Question

Quadratic Reciprocity (3) 1) Compute (We know many different ways to complete this computation: brute force, Euler's Criterion Gauss' Lemma, Quadratic Reciprocity. Be sure to understand how to apply all of these methods.) 3 aE {1,2} aE{1,2,3,4} aE{1,2,3,4,5, 6}, a E 1,2,3,5,6, 7}, a E {2,3,7,8, 15, 17,24, 26}, a€ {2,3,5, 24,73, 221, 225, 627}, p=3 p = 5 p7 p 11 = p= 29 p 1237

Answer 1

1^2 identicalto 1(3), quadratic residue (in short q. r) 2^3-1/2 identicalto -1(3) not q. r i^2 identicalto 1(5) q. r (1/5) = 1 2^5-1/2 identicalto -1(5) not q. r (2/5) = -1 3^5-1/2 identicalto -1(5) not q. r (3/5) = -1 4^5-1/2 identicalto 1(5) q. r (4/5) = 1 1^2 identicalto 1(7), (1/7) = 1 2^7-1/2 identicalto 1(7) (2/7) = 1 3^7-1/2 identicalto -1(7), (3/7) = -1 4^7-1/2 identicalto 1(7), (4/7) = 1 5^7-1/2 = -d(7) (5/7) = -1 6^7-1/2 identicalto -1(7) (6/7) = -1