Quadratic Residues and Reciprocity Is 3 congruent to the square of - - PowerPoint PPT Presentation

Quadratic Residues and Reciprocity

• Is 3 congruent to the square of some number

modulo 7 ?

• 02 ≡ 0 (mod 7), 12 ≡ 1 (mod 7), 22 ≡ 4 (mod 7), 32 ≡ 2

(mod 7), 42 ≡ 2 (mod 7), 52 ≡ 4 (mod 7), 62 ≡ 1 (mod 7) - No

• Does the congruence x2 ≡ -1 (mod 13) have a

solution?

• 02 ≡ 0 (mod 13), 12 ≡ 1 (mod 13), 22 ≡ 4 (mod 13), 32

≡ 9 (mod 13), 42 ≡ 3 (mod 13), 52 ≡ 12 (mod 13), 62 ≡ 10 (mod 13), 72 ≡ 10 (mod 13), 82 ≡ 12 (mod 13), 92 ≡ 3 (mod 13), 102 ≡ 9 (mod 13), 112 ≡ 4 (mod 13), 122 ≡ 1 (mod 13) -Yes (5 and 8)

• 02 ≡ 0 (mod 5), 12 ≡ 1 (mod 5), 22 ≡ 4 (mod 5), 32 ≡ 4

(mod 5), 42 ≡ 1 (mod 5) – Modulo 5

• 02 ≡ 0 (mod 7), 12 ≡ 1 (mod 7), 22 ≡ 4 (mod 7), 32 ≡ 2

(mod 7), 42 ≡ 2 (mod 7), 52 ≡ 4 (mod 7), 62 ≡ 1 (mod 7) – Modulo 7

• 02 ≡ 0 (mod 11), 12 ≡ 1 (mod 11), 22 ≡ 4 (mod 11), 32 ≡ 9

(mod 11), 42 ≡ 5 (mod 11), 52 ≡ 3 (mod 11), 62 ≡ 3 (mod 11), 72 ≡ 5 (mod 11), 82 ≡ 9 (mod 11), 92 ≡ 4 (mod 11), 102 ≡ 1 (mod 11) – Modulo 11

• 02 ≡ 0 (mod 13), 12 ≡ 1 (mod 13), 22 ≡ 4 (mod 13), 32 ≡ 9

(mod 13), 42 ≡ 3 (mod 13), 52 ≡ 12 (mod 13), 62 ≡ 10 (mod 13), 72 ≡ 10 (mod 13), 82 ≡ 12 (mod 13), 92 ≡ 3 (mod 13), 102 ≡ 9 (mod 13), 112 ≡ 4 (mod 13), 122 ≡ 1 (mod 13) – Modulo 13

• Square of the number ‘b’ and the square of the number

‘(p-b)’ are the same modulo p (p-b)2 ≡ b2 (mod p)

• Quadratic residue (QR) modulo p – A nonzero

number that is congruent to a square modulo p.

• Nonresidue (NR) modulo p - A number that is not

congruent to a square modulo p.

• Example:

– Module 5 (QR – 1,4 and NR – 2,3) – Modulo 7 (QR – 1,2,4 and NR – 3,5,6) – Modulo 11 (QR – 1,3,4,5,9 and NR – 2,6,7,8,10) – Modulo 13 (QR – 1,3,4,9,10,12 and NR – 2,4,6,7,8,11)

Theorem: Let p be an odd prime. Then there are exactly (p-1)/2 QR modulo p and (p-1)/2 NR modulo p.

• 5x2 – 6x + 2 ≡ 0 (mod 13)

– x = 10, 12

• x2 + 7x + 10 ≡ 0 (mod 11)

– x = 6, 9

• 3x2 + 9x + 7 ≡ 0 (mod 13)

– x = 4, 6

Integer 2 is NR of modulo 13 Example:

Theorem (QR multiplication rule): Let p be an odd prime. Then (i) QR * QR = QR (ii) QR * NR = NR (iii)NR * NR = QR Example: NR * NR (i) 3 * 5 ≡ 1 (mod 7) QR (ii) 6 * 7 ≡ 9 (mod 11) QR (iii)5 * 11 ≡ 3 (mod 13) QR (iv)7 * 11 ≡ 12 (mod 13) QR Example: QR * NR (i) 2 * 5 ≡ 3 (mod 7) NR (ii) 5 * 6 ≡ 8 (mod 11) NR (iii)4 * 5 ≡ 7 (mod 13) NR (iv)10 * 7 ≡ 5 (mod 13) NR Legendre Symbol of a modulo p is (a/p) = 1 if a is QR modulo p

• 1 if a is NR modulo p

Example: (3/13) = 1, (11/13) = -1 (75/97) = QR or NR ? Theorem: (a/p) (b/p) = (ab/p)

Rather than taking a particular prime p and listing the a’s that are QRs and NRs, we instead fix an a and ask for which primes p is a a QR.

Example 1: Example 2: (13/17) = (17/13), (17/13) = (4/13) = 1 (7/19) = - (19/7), (19/7) = (5/7) Since, 5 ≡ 1 (mod 4) and 7 ≡ 3 (mod 4), (5/7) = (7/5), (7/5) = (2/5) = -1, Hence, (7/19) = 1 Example 3: (713/1009) = ?

(713/1009) = (23/1009)(31/1009), (23/1009) = (1009/23) = (20/23) and (31/1009) = (1009/31) = (17/31). (20/23) = (4/23)(5/23) = (5/23) = (23/5) = (3/5) = (5/3) = (2/3) = -1, So, (23/1009) = -1 (17/31) = (31/17) = (14/17) = (2/17) (7/17) = (7/17) = (17/7) = (3/7) = -(7/3) = - (4/3) = - (2*2/3) = -1, So, (31/1009) = -1. Finally, (713/1009) = 1

There are infinitely many primes that are congruent to 1 modulo 4 Example 4

P1= 5, A = (2P1)2 + 1 = 101=P2, A = (2P1P2)2 + 1 = 1020101 = P3 A = (2P1P2P3)2 + 1 = 1061522231810040101 = 53 · 1613 · 12417062216309: