Chapter 4

Question 1

quadratic residue modulo p

Answer 1

Let p be a prime and a ∈ ℤ with a ̸≡ 0 mod p. Then a is a quadratic residue modulo p if there exists x ∈ ℤ such that x²≡amodp

Question 2

quadratic non-residue modulo p

Answer 2

if there are no such x s.t. x²≡amodp

Question 3

is 0 a quadratic residue or non-residue?

Answer 3

Neither

Question 4

Legendre symbol (a / p)

Answer 4

Let p be a prime and a ∈ ℤ. The Legendre Symbol (a / p) is defined by
(a/p) = { 1 if a is a quadratic residue modulo p, -1 if a is a quadratic non-residue modulo p, 0 if a ≡ 0 modp

Question 5

if a ≡ b mod p then (a/p)=

Answer 5

(b/p)

Question 6

How to calculate quadratic residues and non-residues

Answer 6

1. write a table with the CSR modulo p (shift so that its either side of 0, remove 0)
2. find x² for the CSR
3. find what each x² is congruent to in the CSR
4. the numbers in this bottom row are quadratic residues modulo p, any which are absent are non-residues modulo p

Question 7

Lemma IV.2.1 number of quadratic residues and non-residues

Answer 7

Let p be an odd prime. Then there are precisely 1/2(p-1) quadratic residues modulo p and precisely 1/2(p-1) quadratic non-residues modulo p.

Question 8

Theorem (EULERS CRITERION)

Answer 8

Let p bee an odd prime and a ∈ ℤ. Then,

(a/p) ≡ a¹/²⁽ᵖ⁻¹⁾ mod p

Question 9

corollary IV.3.1

Answer 9

Let p be an odd prime and a,b ∈ ℤ. Then,

ab/p) = (a/p)(b/p

Question 10

corollary IV.3.2

Answer 10

Let p be an odd prime. Then

(-1/p) = (-1)¹/²⁽ᵖ⁻¹⁾ = { 1 if p ≡ 1 mod 4, -1 if p ≡ 3 mod 4

Question 11

least residue

Answer 11

Let p be an odd prime and a ∈ ℤ. The least residue modulo p of a is the unique integer b such that

(i) -1/2(p-1) ≤ b ≤ 1/2(p-1)
(ii) b≡amodp

Question 12

How to calculate least residue

Answer 12

find what b is congruent to in the CSR either side of 0

Question 13

Gauss’ Lemma

Answer 13

Let p be an odd prime, a ∈ ℤ and suppose that a ̸≡ 0 mod p. Consider the following integers:
a, 2a, 3a, … , 1/2(p-1)a
and let µ be the number of these integers that have negative least residue. Then
(a/p) = (-1)^µ

Question 14

How to use gauss’ lemma to evaluate (a/p)

Answer 14

1. write out a table with the CSR (only positive side of 0)
2. find the least residues (calculate first a.1, then just add a, if new elements >1/2(p-1) then substract p)
3. count the number of -ve residues (µ)
4. by gauss’ lemma (a/p)= (-1)^µ

Question 15

Theorem IV.5.1 Quadratic character of 2

Answer 15

Let p be an odd prime. Then (2/p) = {1 if p ≡ ±1 mod 8, -1 if p≡ ±3 mod 8
(p is congruent only to ±1 and ±3 from 0,±1,±2,±3,4 since p is odd)

Question 16

lattice point

Answer 16

and element of ℝ² that has integer co-ordinates

Question 17

Lemma IV.6.2. For geometric version of gauss’ lemma (1)

Answer 17

suppose that 1≤ l ≤1/2(p-1) and that la has negative least residue modulo p. Then there exists a member of Λ whose co-ordinate is l

Question 18

Lemma IV.6.3 For geometric version of gauss’ lemma (2)

Answer 18

Suppose (l,y)∈ ∈ Λ. Then 1 ≤ l ≤. Then 1 ≤ l ≤ 1/2(p-1) and la has negative least residue.

Question 19

Λ =

Answer 19

Λ = {(x,y) ∈ R | (x,y) is a lattice point and -1/2p < ax - py < 0

Question 20

|Λ| =

Answer 20

µ

Question 21

Geometric representation of µ

Answer 21

µ is the number of lattice points in R that are striclty between the lines ax-py=0 and ax -py = 1/2p

Question 22

Gauss’ Law of Quadratic Reciprocity

Answer 22

Suppose that p and q are distinct odd primes. Then (q/p) = (p/q) unless p≡q≡3mod4 then (p/q)=-(q/p)

Question 23

Idea of proof for gauss’ law of quadratic reciprocity

Answer 23

1. find a region A in the rectangle R (0 to 1/2 p 1/2 q) where A is defined by -1/2p < qx - py < 0
2. Let mu be the number of lattice points in A
3. find a region N in the rectangle R (0 to 1/2 q 1/2 p) where B is defined by -1/2q < px - qy < 0
4. Let sigma be the number of lattice points in B
5. apply a map (x,y)->(y,x) on B to get a region C in R (send lattice points to lattice points)
6. combine these regions.
7. There is a bijection from the set of lattice points in the remaining two regions
8. count the number of lattice points in R (note the two remaining regions = 2k points since bijection) and manipulate.