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Example: For what odd primes p is 11 a quadratic residue modulo p? Solution: This is really asking "when is (11 | p) =1?" First, 11 = 3 (mod 4). To use LQR, consider two cases p = 1 or 3 (mod 4): p=1 We have 1 = (11 | p) = (p | 11), so p is a quadratic residue modulo 11. By brute force: 121, 224, 3² = 9, 4² = 5, 5² = 3 (mod 11) so the quadratic residues mod 11 are 1,3,4,5,9. Using CRT for p = 1 (mod 4) & p = 1,3,4,5,9 (mod 11). p = 1 (mod 4) & p = 1 (mod 11 gives p 1 (mod 44). p = 1 (mod 4) & p = 3 (mod 11) gives p25 (mod 44). p = 1 (mod 4) & p = 4 (mod 11) gives p=37 (mod 44). p = 1 (mod 4) & p = 5 (mod 11) gives p 5 (mod 44). p = 1 (mod 4) & p=9 (mod 11) gives p 9 (mod 44). So p =1,5,9,25,37 (mod 44).

Example: For what odd primes p is 11 a quadratic residue modulo p? Solution: This is really asking "when is (11 | p) =1?" First, 11 = 3 (mod 4). To use LQR, consider two cases p = 1 or 3 (mod 4): p=1 We have 1 = (11 | p) = (p | 11), so p is a quadratic residue modulo 11. By brute force: 121, 224, 3² = 9, 4² = 5, 5² = 3 (mod 11) so the quadratic residues mod 11 are 1,3,4,5,9. Using CRT for p = 1 (mod 4) & p = 1,3,4,5,9 (mod 11). p = 1 (mod 4) & p = 1 (mod 11 gives p 1 (mod 44). p = 1 (mod 4) & p = 3 (mod 11) gives p25 (mod 44). p = 1 (mod 4) & p = 4 (mod 11) gives p=37 (mod 44). p = 1 (mod 4) & p = 5 (mod 11) gives p 5 (mod 44). p = 1 (mod 4) & p=9 (mod 11) gives p 9 (mod 44). So p =1,5,9,25,37 (mod 44).

Algebra & Trigonometry with Analytic Geometry
Algebra & Trigonometry with Analytic Geometry
13th Edition
ISBN:9781133382119
Author:Swokowski
Publisher:Swokowski
Chapter1: Fundamental Concepts Of Algebra
Section1.2: Exponents And Radicals
Problem 90E
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Example: For what odd primes p is 11 a quadratic residue modulo p? Solution: This is really asking "when is (11 | p) =1?" First, 11 = 3 (mod 4). To use LQR, consider two cases p = 1 or 3 (mod 4): p=1 We have 1 = (11 | p) = (p | 11), so p is a quadratic residue modulo 11. By brute force: 121, 224, 3² = 9, 4² = 5, 5² = 3 (mod 11) so the quadratic residues mod 11 are 1,3,4,5,9. Using CRT for p = 1 (mod 4) & p = 1,3,4,5,9 (mod 11). p = 1 (mod 4) & p = 1 (mod 11 gives p 1 (mod 44). p = 1 (mod 4) & p = 3 (mod 11) gives p25 (mod 44). p = 1 (mod 4) & p = 4 (mod 11) gives p=37 (mod 44). p = 1 (mod 4) & p = 5 (mod 11) gives p 5 (mod 44). p = 1 (mod 4) & p=9 (mod 11) gives p 9 (mod 44). So p =1,5,9,25,37 (mod 44).
Transcribed Image Text:Example: For what odd primes p is 11 a quadratic residue modulo p? Solution: This is really asking "when is (11 | p) =1?" First, 11 = 3 (mod 4). To use LQR, consider two cases p = 1 or 3 (mod 4): p=1 We have 1 = (11 | p) = (p | 11), so p is a quadratic residue modulo 11. By brute force: 121, 224, 3² = 9, 4² = 5, 5² = 3 (mod 11) so the quadratic residues mod 11 are 1,3,4,5,9. Using CRT for p = 1 (mod 4) & p = 1,3,4,5,9 (mod 11). p = 1 (mod 4) & p = 1 (mod 11 gives p 1 (mod 44). p = 1 (mod 4) & p = 3 (mod 11) gives p25 (mod 44). p = 1 (mod 4) & p = 4 (mod 11) gives p=37 (mod 44). p = 1 (mod 4) & p = 5 (mod 11) gives p 5 (mod 44). p = 1 (mod 4) & p=9 (mod 11) gives p 9 (mod 44). So p =1,5,9,25,37 (mod 44).
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