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Check if 53 is a primitive root of Z151 by: (a) Theorem 6.8 Suppose that P element modulo Р (6) Euler's Criterion > 2 is a prime and a E Zp". Then a is a primitive (p-1)/a_ ² = 1 (mod p) for all primes a such that a I (p-1- of and only if a

Check if 53 is a primitive root of Z151 by: (a) Theorem 6.8 Suppose that P element modulo Р (6) Euler's Criterion > 2 is a prime and a E Zp". Then a is a primitive (p-1)/a_ ² = 1 (mod p) for all primes a such that a I (p-1- of and only if a

Elements Of Modern Algebra
Elements Of Modern Algebra
8th Edition
ISBN:9781285463230
Author:Gilbert, Linda, Jimmie
Publisher:Gilbert, Linda, Jimmie
Chapter2: The Integers
Section2.4: Prime Factors And Greatest Common Divisor
Problem 19E: Prove that if n is a positive integer greater than 1 such that n is not a prime, then n has a...
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Check if 53 is a primitive root of Z₁51 by: (a) Theorem 6.8 - Suppose that p > 2 is a prime and a E Zp". Then a is a primitive (p-1)/a element modulo ² = 1 (mod p) for all primes a such that a I (p-1). P of and only if a (6) Euler's Criterion
Transcribed Image Text:Check if 53 is a primitive root of Z₁51 by: (a) Theorem 6.8 - Suppose that p > 2 is a prime and a E Zp". Then a is a primitive (p-1)/a element modulo ² = 1 (mod p) for all primes a such that a I (p-1). P of and only if a (6) Euler's Criterion
THEOREM 6.9 (Euler's Criterion) Let p be an odd prime. Then a is a quadratic residue modulo p if and only if a(p-1)/2 = 1 (mod p). PROOF First, suppose a = y² (mod p). Recall from Corollary 6.6 that if p is prime, then ap−¹ = 1 (mod p) for any a #0 (mod p). Thus we have a(p-1)/2 = = = (y²)(p-1)/2 (mod p) yp-¹ (mod p) 1 (mod p). Conversely, suppose a(p-1)/2 = 1 (mod p). Let b be a primitive element modulo p. Then a = b¹ (mod p) for some positive integer i. Then we have a(p-1)/2 = (bi) (p-1)/2 (mod p) = fi(p-1)/2 (mod p). Since b has order p - 1, it must be the case that p - 1 divides i(p-1)/2. Hence, i is even, and then the square roots of a are ±b¹/2 mod p. I
Transcribed Image Text:THEOREM 6.9 (Euler's Criterion) Let p be an odd prime. Then a is a quadratic residue modulo p if and only if a(p-1)/2 = 1 (mod p). PROOF First, suppose a = y² (mod p). Recall from Corollary 6.6 that if p is prime, then ap−¹ = 1 (mod p) for any a #0 (mod p). Thus we have a(p-1)/2 = = = (y²)(p-1)/2 (mod p) yp-¹ (mod p) 1 (mod p). Conversely, suppose a(p-1)/2 = 1 (mod p). Let b be a primitive element modulo p. Then a = b¹ (mod p) for some positive integer i. Then we have a(p-1)/2 = (bi) (p-1)/2 (mod p) = fi(p-1)/2 (mod p). Since b has order p - 1, it must be the case that p - 1 divides i(p-1)/2. Hence, i is even, and then the square roots of a are ±b¹/2 mod p. I
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