The preceding part of the problem proves the reverse implication unless n = p2 where p is a prime. Prove that if p > 2 is prime then (p2− 1)! is congruent to 0 mod p2, and find k ∈ {0, 1, 2, 3} such that (22 − 1)! is congruent to k mod 4.
The preceding part of the problem proves the reverse implication unless n = p2 where p is a prime. Prove that if p > 2 is prime then (p2− 1)! is congruent to 0 mod p2, and find k ∈ {0, 1, 2, 3} such that (22 − 1)! is congruent to k mod 4.
Elements Of Modern Algebra
8th Edition
ISBN:9781285463230
Author:Gilbert, Linda, Jimmie
Publisher:Gilbert, Linda, Jimmie
Chapter2: The Integers
Section2.5: Congruence Of Integers
Problem 58E: a. Prove that 10n(1)n(mod11) for every positive integer n. b. Prove that a positive integer z is...
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One number-theoretic result mentioned in the course was Wilson’s
Theorem: If p is a prime then (p − 1)! is congruent to −1 mod p. — The purpose of this exercise is to show the reverse implication.
(a) Suppose n > 1 is a composite integer ab where a and b are unequal integers both greater than 1. Prove that (n − 1)! is congruent to 0 mod n. [Hint: Why are both factors less than n/2?]
(b) The preceding part of the problem proves the reverse implication unless n = p2 where p is a prime. Prove that if p > 2 is prime then (p2− 1)! is congruent to 0 mod p2, and find k ∈ {0, 1, 2, 3} such that (22 − 1)! is congruent to k mod 4.
Please answer part b of this question.
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