3, 4, 5}. = − − − − − Then, LR (21) 1 since 21 1(mod 11) 11 =− ≡−

Jennifer Li (Louisiana State University) Quadratic Reciprocity May 8, 2015 21 / 79 n (p 1) o Consider set T a, 2a, 3a, , − a . = ··· 2

Denote by µ the number of least residues of T which are negative.

Negative Least Residues Modulo p

Let a Z and suppose p - a. ∈

Jennifer Li (Louisiana State University) Quadratic Reciprocity May 8, 2015 22 / 79 Denote by µ the number of least residues of T which are negative.

Negative Least Residues Modulo p

Let a Z and suppose p - a. ∈ n (p 1) o Consider set T a, 2a, 3a, , − a . = ··· 2

Jennifer Li (Louisiana State University) Quadratic Reciprocity May 8, 2015 22 / 79 Negative Least Residues Modulo p

Let a Z and suppose p - a. ∈ n (p 1) o Consider set T a, 2a, 3a, , − a . = ··· 2

Denote by µ the number of least residues of T which are negative.

Jennifer Li (Louisiana State University) Quadratic Reciprocity May 8, 2015 22 / 79 Let p 7 and a 4. = = (7 1) Then − 3. 2 = T { 4, 2(4) 3(4) } { 4, 8, 12 }. = = S { 3, 2, 1, 1, 2, 3 }. = − − − Least residues: LR (4) 3 7 = − LR (8) 1 7 = LR (12) 2 7 = − µ 2. ⇒ =

Negative Least Residues Modulo p : Example

n (p 1) (p 3) (p 1)o S − , − , , 1, 1, 2, , − and = − 2 − 2 ··· − ··· 2 n (p 1) o T a, 2a, 3a, , − a = ··· 2

Jennifer Li (Louisiana State University) Quadratic Reciprocity May 8, 2015 23 / 79 (7 1) Then − 3. 2 = T { 4, 2(4) 3(4) } { 4, 8, 12 }. = = S { 3, 2, 1, 1, 2, 3 }. = − − − Least residues: LR (4) 3 7 = − LR (8) 1 7 = LR (12) 2 7 = − µ 2. ⇒ =

Negative Least Residues Modulo p : Example

n (p 1) (p 3) (p 1)o S − , − , , 1, 1, 2, , − and = − 2 − 2 ··· − ··· 2 n (p 1) o T a, 2a, 3a, , − a = ··· 2

Let p 7 and a 4. = =

Jennifer Li (Louisiana State University) Quadratic Reciprocity May 8, 2015 23 / 79 T { 4, 2(4) 3(4) } { 4, 8, 12 }. = = S { 3, 2, 1, 1, 2, 3 }. = − − − Least residues: LR (4) 3 7 = − LR (8) 1 7 = LR (12) 2 7 = − µ 2. ⇒ =

Negative Least Residues Modulo p : Example

n (p 1) (p 3) (p 1)o S − , − , , 1, 1, 2, , − and = − 2 − 2 ··· − ··· 2 n (p 1) o T a, 2a, 3a, , − a = ··· 2

Let p 7 and a 4. = = (7 1) Then − 3. 2 =

Jennifer Li (Louisiana State University) Quadratic Reciprocity May 8, 2015 23 / 79 S { 3, 2, 1, 1, 2, 3 }. = − − − Least residues: LR (4) 3 7 = − LR (8) 1 7 = LR (12) 2 7 = − µ 2. ⇒ =

Negative Least Residues Modulo p : Example

n (p 1) (p 3) (p 1)o S − , − , , 1, 1, 2, , − and = − 2 − 2 ··· − ··· 2 n (p 1) o T a, 2a, 3a, , − a = ··· 2

Let p 7 and a 4. = = (7 1) Then − 3. 2 = T { 4, 2(4) 3(4) } { 4, 8, 12 }. = =

Jennifer Li (Louisiana State University) Quadratic Reciprocity May 8, 2015 23 / 79 Least residues: LR (4) 3 7 = − LR (8) 1 7 = LR (12) 2 7 = − µ 2. ⇒ =

Negative Least Residues Modulo p : Example

n (p 1) (p 3) (p 1)o S − , − , , 1, 1, 2, , − and = − 2 − 2 ··· − ··· 2 n (p 1) o T a, 2a, 3a, , − a = ··· 2

Let p 7 and a 4. = = (7 1) Then − 3. 2 = T { 4, 2(4) 3(4) } { 4, 8, 12 }. = = S { 3, 2, 1, 1, 2, 3 }. = − − −

Jennifer Li (Louisiana State University) Quadratic Reciprocity May 8, 2015 23 / 79 LR (4) 3 7 = − LR (8) 1 7 = LR (12) 2 7 = − µ 2. ⇒ =

Negative Least Residues Modulo p : Example

n (p 1) (p 3) (p 1)o S − , − , , 1, 1, 2, , − and = − 2 − 2 ··· − ··· 2 n (p 1) o T a, 2a, 3a, , − a = ··· 2

Let p 7 and a 4. = = (7 1) Then − 3. 2 = T { 4, 2(4) 3(4) } { 4, 8, 12 }. = = S { 3, 2, 1, 1, 2, 3 }. = − − − Least residues:

Jennifer Li (Louisiana State University) Quadratic Reciprocity May 8, 2015 23 / 79 LR (12) 2 7 = − µ 2. ⇒ =

Negative Least Residues Modulo p : Example

n (p 1) (p 3) (p 1)o S − , − , , 1, 1, 2, , − and = − 2 − 2 ··· − ··· 2 n (p 1) o T a, 2a, 3a, , − a = ··· 2

Let p 7 and a 4. = = (7 1) Then − 3. 2 = T { 4, 2(4) 3(4) } { 4, 8, 12 }. = = S { 3, 2, 1, 1, 2, 3 }. = − − − Least residues: LR (4) 3 7 = − LR (8) 1 7 =

Jennifer Li (Louisiana State University) Quadratic Reciprocity May 8, 2015 23 / 79 µ 2. ⇒ =

Negative Least Residues Modulo p : Example

n (p 1) (p 3) (p 1)o S − , − , , 1, 1, 2, , − and = − 2 − 2 ··· − ··· 2 n (p 1) o T a, 2a, 3a, , − a = ··· 2

Let p 7 and a 4. = = (7 1) Then − 3. 2 = T { 4, 2(4) 3(4) } { 4, 8, 12 }. = = S { 3, 2, 1, 1, 2, 3 }. = − − − Least residues: LR (4) 3 7 = − LR (8) 1 7 = LR (12) 2 7 = −

Jennifer Li (Louisiana State University) Quadratic Reciprocity May 8, 2015 23 / 79 Negative Least Residues Modulo p : Example

n (p 1) (p 3) (p 1)o S − , − , , 1, 1, 2, , − and = − 2 − 2 ··· − ··· 2 n (p 1) o T a, 2a, 3a, , − a = ··· 2

Let p 7 and a 4. = = (7 1) Then − 3. 2 = T { 4, 2(4) 3(4) } { 4, 8, 12 }. = = S { 3, 2, 1, 1, 2, 3 }. = − − − Least residues: LR (4) 3 7 = − LR (8) 1 7 = LR (12) 2 7 = − µ 2. ⇒ = Jennifer Li (Louisiana State University) Quadratic Reciprocity May 8, 2015 23 / 79 Gauss’ Lemma

(a/p) ( 1)µ = −

Jennifer Li (Louisiana State University) Quadratic Reciprocity May 8, 2015 24 / 79 Is x2 a(mod p) solvable? ≡

Theorem. Let p be prime and suppose p - a. Then,

2 (p 1)/2 x a(mod p) is solvable iff a − 1(mod p). ≡ ≡

Jennifer Li (Louisiana State University) Quadratic Reciprocity May 8, 2015 25 / 79 ( ) Suppose a is a quadratic residue modulo p. ⇒ Let x Z such that x2 a(mod p). ∈ ≡ Since p - a, it follows that p - x.

(p 1)/2 2 (p 1)/2 a − (x ) − (mod p) ≡ p 1 x − (mod p) ≡ 1 (mod p) ≡

Is x2 a(mod p) solvable? ≡

2 (p 1)/2 x a(mod p) is solvable iff a − 1(mod p). ≡ ≡ Proof.

Jennifer Li (Louisiana State University) Quadratic Reciprocity May 8, 2015 26 / 79 Let x Z such that x2 a(mod p). ∈ ≡ Since p - a, it follows that p - x.

(p 1)/2 2 (p 1)/2 a − (x ) − (mod p) ≡ p 1 x − (mod p) ≡ 1 (mod p) ≡

Is x2 a(mod p) solvable? ≡

2 (p 1)/2 x a(mod p) is solvable iff a − 1(mod p). ≡ ≡ Proof. ( ) Suppose a is a quadratic residue modulo p. ⇒

Jennifer Li (Louisiana State University) Quadratic Reciprocity May 8, 2015 26 / 79 Since p - a, it follows that p - x.

(p 1)/2 2 (p 1)/2 a − (x ) − (mod p) ≡ p 1 x − (mod p) ≡ 1 (mod p) ≡

Is x2 a(mod p) solvable? ≡

2 (p 1)/2 x a(mod p) is solvable iff a − 1(mod p). ≡ ≡ Proof. ( ) Suppose a is a quadratic residue modulo p. ⇒ Let x Z such that x2 a(mod p). ∈ ≡

Jennifer Li (Louisiana State University) Quadratic Reciprocity May 8, 2015 26 / 79 (p 1)/2 2 (p 1)/2 a − (x ) − (mod p) ≡ p 1 x − (mod p) ≡ 1 (mod p) ≡

Is x2 a(mod p) solvable? ≡

2 (p 1)/2 x a(mod p) is solvable iff a − 1(mod p). ≡ ≡ Proof. ( ) Suppose a is a quadratic residue modulo p. ⇒ Let x Z such that x2 a(mod p). ∈ ≡ Since p - a, it follows that p - x.

Jennifer Li (Louisiana State University) Quadratic Reciprocity May 8, 2015 26 / 79 Is x2 a(mod p) solvable? ≡

2 (p 1)/2 x a(mod p) is solvable iff a − 1(mod p). ≡ ≡ Proof. ( ) Suppose a is a quadratic residue modulo p. ⇒ Let x Z such that x2 a(mod p). ∈ ≡ Since p - a, it follows that p - x.

(p 1)/2 2 (p 1)/2 a − (x ) − (mod p) ≡ p 1 x − (mod p) ≡ 1 (mod p) ≡

Jennifer Li (Louisiana State University) Quadratic Reciprocity May 8, 2015 26 / 79 (p 1)/2 ( ) Suppose a − 1(mod p). ⇐ ≡ Let b be a primitive root modulo p. Then r Z such that ∃ ∈ br a(mod p) ≡ So

(p 1)/2 r(p 1)/2 a − (mod p) b − ≡ 1(mod p) ≡

Is x2 a(mod p) solvable? ≡

2 (p 1)/2 x a(mod p) is solvable iff a − 1(mod p). ≡ ≡ Proof (continued).

Jennifer Li (Louisiana State University) Quadratic Reciprocity May 8, 2015 27 / 79 Let b be a primitive root modulo p. Then r Z such that ∃ ∈ br a(mod p) ≡ So

(p 1)/2 r(p 1)/2 a − (mod p) b − ≡ 1(mod p) ≡

Is x2 a(mod p) solvable? ≡

2 (p 1)/2 x a(mod p) is solvable iff a − 1(mod p). ≡ ≡ Proof (continued). (p 1)/2 ( ) Suppose a − 1(mod p). ⇐ ≡

Jennifer Li (Louisiana State University) Quadratic Reciprocity May 8, 2015 27 / 79 Then r Z such that ∃ ∈ br a(mod p) ≡ So

(p 1)/2 r(p 1)/2 a − (mod p) b − ≡ 1(mod p) ≡

Is x2 a(mod p) solvable? ≡

2 (p 1)/2 x a(mod p) is solvable iff a − 1(mod p). ≡ ≡ Proof (continued). (p 1)/2 ( ) Suppose a − 1(mod p). ⇐ ≡ Let b be a primitive root modulo p.

Jennifer Li (Louisiana State University) Quadratic Reciprocity May 8, 2015 27 / 79 So

(p 1)/2 r(p 1)/2 a − (mod p) b − ≡ 1(mod p) ≡

Is x2 a(mod p) solvable? ≡

2 (p 1)/2 x a(mod p) is solvable iff a − 1(mod p). ≡ ≡ Proof (continued). (p 1)/2 ( ) Suppose a − 1(mod p). ⇐ ≡ Let b be a primitive root modulo p. Then r Z such that ∃ ∈ br a(mod p) ≡

Jennifer Li (Louisiana State University) Quadratic Reciprocity May 8, 2015 27 / 79 Is x2 a(mod p) solvable? ≡

2 (p 1)/2 x a(mod p) is solvable iff a − 1(mod p). ≡ ≡ Proof (continued). (p 1)/2 ( ) Suppose a − 1(mod p). ⇐ ≡ Let b be a primitive root modulo p. Then r Z such that ∃ ∈ br a(mod p) ≡ So

(p 1)/2 r(p 1)/2 a − (mod p) b − ≡ 1(mod p) ≡

Jennifer Li (Louisiana State University) Quadratic Reciprocity May 8, 2015 27 / 79 (p 1) ( ) By Theorem we have (p 1) r − ⇐ − | 2 r Then t Z, so r 2t . 2 = ∈ = If x b t, then =

x 2 b 2t (mod p) ≡ b r (mod p) ≡ a(mod p) ≡

Is x2 a(mod p) solvable? ≡

2 (p 1)/2 x a(mod p) is solvable iff a − 1(mod p). ≡ ≡ Proof (continued).

Jennifer Li (Louisiana State University) Quadratic Reciprocity May 8, 2015 28 / 79 r Then t Z, so r 2t . 2 = ∈ = If x b t, then =

x 2 b 2t (mod p) ≡ b r (mod p) ≡ a(mod p) ≡

Is x2 a(mod p) solvable? ≡

2 (p 1)/2 x a(mod p) is solvable iff a − 1(mod p). ≡ ≡ Proof (continued). (p 1) ( ) By Theorem we have (p 1) r − ⇐ − | 2

Jennifer Li (Louisiana State University) Quadratic Reciprocity May 8, 2015 28 / 79 If x b t, then =

x 2 b 2t (mod p) ≡ b r (mod p) ≡ a(mod p) ≡

Is x2 a(mod p) solvable? ≡

2 (p 1)/2 x a(mod p) is solvable iff a − 1(mod p). ≡ ≡ Proof (continued). (p 1) ( ) By Theorem we have (p 1) r − ⇐ − | 2 r Then t Z, so r 2t . 2 = ∈ =

Jennifer Li (Louisiana State University) Quadratic Reciprocity May 8, 2015 28 / 79 Is x2 a(mod p) solvable? ≡

2 (p 1)/2 x a(mod p) is solvable iff a − 1(mod p). ≡ ≡ Proof (continued). (p 1) ( ) By Theorem we have (p 1) r − ⇐ − | 2 r Then t Z, so r 2t . 2 = ∈ = If x b t, then =

x 2 b 2t (mod p) ≡ b r (mod p) ≡ a(mod p) ≡

Jennifer Li (Louisiana State University) Quadratic Reciprocity May 8, 2015 28 / 79 Proposition. Let m 2epe1 pek be the prime decomposition of m. = 1 ··· k Suppose gcd(a,m) 1. = Then x2 a(mod m) is solvable iff: ≡ I. (a) e 2 a 1(mod 4) = ⇒ ≡ (b) e 3 a 1(mod 8) ≥ ⇒ ≡ II. For each index i, we have

(p 1)/2 a i− 1(mod p ) ≡ i

Is x2 a(mod m) solvable? ≡

Let m Z. ∈

Jennifer Li (Louisiana State University) Quadratic Reciprocity May 8, 2015 29 / 79 Then x2 a(mod m) is solvable iff: ≡ I. (a) e 2 a 1(mod 4) = ⇒ ≡ (b) e 3 a 1(mod 8) ≥ ⇒ ≡ II. For each index i, we have

(p 1)/2 a i− 1(mod p ) ≡ i

Is x2 a(mod m) solvable? ≡

Let m Z. ∈ Proposition. Let m 2epe1 pek be the prime decomposition of m. = 1 ··· k Suppose gcd(a,m) 1. =

Jennifer Li (Louisiana State University) Quadratic Reciprocity May 8, 2015 29 / 79 I. (a) e 2 a 1(mod 4) = ⇒ ≡ (b) e 3 a 1(mod 8) ≥ ⇒ ≡ II. For each index i, we have

(p 1)/2 a i− 1(mod p ) ≡ i

Is x2 a(mod m) solvable? ≡

Let m Z. ∈ Proposition. Let m 2epe1 pek be the prime decomposition of m. = 1 ··· k Suppose gcd(a,m) 1. = Then x2 a(mod m) is solvable iff: ≡

Jennifer Li (Louisiana State University) Quadratic Reciprocity May 8, 2015 29 / 79 (b) e 3 a 1(mod 8) ≥ ⇒ ≡ II. For each index i, we have

(p 1)/2 a i− 1(mod p ) ≡ i

Is x2 a(mod m) solvable? ≡

Let m Z. ∈ Proposition. Let m 2epe1 pek be the prime decomposition of m. = 1 ··· k Suppose gcd(a,m) 1. = Then x2 a(mod m) is solvable iff: ≡ I. (a) e 2 a 1(mod 4) = ⇒ ≡

Jennifer Li (Louisiana State University) Quadratic Reciprocity May 8, 2015 29 / 79 II. For each index i, we have

(p 1)/2 a i− 1(mod p ) ≡ i

Is x2 a(mod m) solvable? ≡

Let m Z. ∈ Proposition. Let m 2epe1 pek be the prime decomposition of m. = 1 ··· k Suppose gcd(a,m) 1. = Then x2 a(mod m) is solvable iff: ≡ I. (a) e 2 a 1(mod 4) = ⇒ ≡ (b) e 3 a 1(mod 8) ≥ ⇒ ≡

Jennifer Li (Louisiana State University) Quadratic Reciprocity May 8, 2015 29 / 79 Is x2 a(mod m) solvable? ≡

Let m Z. ∈ Proposition. Let m 2epe1 pek be the prime decomposition of m. = 1 ··· k Suppose gcd(a,m) 1. = Then x2 a(mod m) is solvable iff: ≡ I. (a) e 2 a 1(mod 4) = ⇒ ≡ (b) e 3 a 1(mod 8) ≥ ⇒ ≡ II. For each index i, we have

(p 1)/2 a i− 1(mod p ) ≡ i

Jennifer Li (Louisiana State University) Quadratic Reciprocity May 8, 2015 29 / 79 Let p and q be distinct odd primes. Then,

(p 1)/2 (a) ( 1/p) ( 1) − − = − (p2 1)/8 (b) (2/p) ( 1) − = − ((p 1)/2)((q 1)/2) (c) (p/q) (q/p) ( 1) − − · = −

The Law of Quadratic Reciprocity

Jennifer Li (Louisiana State University) Quadratic Reciprocity May 8, 2015 30 / 79 (p 1)/2 (a) ( 1/p) ( 1) − − = − (p2 1)/8 (b) (2/p) ( 1) − = − ((p 1)/2)((q 1)/2) (c) (p/q) (q/p) ( 1) − − · = −

The Law of Quadratic Reciprocity

Let p and q be distinct odd primes. Then,

Jennifer Li (Louisiana State University) Quadratic Reciprocity May 8, 2015 30 / 79 The Law of Quadratic Reciprocity

Let p and q be distinct odd primes. Then,

(p 1)/2 (a) ( 1/p) ( 1) − − = − (p2 1)/8 (b) (2/p) ( 1) − = − ((p 1)/2)((q 1)/2) (c) (p/q) (q/p) ( 1) − − · = −

Jennifer Li (Louisiana State University) Quadratic Reciprocity May 8, 2015 30 / 79 By deﬁnition of the Legendre symbol,

(p 1)/2 ( 1/p) ( 1) − (mod p) − ≡ − But the only possible values of either side are 1 and 1. + − (p 1)/2 ( 1/p) ( 1) − ⇒ − = −

The Law of Quadratic Reciprocity (a)

Consider (a):

(p 1)/2 ( 1/p) ( 1) − − = −

Proof.

Jennifer Li (Louisiana State University) Quadratic Reciprocity May 8, 2015 31 / 79 But the only possible values of either side are 1 and 1. + − (p 1)/2 ( 1/p) ( 1) − ⇒ − = −

The Law of Quadratic Reciprocity (a)

Consider (a):

(p 1)/2 ( 1/p) ( 1) − − = −

Proof. By deﬁnition of the Legendre symbol,

(p 1)/2 ( 1/p) ( 1) − (mod p) − ≡ −

Jennifer Li (Louisiana State University) Quadratic Reciprocity May 8, 2015 31 / 79 (p 1)/2 ( 1/p) ( 1) − ⇒ − = −

The Law of Quadratic Reciprocity (a)

Consider (a):

(p 1)/2 ( 1/p) ( 1) − − = −

Proof. By deﬁnition of the Legendre symbol,

(p 1)/2 ( 1/p) ( 1) − (mod p) − ≡ − But the only possible values of either side are 1 and 1. + −

Jennifer Li (Louisiana State University) Quadratic Reciprocity May 8, 2015 31 / 79 The Law of Quadratic Reciprocity (a)

Consider (a):

(p 1)/2 ( 1/p) ( 1) − − = −

Proof. By deﬁnition of the Legendre symbol,

(p 1)/2 ( 1/p) ( 1) − (mod p) − ≡ − But the only possible values of either side are 1 and 1. + − (p 1)/2 ( 1/p) ( 1) − ⇒ − = −

Jennifer Li (Louisiana State University) Quadratic Reciprocity May 8, 2015 31 / 79 In this case, a 2. = Then µ is the number of integers in the set

S { 2, 4, , p 1 } = ··· − which have negative least residues modulo p. ·p 1 ¸ But this is the same as the number of even integers in + , p 1 . 2 −

The Law of Quadratic Reciprocity (b)

Consider (b):

(p2 1)/8 (2/p) ( 1) − = −

Proof. (Continued)

Jennifer Li (Louisiana State University) Quadratic Reciprocity May 8, 2015 32 / 79 Then µ is the number of integers in the set

S { 2, 4, , p 1 } = ··· − which have negative least residues modulo p. ·p 1 ¸ But this is the same as the number of even integers in + , p 1 . 2 −

The Law of Quadratic Reciprocity (b)

Consider (b):

(p2 1)/8 (2/p) ( 1) − = −

Proof. (Continued) In this case, a 2. =

Jennifer Li (Louisiana State University) Quadratic Reciprocity May 8, 2015 32 / 79 ·p 1 ¸ But this is the same as the number of even integers in + , p 1 . 2 −

The Law of Quadratic Reciprocity (b)

Consider (b):

(p2 1)/8 (2/p) ( 1) − = −

Proof. (Continued) In this case, a 2. = Then µ is the number of integers in the set

S { 2, 4, , p 1 } = ··· − which have negative least residues modulo p.

Jennifer Li (Louisiana State University) Quadratic Reciprocity May 8, 2015 32 / 79 The Law of Quadratic Reciprocity (b)

Consider (b):

(p2 1)/8 (2/p) ( 1) − = −

Proof. (Continued) In this case, a 2. = Then µ is the number of integers in the set

S { 2, 4, , p 1 } = ··· − which have negative least residues modulo p. ·p 1 ¸ But this is the same as the number of even integers in + , p 1 . 2 −

Jennifer Li (Louisiana State University) Quadratic Reciprocity May 8, 2015 32 / 79 But, p2 1 − ( 1) 8 1 when p 1(mod 8) and − = ≡ ± p2 1 − ( 1) 8 1 when p 3(mod 8). − = − ≡ ±

The Law of Quadratic Reciprocity (b)

Consider (b):

(p2 1)/8 (2/p) ( 1) − = − Proof. Then,

Jennifer Li (Louisiana State University) Quadratic Reciprocity May 8, 2015 33 / 79 But, p2 1 − ( 1) 8 1 when p 1(mod 8) and − = ≡ ± p2 1 − ( 1) 8 1 when p 3(mod 8). − = − ≡ ±

The Law of Quadratic Reciprocity (b)

Consider (b):

(p2 1)/8 (2/p) ( 1) − = − Proof. Then,

Jennifer Li (Louisiana State University) Quadratic Reciprocity May 8, 2015 33 / 79 The Law of Quadratic Reciprocity (b)

Consider (b):

(p2 1)/8 (2/p) ( 1) − = − Proof. Then,

But, p2 1 − ( 1) 8 1 when p 1(mod 8) and − = ≡ ± p2 1 − ( 1) 8 1 when p 3(mod 8). − = − ≡ ± Jennifer Li (Louisiana State University) Quadratic Reciprocity May 8, 2015 33 / 79 Note: Every odd number is either of form 4k 1 or 4k 3. + + (I) If q or p is of form 4k 1: + (4k 1 1)/2 4k/2 2k + − = = ((p 1)/2) or(( q 1)/2) is even. ⇒ − − ((p 1)/2)(( q 1)/2) 0(mod 2) ⇒ − − ≡ (p/q)(q/p) 1. ⇒ = (p/q) and (q/p) have the same signs. ⇒ (p/q) (q/p). ⇒ =

The Law of Quadratic Reciprocity (c)

Consider (c):

((p 1)/2)((q 1)/2) (p/q) (q/p) ( 1) − − · = −

Jennifer Li (Louisiana State University) Quadratic Reciprocity May 8, 2015 34 / 79 (I) If q or p is of form 4k 1: + (4k 1 1)/2 4k/2 2k + − = = ((p 1)/2) or(( q 1)/2) is even. ⇒ − − ((p 1)/2)(( q 1)/2) 0(mod 2) ⇒ − − ≡ (p/q)(q/p) 1. ⇒ = (p/q) and (q/p) have the same signs. ⇒ (p/q) (q/p). ⇒ =

The Law of Quadratic Reciprocity (c)

Consider (c):

((p 1)/2)((q 1)/2) (p/q) (q/p) ( 1) − − · = −

Note: Every odd number is either of form 4k 1 or 4k 3. + +

Jennifer Li (Louisiana State University) Quadratic Reciprocity May 8, 2015 34 / 79 ((p 1)/2) or(( q 1)/2) is even. ⇒ − − ((p 1)/2)(( q 1)/2) 0(mod 2) ⇒ − − ≡ (p/q)(q/p) 1. ⇒ = (p/q) and (q/p) have the same signs. ⇒ (p/q) (q/p). ⇒ =

The Law of Quadratic Reciprocity (c)

Consider (c):

((p 1)/2)((q 1)/2) (p/q) (q/p) ( 1) − − · = −

Note: Every odd number is either of form 4k 1 or 4k 3. + + (I) If q or p is of form 4k 1: + (4k 1 1)/2 4k/2 2k + − = =

Jennifer Li (Louisiana State University) Quadratic Reciprocity May 8, 2015 34 / 79 ((p 1)/2)(( q 1)/2) 0(mod 2) ⇒ − − ≡ (p/q)(q/p) 1. ⇒ = (p/q) and (q/p) have the same signs. ⇒ (p/q) (q/p). ⇒ =

The Law of Quadratic Reciprocity (c)

Consider (c):

((p 1)/2)((q 1)/2) (p/q) (q/p) ( 1) − − · = −

Note: Every odd number is either of form 4k 1 or 4k 3. + + (I) If q or p is of form 4k 1: + (4k 1 1)/2 4k/2 2k + − = = ((p 1)/2) or(( q 1)/2) is even. ⇒ − −

Jennifer Li (Louisiana State University) Quadratic Reciprocity May 8, 2015 34 / 79 (p/q)(q/p) 1. ⇒ = (p/q) and (q/p) have the same signs. ⇒ (p/q) (q/p). ⇒ =

The Law of Quadratic Reciprocity (c)

Consider (c):

((p 1)/2)((q 1)/2) (p/q) (q/p) ( 1) − − · = −

Note: Every odd number is either of form 4k 1 or 4k 3. + + (I) If q or p is of form 4k 1: + (4k 1 1)/2 4k/2 2k + − = = ((p 1)/2) or(( q 1)/2) is even. ⇒ − − ((p 1)/2)(( q 1)/2) 0(mod 2) ⇒ − − ≡

Jennifer Li (Louisiana State University) Quadratic Reciprocity May 8, 2015 34 / 79 (p/q) and (q/p) have the same signs. ⇒ (p/q) (q/p). ⇒ =

The Law of Quadratic Reciprocity (c)

Consider (c):

((p 1)/2)((q 1)/2) (p/q) (q/p) ( 1) − − · = −

Jennifer Li (Louisiana State University) Quadratic Reciprocity May 8, 2015 34 / 79 (p/q) (q/p). ⇒ =

The Law of Quadratic Reciprocity (c)

Consider (c):

((p 1)/2)((q 1)/2) (p/q) (q/p) ( 1) − − · = −

Jennifer Li (Louisiana State University) Quadratic Reciprocity May 8, 2015 34 / 79 The Law of Quadratic Reciprocity (c)

Consider (c):

((p 1)/2)((q 1)/2) (p/q) (q/p) ( 1) − − · = −

Note: Every odd number is either of form 4k 1 or 4k 3. + + (I) If q or p is of form 4k 1: + (4k 1 1)/2 4k/2 2k + − = = ((p 1)/2) or(( q 1)/2) is even. ⇒ − − ((p 1)/2)(( q 1)/2) 0(mod 2) ⇒ − − ≡ (p/q)(q/p) 1. ⇒ = (p/q) and (q/p) have the same signs. ⇒ (p/q) (q/p). ⇒ = Jennifer Li (Louisiana State University) Quadratic Reciprocity May 8, 2015 34 / 79 (II) If q and p are of form 4k 3: + (4k 3 1)/2 (4k 2)/2 2k 1 + − = + = + ((p 1)/2) and(( q 1)/2) are odd. ⇒ − − ((p 1)/2)(( q 1)/2) 1(mod 2) ⇒ − − ≡ (p/q)(q/p) 1. ⇒ = − (p/q) and (q/p) have opposite signs. ⇒

The Law of Quadratic Reciprocity (c)

Consider (c):

((p 1)/2)((q 1)/2) (p/q) (q/p) ( 1) − − · = −

Note: Every odd number is either of form 4k 1 or 4k 3. + +

Jennifer Li (Louisiana State University) Quadratic Reciprocity May 8, 2015 35 / 79 ((p 1)/2) and(( q 1)/2) are odd. ⇒ − − ((p 1)/2)(( q 1)/2) 1(mod 2) ⇒ − − ≡ (p/q)(q/p) 1. ⇒ = − (p/q) and (q/p) have opposite signs. ⇒

The Law of Quadratic Reciprocity (c)

Consider (c):

((p 1)/2)((q 1)/2) (p/q) (q/p) ( 1) − − · = −

Note: Every odd number is either of form 4k 1 or 4k 3. + + (II) If q and p are of form 4k 3: + (4k 3 1)/2 (4k 2)/2 2k 1 + − = + = +

Jennifer Li (Louisiana State University) Quadratic Reciprocity May 8, 2015 35 / 79 ((p 1)/2)(( q 1)/2) 1(mod 2) ⇒ − − ≡ (p/q)(q/p) 1. ⇒ = − (p/q) and (q/p) have opposite signs. ⇒

The Law of Quadratic Reciprocity (c)

Consider (c):

((p 1)/2)((q 1)/2) (p/q) (q/p) ( 1) − − · = −

Note: Every odd number is either of form 4k 1 or 4k 3. + + (II) If q and p are of form 4k 3: + (4k 3 1)/2 (4k 2)/2 2k 1 + − = + = + ((p 1)/2) and(( q 1)/2) are odd. ⇒ − −

Jennifer Li (Louisiana State University) Quadratic Reciprocity May 8, 2015 35 / 79 (p/q)(q/p) 1. ⇒ = − (p/q) and (q/p) have opposite signs. ⇒

The Law of Quadratic Reciprocity (c)

Consider (c):

((p 1)/2)((q 1)/2) (p/q) (q/p) ( 1) − − · = −

Note: Every odd number is either of form 4k 1 or 4k 3. + + (II) If q and p are of form 4k 3: + (4k 3 1)/2 (4k 2)/2 2k 1 + − = + = + ((p 1)/2) and(( q 1)/2) are odd. ⇒ − − ((p 1)/2)(( q 1)/2) 1(mod 2) ⇒ − − ≡

Jennifer Li (Louisiana State University) Quadratic Reciprocity May 8, 2015 35 / 79 (p/q) and (q/p) have opposite signs. ⇒

The Law of Quadratic Reciprocity (c)

Consider (c):

((p 1)/2)((q 1)/2) (p/q) (q/p) ( 1) − − · = −

Note: Every odd number is either of form 4k 1 or 4k 3. + + (II) If q and p are of form 4k 3: + (4k 3 1)/2 (4k 2)/2 2k 1 + − = + = + ((p 1)/2) and(( q 1)/2) are odd. ⇒ − − ((p 1)/2)(( q 1)/2) 1(mod 2) ⇒ − − ≡ (p/q)(q/p) 1. ⇒ = −

Jennifer Li (Louisiana State University) Quadratic Reciprocity May 8, 2015 35 / 79 The Law of Quadratic Reciprocity (c)

Consider (c):

((p 1)/2)((q 1)/2) (p/q) (q/p) ( 1) − − · = −

Jennifer Li (Louisiana State University) Quadratic Reciprocity May 8, 2015 35 / 79 Another Statement

(I) If p or q is of form4 k 1, then + (p/q) (q/p) = (II) If p and q are of form4 k 3, then + (p/q) (q/p) = −

Jennifer Li (Louisiana State University) Quadratic Reciprocity May 8, 2015 36 / 79 p 79 and q 101 are odd primes. = = 101 4(25) 1 = + µ79 1¶ µ101 1¶ − − 0(mod 2) ⇒ 2 2 ≡

(79/101)(101/79) ( 1)even # 1 = − = (79/101) (101/79) ⇒ =

Quadratic Reciprocity: An Example

Calculate (79/101).

Jennifer Li (Louisiana State University) Quadratic Reciprocity May 8, 2015 37 / 79 101 4(25) 1 = + µ79 1¶ µ101 1¶ − − 0(mod 2) ⇒ 2 2 ≡

(79/101)(101/79) ( 1)even # 1 = − = (79/101) (101/79) ⇒ =

Quadratic Reciprocity: An Example

Calculate (79/101).

p 79 and q 101 are odd primes. = =

Jennifer Li (Louisiana State University) Quadratic Reciprocity May 8, 2015 37 / 79 µ79 1¶ µ101 1¶ − − 0(mod 2) ⇒ 2 2 ≡

(79/101)(101/79) ( 1)even # 1 = − = (79/101) (101/79) ⇒ =

Quadratic Reciprocity: An Example

Calculate (79/101).

p 79 and q 101 are odd primes. = = 101 4(25) 1 = +

Jennifer Li (Louisiana State University) Quadratic Reciprocity May 8, 2015 37 / 79 (79/101)(101/79) ( 1)even # 1 = − = (79/101) (101/79) ⇒ =

Quadratic Reciprocity: An Example

Calculate (79/101).

p 79 and q 101 are odd primes. = = 101 4(25) 1 = + µ79 1¶ µ101 1¶ − − 0(mod 2) ⇒ 2 2 ≡

Jennifer Li (Louisiana State University) Quadratic Reciprocity May 8, 2015 37 / 79 (79/101) (101/79) ⇒ =

Quadratic Reciprocity: An Example

Calculate (79/101).

p 79 and q 101 are odd primes. = = 101 4(25) 1 = + µ79 1¶ µ101 1¶ − − 0(mod 2) ⇒ 2 2 ≡

(79/101)(101/79) ( 1)even # 1 = − =

Jennifer Li (Louisiana State University) Quadratic Reciprocity May 8, 2015 37 / 79 Quadratic Reciprocity: An Example

Calculate (79/101).

p 79 and q 101 are odd primes. = = 101 4(25) 1 = + µ79 1¶ µ101 1¶ − − 0(mod 2) ⇒ 2 2 ≡

(79/101)(101/79) ( 1)even # 1 = − = (79/101) (101/79) ⇒ =

Jennifer Li (Louisiana State University) Quadratic Reciprocity May 8, 2015 37 / 79 Proposition (c) (101/79) (22/79) 101 22(mod 79) ⇒ = ≡ Proposition (b) (22/79) (2/79)(11/79) 22 = 2 11 ⇒ = · Law of Quad. Reciprocity (b) (2/79) 1 79 7(mod 8) ⇒ = ≡ (79/101) (11/79) =

Quadratic Reciprocity: An Example

Calculate (79/101).

Jennifer Li (Louisiana State University) Quadratic Reciprocity May 8, 2015 38 / 79 Proposition (b) (22/79) (2/79)(11/79) 22 = 2 11 ⇒ = · Law of Quad. Reciprocity (b) (2/79) 1 79 7(mod 8) ⇒ = ≡ (79/101) (11/79) =

Quadratic Reciprocity: An Example

Calculate (79/101).

Proposition (c) (101/79) (22/79) 101 22(mod 79) ⇒ = ≡

Jennifer Li (Louisiana State University) Quadratic Reciprocity May 8, 2015 38 / 79 Law of Quad. Reciprocity (b) (2/79) 1 79 7(mod 8) ⇒ = ≡ (79/101) (11/79) =

Quadratic Reciprocity: An Example

Calculate (79/101).

Proposition (c) (101/79) (22/79) 101 22(mod 79) ⇒ = ≡ Proposition (b) (22/79) (2/79)(11/79) 22 = 2 11 ⇒ = ·

Jennifer Li (Louisiana State University) Quadratic Reciprocity May 8, 2015 38 / 79 (79/101) (11/79) =

Quadratic Reciprocity: An Example

Calculate (79/101).

Proposition (c) (101/79) (22/79) 101 22(mod 79) ⇒ = ≡ Proposition (b) (22/79) (2/79)(11/79) 22 = 2 11 ⇒ = · Law of Quad. Reciprocity (b) (2/79) 1 79 7(mod 8) ⇒ = ≡

Jennifer Li (Louisiana State University) Quadratic Reciprocity May 8, 2015 38 / 79 Quadratic Reciprocity: An Example

Calculate (79/101).

Proposition (c) (101/79) (22/79) 101 22(mod 79) ⇒ = ≡ Proposition (b) (22/79) (2/79)(11/79) 22 = 2 11 ⇒ = · Law of Quad. Reciprocity (b) (2/79) 1 79 7(mod 8) ⇒ = ≡ (79/101) (11/79) =

Jennifer Li (Louisiana State University) Quadratic Reciprocity May 8, 2015 38 / 79 Note: p q 3(mod 4) ≡ ≡ 11 4(2) 3 and 79 4(19) 3. = + = +

µ11 1¶ µ79 1¶ − − 1(mod 2) ⇒ 2 2 ≡

(79/101)(101/79) ( 1)odd# 1 = − = − (11/79) (79/11) ⇒ = −

Quadratic Reciprocity: An Example

Calculate (79/101).

Consider: q 0 11 and p 0 79, two odd primes. = =

Jennifer Li (Louisiana State University) Quadratic Reciprocity May 8, 2015 39 / 79 µ11 1¶ µ79 1¶ − − 1(mod 2) ⇒ 2 2 ≡

(79/101)(101/79) ( 1)odd# 1 = − = − (11/79) (79/11) ⇒ = −

Quadratic Reciprocity: An Example

Calculate (79/101).

Consider: q 0 11 and p 0 79, two odd primes. = = Note: p q 3(mod 4) ≡ ≡ 11 4(2) 3 and 79 4(19) 3. = + = +

Jennifer Li (Louisiana State University) Quadratic Reciprocity May 8, 2015 39 / 79 (79/101)(101/79) ( 1)odd# 1 = − = − (11/79) (79/11) ⇒ = −

Quadratic Reciprocity: An Example

Calculate (79/101).

Consider: q 0 11 and p 0 79, two odd primes. = = Note: p q 3(mod 4) ≡ ≡ 11 4(2) 3 and 79 4(19) 3. = + = +

µ11 1¶ µ79 1¶ − − 1(mod 2) ⇒ 2 2 ≡

Jennifer Li (Louisiana State University) Quadratic Reciprocity May 8, 2015 39 / 79 (11/79) (79/11) ⇒ = −

Quadratic Reciprocity: An Example

Calculate (79/101).

Consider: q 0 11 and p 0 79, two odd primes. = = Note: p q 3(mod 4) ≡ ≡ 11 4(2) 3 and 79 4(19) 3. = + = +

µ11 1¶ µ79 1¶ − − 1(mod 2) ⇒ 2 2 ≡

(79/101)(101/79) ( 1)odd# 1 = − = −

Jennifer Li (Louisiana State University) Quadratic Reciprocity May 8, 2015 39 / 79 Quadratic Reciprocity: An Example

Calculate (79/101).

Consider: q 0 11 and p 0 79, two odd primes. = = Note: p q 3(mod 4) ≡ ≡ 11 4(2) 3 and 79 4(19) 3. = + = +

µ11 1¶ µ79 1¶ − − 1(mod 2) ⇒ 2 2 ≡

(79/101)(101/79) ( 1)odd# 1 = − = − (11/79) (79/11) ⇒ = −

Jennifer Li (Louisiana State University) Quadratic Reciprocity May 8, 2015 39 / 79 Law of Quad. Reciprocity (b) (2/11) 1 11 3(mod 8) ⇒ = − ≡

Quadratic Reciprocity: An Example

Calculate (79/101).

Proposition (c) (79/11) (2/11) 79 2(mod 11) ⇒ − = − ≡

Jennifer Li (Louisiana State University) Quadratic Reciprocity May 8, 2015 40 / 79 Quadratic Reciprocity: An Example

Calculate (79/101).

Proposition (c) (79/11) (2/11) 79 2(mod 11) ⇒ − = − ≡ Law of Quad. Reciprocity (b) (2/11) 1 11 3(mod 8) ⇒ = − ≡

Jennifer Li (Louisiana State University) Quadratic Reciprocity May 8, 2015 40 / 79 ∴ 79 is a quadratic residue modulo 101.

Quadratic Reciprocity: An Example

Calculate (79/101).

Thus, we have

(79/101) (101/79) = (22/79) = (2/79) (11/79) = 1 (11/79) = · (2/11) = − 1 1 = − · − 1 =

Jennifer Li (Louisiana State University) Quadratic Reciprocity May 8, 2015 41 / 79 Quadratic Reciprocity: An Example

Calculate (79/101).

Thus, we have

(79/101) (101/79) = (22/79) = (2/79) (11/79) = 1 (11/79) = · (2/11) = − 1 1 = − · − 1 = ∴ 79 is a quadratic residue modulo 101.

Jennifer Li (Louisiana State University) Quadratic Reciprocity May 8, 2015 41 / 79 Quadratic Reciprocity: An Example

It’s true!

332 79(mod 101) ≡

Jennifer Li (Louisiana State University) Quadratic Reciprocity May 8, 2015 42 / 79 Difﬁcult!

Euler and Legendre???

History

Jennifer Li (Louisiana State University) Quadratic Reciprocity May 8, 2015 43 / 79 Euler and Legendre???

History

Difﬁcult!

Jennifer Li (Louisiana State University) Quadratic Reciprocity May 8, 2015 43 / 79 History

Difﬁcult!

Euler and Legendre???

Jennifer Li (Louisiana State University) Quadratic Reciprocity May 8, 2015 43 / 79 Gauss gave SIX different proofs!

History

Jennifer Li (Louisiana State University) Quadratic Reciprocity May 8, 2015 44 / 79 History

Gauss gave SIX different proofs!

Jennifer Li (Louisiana State University) Quadratic Reciprocity May 8, 2015 44 / 79 The Law of Quadratic Reciprocity: A Geometric Proof

Consider part (c) of The Law of Quadratic Reciprocity:

If p and q are distinct odd primes, then (a) p 1(mod 4)or q 1(mod 4) (p/q) (q/p) ≡ ≡ ⇒ = (b) p 3(mod 4) and q 3(mod 4) (p/q) (q/p) ≡ ≡ ⇒ 6=

Jennifer Li (Louisiana State University) Quadratic Reciprocity May 8, 2015 45 / 79 Let µ number of integers in S with negative least residues modulo p. 1 = Let µ number of integers in T with negative least residues modulo q. 2 =

The Law of Quadratic Reciprocity: A Geometric Proof

½ (p 1) ¾ ½ (q 1) ¾ Let S q, 2q, , − q and T p, 2p, , − p = ··· 2 = ··· 2

Jennifer Li (Louisiana State University) Quadratic Reciprocity May 8, 2015 46 / 79 Let µ number of integers in T with negative least residues modulo q. 2 =

The Law of Quadratic Reciprocity: A Geometric Proof

½ (p 1) ¾ ½ (q 1) ¾ Let S q, 2q, , − q and T p, 2p, , − p = ··· 2 = ··· 2

Let µ number of integers in S with negative least residues modulo p. 1 =

Jennifer Li (Louisiana State University) Quadratic Reciprocity May 8, 2015 46 / 79 The Law of Quadratic Reciprocity: A Geometric Proof

½ (p 1) ¾ ½ (q 1) ¾ Let S q, 2q, , − q and T p, 2p, , − p = ··· 2 = ··· 2

Let µ number of integers in S with negative least residues modulo p. 1 = Let µ number of integers in T with negative least residues modulo q. 2 =

Jennifer Li (Louisiana State University) Quadratic Reciprocity May 8, 2015 46 / 79 (p/q) ( 1)µ2 = − (q/p) ( 1)µ1 = −

(p/q) (q/p) iff (p/q)(q/p) 1 = = iff ( 1)µ1 ( 1)µ2 1 − · − = µ µ iff ( 1) 1+ 2 1 − = WTS: µ µ is odd iff p 3(mod 4) and q 3(mod 4). 1 + 2 ≡ ≡

The Law of Quadratic Reciprocity: A Geometric Proof

Proof. By Gauss’ Lemma:

Jennifer Li (Louisiana State University) Quadratic Reciprocity May 8, 2015 47 / 79 (p/q) (q/p) iff (p/q)(q/p) 1 = = iff ( 1)µ1 ( 1)µ2 1 − · − = µ µ iff ( 1) 1+ 2 1 − = WTS: µ µ is odd iff p 3(mod 4) and q 3(mod 4). 1 + 2 ≡ ≡

The Law of Quadratic Reciprocity: A Geometric Proof

Proof. By Gauss’ Lemma:

(p/q) ( 1)µ2 = − (q/p) ( 1)µ1 = −

Jennifer Li (Louisiana State University) Quadratic Reciprocity May 8, 2015 47 / 79 WTS: µ µ is odd iff p 3(mod 4) and q 3(mod 4). 1 + 2 ≡ ≡

The Law of Quadratic Reciprocity: A Geometric Proof

Proof. By Gauss’ Lemma:

(p/q) ( 1)µ2 = − (q/p) ( 1)µ1 = −

(p/q) (q/p) iff (p/q)(q/p) 1 = = iff ( 1)µ1 ( 1)µ2 1 − · − = µ µ iff ( 1) 1+ 2 1 − =

Jennifer Li (Louisiana State University) Quadratic Reciprocity May 8, 2015 47 / 79 The Law of Quadratic Reciprocity: A Geometric Proof

Proof. By Gauss’ Lemma:

(p/q) ( 1)µ2 = − (q/p) ( 1)µ1 = −

(p/q) (q/p) iff (p/q)(q/p) 1 = = iff ( 1)µ1 ( 1)µ2 1 − · − = µ µ iff ( 1) 1+ 2 1 − = WTS: µ µ is odd iff p 3(mod 4) and q 3(mod 4). 1 + 2 ≡ ≡

Jennifer Li (Louisiana State University) Quadratic Reciprocity May 8, 2015 47 / 79 The Law of Quadratic Reciprocity: A Geometric Proof

Lattice point:

Jennifer Li (Louisiana State University) Quadratic Reciprocity May 8, 2015 48 / 79 (1) Total µ µ lattice points inside hexagon. 1 + 2

(2) Odd number of lattice points in hexagon iff p q 3(mod 4). ≡ ≡

The Law of Quadratic Reciprocity: A Geometric Proof

Proof idea: Count lattice points inside a hexagon in two ways:

Jennifer Li (Louisiana State University) Quadratic Reciprocity May 8, 2015 49 / 79 (2) Odd number of lattice points in hexagon iff p q 3(mod 4). ≡ ≡

The Law of Quadratic Reciprocity: A Geometric Proof

Proof idea: Count lattice points inside a hexagon in two ways: (1) Total µ µ lattice points inside hexagon. 1 + 2

Jennifer Li (Louisiana State University) Quadratic Reciprocity May 8, 2015 49 / 79 The Law of Quadratic Reciprocity: A Geometric Proof

Proof idea: Count lattice points inside a hexagon in two ways: (1) Total µ µ lattice points inside hexagon. 1 + 2

(2) Odd number of lattice points in hexagon iff p q 3(mod 4). ≡ ≡

Jennifer Li (Louisiana State University) Quadratic Reciprocity May 8, 2015 49 / 79 The Law of Quadratic Reciprocity: A Geometric Proof

Consider hexagon H:

Jennifer Li (Louisiana State University) Quadratic Reciprocity May 8, 2015 50 / 79 The Law of Quadratic Reciprocity: A Geometric Proof

Jennifer Li (Louisiana State University) Quadratic Reciprocity May 8, 2015 51 / 79 The Law of Quadratic Reciprocity: A Geometric Proof

Jennifer Li (Louisiana State University) Quadratic Reciprocity May 8, 2015 52 / 79 The Law of Quadratic Reciprocity: A Geometric Proof

Jennifer Li (Louisiana State University) Quadratic Reciprocity May 8, 2015 53 / 79 The Law of Quadratic Reciprocity: A Geometric Proof

For any (x,y) H, the following must be satisﬁed: ∈ p (1) 0 x < < 2

Jennifer Li (Louisiana State University) Quadratic Reciprocity May 8, 2015 54 / 79 The Law of Quadratic Reciprocity: A Geometric Proof

For any (x,y) H, the following must be satisﬁed: ∈ q (2) 0 y < < 2

Jennifer Li (Louisiana State University) Quadratic Reciprocity May 8, 2015 55 / 79 The Law of Quadratic Reciprocity: A Geometric Proof

For any (x,y) H, the following must be satisﬁed: ∈ q 1 (3) y x < p + 2

Jennifer Li (Louisiana State University) Quadratic Reciprocity May 8, 2015 56 / 79 The Law of Quadratic Reciprocity: A Geometric Proof

For any (x,y) H, the following must be satisﬁed: ∈ q q (4) y x > p + 2p

Jennifer Li (Louisiana State University) Quadratic Reciprocity May 8, 2015 57 / 79 The Law of Quadratic Reciprocity: A Geometric Proof

Suppose (m,n) H. ∈ µp 1 q 1 ¶ Then + m, + n H. 2 − 2 − ∈

Jennifer Li (Louisiana State University) Quadratic Reciprocity May 8, 2015 58 / 79 µp 1 q 1 ¶ (m1,n1) + m1 , + n1 ←→ 2 − 2 − µp 1 q 1 ¶ (m2,n2) + m2 , + n2 ←→ 2 − 2 − . . µp 1 q 1 ¶ (m ,n ) + m , + n k k ←→ 2 − k 2 − k

The Law of Quadratic Reciprocity: A Geometric Proof

Then we can pair these points:

Jennifer Li (Louisiana State University) Quadratic Reciprocity May 8, 2015 59 / 79 The Law of Quadratic Reciprocity: A Geometric Proof

Then we can pair these points:

µp 1 q 1 ¶ (m1,n1) + m1 , + n1 ←→ 2 − 2 − µp 1 q 1 ¶ (m2,n2) + m2 , + n2 ←→ 2 − 2 − . . µp 1 q 1 ¶ (m ,n ) + m , + n k k ←→ 2 − k 2 − k

Jennifer Li (Louisiana State University) Quadratic Reciprocity May 8, 2015 59 / 79 µp 1 q 1 ¶ p 1 q 1 (m,n) + m , + n iff m + and n + = 2 − 2 − = 4 = 4

The Law of Quadratic Reciprocity: A Geometric Proof

BUT...

Jennifer Li (Louisiana State University) Quadratic Reciprocity May 8, 2015 60 / 79 The Law of Quadratic Reciprocity: A Geometric Proof

BUT...

µp 1 q 1 ¶ p 1 q 1 (m,n) + m , + n iff m + and n + = 2 − 2 − = 4 = 4

Jennifer Li (Louisiana State University) Quadratic Reciprocity May 8, 2015 60 / 79 The Law of Quadratic Reciprocity: A Geometric Proof

The point (m,n) is without a pair!

µp 1 q 1 ¶ (m1,n1) + m1 , + n1 ←→ 2 − 2 − µp 1 q 1 ¶ (m2,n2) + m2 , + n2 ←→ 2 − 2 − . . µp 1 q 1 ¶ (m,n) + m , + n = 2 − 2 − . . µp 1 q 1 ¶ (m ,n ) + m , + n k k ←→ 2 − k 2 − k

Jennifer Li (Louisiana State University) Quadratic Reciprocity May 8, 2015 61 / 79 Law of Quadratic Reciprocity: A Geometric Proof

Jennifer Li (Louisiana State University) Quadratic Reciprocity May 8, 2015 62 / 79 p 1 q 1 Suppose + s and + t. 4 = 4 =

Then p 1 4s and q 1 4t. + = + = p 3(mod 4) and q 3(mod 4) ⇒ ≡ ≡ µp 1 q 1¶ + , + is a lattice point iff p q 3(mod 4) ∴ 4 4 ≡ ≡

The Law of Quadratic Reciprocity: A Geometric Proof

By deﬁnition:

µp 1 q 1¶ p 1 q 1 (m,n) + , + is a lattice point iff + Z and + Z = 4 4 4 ∈ 4 ∈

Jennifer Li (Louisiana State University) Quadratic Reciprocity May 8, 2015 63 / 79 Then p 1 4s and q 1 4t. + = + = p 3(mod 4) and q 3(mod 4) ⇒ ≡ ≡ µp 1 q 1¶ + , + is a lattice point iff p q 3(mod 4) ∴ 4 4 ≡ ≡

The Law of Quadratic Reciprocity: A Geometric Proof

By deﬁnition:

µp 1 q 1¶ p 1 q 1 (m,n) + , + is a lattice point iff + Z and + Z = 4 4 4 ∈ 4 ∈

p 1 q 1 Suppose + s and + t. 4 = 4 =

Jennifer Li (Louisiana State University) Quadratic Reciprocity May 8, 2015 63 / 79 p 3(mod 4) and q 3(mod 4) ⇒ ≡ ≡ µp 1 q 1¶ + , + is a lattice point iff p q 3(mod 4) ∴ 4 4 ≡ ≡

The Law of Quadratic Reciprocity: A Geometric Proof

By deﬁnition:

µp 1 q 1¶ p 1 q 1 (m,n) + , + is a lattice point iff + Z and + Z = 4 4 4 ∈ 4 ∈

p 1 q 1 Suppose + s and + t. 4 = 4 =

Then p 1 4s and q 1 4t. + = + =

Jennifer Li (Louisiana State University) Quadratic Reciprocity May 8, 2015 63 / 79 µp 1 q 1¶ + , + is a lattice point iff p q 3(mod 4) ∴ 4 4 ≡ ≡

The Law of Quadratic Reciprocity: A Geometric Proof

By deﬁnition:

µp 1 q 1¶ p 1 q 1 (m,n) + , + is a lattice point iff + Z and + Z = 4 4 4 ∈ 4 ∈

p 1 q 1 Suppose + s and + t. 4 = 4 =

Then p 1 4s and q 1 4t. + = + = p 3(mod 4) and q 3(mod 4) ⇒ ≡ ≡

Jennifer Li (Louisiana State University) Quadratic Reciprocity May 8, 2015 63 / 79 The Law of Quadratic Reciprocity: A Geometric Proof

By deﬁnition:

µp 1 q 1¶ p 1 q 1 (m,n) + , + is a lattice point iff + Z and + Z = 4 4 4 ∈ 4 ∈

p 1 q 1 Suppose + s and + t. 4 = 4 =

Then p 1 4s and q 1 4t. + = + = p 3(mod 4) and q 3(mod 4) ⇒ ≡ ≡ µp 1 q 1¶ + , + is a lattice point iff p q 3(mod 4) ∴ 4 4 ≡ ≡

Jennifer Li (Louisiana State University) Quadratic Reciprocity May 8, 2015 63 / 79 The Law of Quadratic Reciprocity: A Geometric Proof

This proves (2):

Total number of lattice points in H is odd iff p q 3(mod 4) ≡ ≡

Jennifer Li (Louisiana State University) Quadratic Reciprocity May 8, 2015 64 / 79 The Law of Quadratic Reciprocity: A Geometric Proof

This proves (2):

Total number of lattice points in H is odd iff p q 3(mod 4) ≡ ≡

Jennifer Li (Louisiana State University) Quadratic Reciprocity May 8, 2015 64 / 79 The Law of Quadratic Reciprocity: A Geometric Proof

No lattice points on diagonal AD.

Jennifer Li (Louisiana State University) Quadratic Reciprocity May 8, 2015 65 / 79 q q n my x < p < p

Then

q q q q 1 m n my x p − 2p < < p < p + 2 q np qm 0 − 2 < − < np has a negative least residue modulo q. ⇒ If (m,n) lies below AD, then np has negative least residue modulo q.

The Law of Quadratic Reciprocity: A Geometric Proof

Suppose (m,n) lies below diagonal AD.

Jennifer Li (Louisiana State University) Quadratic Reciprocity May 8, 2015 66 / 79 Then

q q q q 1 m n my x p − 2p < < p < p + 2 q np qm 0 − 2 < − < np has a negative least residue modulo q. ⇒ If (m,n) lies below AD, then np has negative least residue modulo q.

The Law of Quadratic Reciprocity: A Geometric Proof

Suppose (m,n) lies below diagonal AD. q q n my x < p < p

Jennifer Li (Louisiana State University) Quadratic Reciprocity May 8, 2015 66 / 79 np has a negative least residue modulo q. ⇒ If (m,n) lies below AD, then np has negative least residue modulo q.

The Law of Quadratic Reciprocity: A Geometric Proof

Suppose (m,n) lies below diagonal AD. q q n my x < p < p

Then

q q q q 1 m n my x p − 2p < < p < p + 2 q np qm 0 − 2 < − <

Jennifer Li (Louisiana State University) Quadratic Reciprocity May 8, 2015 66 / 79 If (m,n) lies below AD, then np has negative least residue modulo q.

The Law of Quadratic Reciprocity: A Geometric Proof

Suppose (m,n) lies below diagonal AD. q q n my x < p < p

Then

q q q q 1 m n my x p − 2p < < p < p + 2 q np qm 0 − 2 < − < np has a negative least residue modulo q. ⇒

Jennifer Li (Louisiana State University) Quadratic Reciprocity May 8, 2015 66 / 79 The Law of Quadratic Reciprocity: A Geometric Proof

Suppose (m,n) lies below diagonal AD. q q n my x < p < p

Then

q q q q 1 m n my x p − 2p < < p < p + 2 q np qm 0 − 2 < − < np has a negative least residue modulo q. ⇒ If (m,n) lies below AD, then np has negative least residue modulo q.

Jennifer Li (Louisiana State University) Quadratic Reciprocity May 8, 2015 66 / 79 Then there exists m such that: q np qm 0 − 2 < − <

(m,n) H and (m,n) lies below diagonal AD. ⇒ ∈ If np has negative least residue modulo q, then (m,n) lies below AD.

The Law of Quadratic Reciprocity: A Geometric Proof

Suppose np has negative least residue modulo q.

Jennifer Li (Louisiana State University) Quadratic Reciprocity May 8, 2015 67 / 79 (m,n) H and (m,n) lies below diagonal AD. ⇒ ∈ If np has negative least residue modulo q, then (m,n) lies below AD.

The Law of Quadratic Reciprocity: A Geometric Proof

Suppose np has negative least residue modulo q. Then there exists m such that: q np qm 0 − 2 < − <

Jennifer Li (Louisiana State University) Quadratic Reciprocity May 8, 2015 67 / 79 If np has negative least residue modulo q, then (m,n) lies below AD.

The Law of Quadratic Reciprocity: A Geometric Proof

Suppose np has negative least residue modulo q. Then there exists m such that: q np qm 0 − 2 < − <

(m,n) H and (m,n) lies below diagonal AD. ⇒ ∈

Jennifer Li (Louisiana State University) Quadratic Reciprocity May 8, 2015 67 / 79 The Law of Quadratic Reciprocity: A Geometric Proof

Suppose np has negative least residue modulo q. Then there exists m such that: q np qm 0 − 2 < − <

(m,n) H and (m,n) lies below diagonal AD. ⇒ ∈ If np has negative least residue modulo q, then (m,n) lies below AD.

Jennifer Li (Louisiana State University) Quadratic Reciprocity May 8, 2015 67 / 79 The Law of Quadratic Reciprocity: A Geometric Proof

µ number of negative least residues modulo q 2 = number of points in H lying below AD. =

Jennifer Li (Louisiana State University) Quadratic Reciprocity May 8, 2015 68 / 79 The Law of Quadratic Reciprocity: A Geometric Proof

µ number of negative least residues modulo p 1 = number of points in H lying above AD. =

Jennifer Li (Louisiana State University) Quadratic Reciprocity May 8, 2015 69 / 79 This proves (1)

The Law of Quadratic Reciprocity: A Geometric Proof

µ µ is total number of lattice points in H. ∴ 1 + 2

Jennifer Li (Louisiana State University) Quadratic Reciprocity May 8, 2015 70 / 79 The Law of Quadratic Reciprocity: A Geometric Proof

µ µ is total number of lattice points in H. ∴ 1 + 2

This proves (1)

Jennifer Li (Louisiana State University) Quadratic Reciprocity May 8, 2015 70 / 79 The Law of Quadratic Reciprocity: A Geometric Proof

µ µ is odd iff p q 3(mod 4) ∴ 1 + 2 ≡ ≡

Jennifer Li (Louisiana State University) Quadratic Reciprocity May 8, 2015 71 / 79 Cauchy Eisenstein Dirichlet Dedekind Kronecker . .

The Law of Quadratic Reciprocity: Other proofs

Jennifer Li (Louisiana State University) Quadratic Reciprocity May 8, 2015 72 / 79 Eisenstein Dirichlet Dedekind Kronecker . .

The Law of Quadratic Reciprocity: Other proofs

Cauchy

Jennifer Li (Louisiana State University) Quadratic Reciprocity May 8, 2015 72 / 79 Dirichlet Dedekind Kronecker . .

The Law of Quadratic Reciprocity: Other proofs

Cauchy Eisenstein

Jennifer Li (Louisiana State University) Quadratic Reciprocity May 8, 2015 72 / 79 Dedekind Kronecker . .

The Law of Quadratic Reciprocity: Other proofs

Cauchy Eisenstein Dirichlet

Jennifer Li (Louisiana State University) Quadratic Reciprocity May 8, 2015 72 / 79 Kronecker . .

The Law of Quadratic Reciprocity: Other proofs

Cauchy Eisenstein Dirichlet Dedekind

Jennifer Li (Louisiana State University) Quadratic Reciprocity May 8, 2015 72 / 79 The Law of Quadratic Reciprocity: Other proofs

Cauchy Eisenstein Dirichlet Dedekind Kronecker . .

Jennifer Li (Louisiana State University) Quadratic Reciprocity May 8, 2015 72 / 79 Is x3 a(mod p) solvable? ≡ cubic reciprocity

Is x4 a(mod p) solvable? ≡ biquadratic reciprocity

Generalizations

Instead of squares, ask same question for n-th powers:

Jennifer Li (Louisiana State University) Quadratic Reciprocity May 8, 2015 73 / 79 cubic reciprocity

Is x4 a(mod p) solvable? ≡ biquadratic reciprocity

Generalizations

Instead of squares, ask same question for n-th powers:

Is x3 a(mod p) solvable? ≡

Jennifer Li (Louisiana State University) Quadratic Reciprocity May 8, 2015 73 / 79 Is x4 a(mod p) solvable? ≡ biquadratic reciprocity

Generalizations

Instead of squares, ask same question for n-th powers:

Is x3 a(mod p) solvable? ≡ cubic reciprocity

Jennifer Li (Louisiana State University) Quadratic Reciprocity May 8, 2015 73 / 79 biquadratic reciprocity

Generalizations

Instead of squares, ask same question for n-th powers:

Is x3 a(mod p) solvable? ≡ cubic reciprocity

Is x4 a(mod p) solvable? ≡

Jennifer Li (Louisiana State University) Quadratic Reciprocity May 8, 2015 73 / 79 Generalizations

Instead of squares, ask same question for n-th powers:

Is x3 a(mod p) solvable? ≡ cubic reciprocity

Is x4 a(mod p) solvable? ≡ biquadratic reciprocity

Jennifer Li (Louisiana State University) Quadratic Reciprocity May 8, 2015 73 / 79 Ring: Z[ω], where ω ( 1 p 3)/2 = − + − Let α Z[ω] and prime π Z[ω]. ∈ ∈ Suppose π - α. Then

N(π) 1 α − 1(mod π) ≡

Cubic Reciprocity

Need n-th root of unity in ﬁeld.

Jennifer Li (Louisiana State University) Quadratic Reciprocity May 8, 2015 74 / 79 Let α Z[ω] and prime π Z[ω]. ∈ ∈ Suppose π - α. Then

N(π) 1 α − 1(mod π) ≡

Cubic Reciprocity

Need n-th root of unity in ﬁeld. Ring: Z[ω], where ω ( 1 p 3)/2 = − + −

Jennifer Li (Louisiana State University) Quadratic Reciprocity May 8, 2015 74 / 79 Cubic Reciprocity

Need n-th root of unity in ﬁeld. Ring: Z[ω], where ω ( 1 p 3)/2 = − + − Let α Z[ω] and prime π Z[ω]. ∈ ∈ Suppose π - α. Then

N(π) 1 α − 1(mod π) ≡

Jennifer Li (Louisiana State University) Quadratic Reciprocity May 8, 2015 74 / 79 (a) (α/π)3 0 if π α. (N(π) =1)/3 | 2 (b) α − (α/π) (π), where (α/π) 1, ω, or ω . ≡ 3 3 =

Cubic Reciprocity

Suppose N(π) 3. 6= The cubic residue character of α modulo π is deﬁned as:

Jennifer Li (Louisiana State University) Quadratic Reciprocity May 8, 2015 75 / 79 Cubic Reciprocity

Suppose N(π) 3. 6= The cubic residue character of α modulo π is deﬁned as:

(a) (α/π)3 0 if π α. (N(π) =1)/3 | 2 (b) α − (α/π) (π), where (α/π) 1, ω, or ω . ≡ 3 3 =

Jennifer Li (Louisiana State University) Quadratic Reciprocity May 8, 2015 75 / 79 The Law of Cubic Reciprocity

Let π1 and π2 be primary. Then

(π /π ) (π /π ) 1 2 3 = 2 1 3

Jennifer Li (Louisiana State University) Quadratic Reciprocity May 8, 2015 76 / 79 n F 22 1 n = + May or may not be prime. Are there inﬁnitely many Fermat primes? Only ﬁve known: F 3 0 = F 5 1 = F 17 2 = F 257 3 = F 65537 4 =

Applications

Fermat numbers: •

Jennifer Li (Louisiana State University) Quadratic Reciprocity May 8, 2015 77 / 79 May or may not be prime. Are there inﬁnitely many Fermat primes? Only ﬁve known: F 3 0 = F 5 1 = F 17 2 = F 257 3 = F 65537 4 =

Applications

Fermat numbers: • n F 22 1 n = +

Jennifer Li (Louisiana State University) Quadratic Reciprocity May 8, 2015 77 / 79 Are there inﬁnitely many Fermat primes? Only ﬁve known: F 3 0 = F 5 1 = F 17 2 = F 257 3 = F 65537 4 =

Applications

Fermat numbers: • n F 22 1 n = + May or may not be prime.

Jennifer Li (Louisiana State University) Quadratic Reciprocity May 8, 2015 77 / 79 Only ﬁve known: F 3 0 = F 5 1 = F 17 2 = F 257 3 = F 65537 4 =

Applications

Fermat numbers: • n F 22 1 n = + May or may not be prime. Are there inﬁnitely many Fermat primes?

Jennifer Li (Louisiana State University) Quadratic Reciprocity May 8, 2015 77 / 79 Applications

Fermat numbers: • n F 22 1 n = + May or may not be prime. Are there inﬁnitely many Fermat primes? Only ﬁve known: F 3 0 = F 5 1 = F 17 2 = F 257 3 = F 65537 4 =

Jennifer Li (Louisiana State University) Quadratic Reciprocity May 8, 2015 77 / 79 Number field: finite degree field extension of Q. Abelian extension: Galois extension with abelian Galois group. Main idea: Abelian extensions of number fields

Applications

Class Field Theory: •

Jennifer Li (Louisiana State University) Quadratic Reciprocity May 8, 2015 78 / 79 Abelian extension: Galois extension with abelian Galois group. Main idea: Abelian extensions of number ﬁelds

Applications

Class Field Theory: • Number field: finite degree field extension of Q.

Jennifer Li (Louisiana State University) Quadratic Reciprocity May 8, 2015 78 / 79 Applications

Class Field Theory: • Number field: finite degree field extension of Q. Abelian extension: Galois extension with abelian Galois group. Main idea: Abelian extensions of number fields

Jennifer Li (Louisiana State University) Quadratic Reciprocity May 8, 2015 78 / 79 Acknowledgements

Thanks to Professor Long and Professor Hoffman!

Jennifer Li (Louisiana State University) Quadratic Reciprocity May 8, 2015 79 / 79