4. For an odd prime p, suppose that (=) = 1 and consider the congruence x² = a mod p. (i) For p = 3 + 4k, i.e., for p = 3 mod 4, show that the solutions to the congruence are x = ±ak+¹. (ii) For p = 5 + 8k, i.e., for p = 5 mod 8, show that the solutions to the congruence are = x = ±2²k+1k+1 tak+¹, or x a 1 Explain how to see which is the correct choice. (iii) Use the results of (i) and (ii) to find the solutions to the congruences x² = 23 mod 751 x² = 15 mod 797

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...
icon
Related questions
Question

[Number Theory] How do you solve question 4? thanks

1. Evaluate the following Legendre symbols:
2. Use Gauss' Lemma,
(i)
to compute
(ii)
(iii)
In (ii) p is an odd prime, p ‡ 7; in (iii) p is an odd prime p ‡ 3. In (ii) express the answer
in terms of a congruence condition on p mod 28, and in (iii) express the answer in terms of
a congruence condition on p mod 24, analogous to
= 1 if p = 1 mod 4 and -1 if p = 3 mod 4.
463
541
= (-1)",
4. For an odd prime p, suppose that
31
8933
104729
3. For an odd prime p≥ 7, show that there are consecutive quadratic residues mod p, i.e.,
that there exists an element a € Up such that
μ = |aPN,
a
(²) - ( ² + ¹) -
=
Hint: First show that at least one of the numbers 2, 5 and 10 is a quadratic residue mod p.
x =
a = 1 and consider the congruence
= 1.
x² = a mod p.
(i) For p = 3 + 4k, i.e., for p = 3 mod 4, show that the solutions to the congruence are
tak+1
or x =
1
(ii) For p = 5 + 8k, i.e., for p = 5 mod 8, show that the solutions to the congruence are
x = ±ak+¹,
+2²k+1 k+1
a
Explain how to see which is the correct choice.
(iii) Use the results of (i) and (ii) to find the solutions to the congruences
x² = 23 mod 751
x² = 15 mod 797
Transcribed Image Text:1. Evaluate the following Legendre symbols: 2. Use Gauss' Lemma, (i) to compute (ii) (iii) In (ii) p is an odd prime, p ‡ 7; in (iii) p is an odd prime p ‡ 3. In (ii) express the answer in terms of a congruence condition on p mod 28, and in (iii) express the answer in terms of a congruence condition on p mod 24, analogous to = 1 if p = 1 mod 4 and -1 if p = 3 mod 4. 463 541 = (-1)", 4. For an odd prime p, suppose that 31 8933 104729 3. For an odd prime p≥ 7, show that there are consecutive quadratic residues mod p, i.e., that there exists an element a € Up such that μ = |aPN, a (²) - ( ² + ¹) - = Hint: First show that at least one of the numbers 2, 5 and 10 is a quadratic residue mod p. x = a = 1 and consider the congruence = 1. x² = a mod p. (i) For p = 3 + 4k, i.e., for p = 3 mod 4, show that the solutions to the congruence are tak+1 or x = 1 (ii) For p = 5 + 8k, i.e., for p = 5 mod 8, show that the solutions to the congruence are x = ±ak+¹, +2²k+1 k+1 a Explain how to see which is the correct choice. (iii) Use the results of (i) and (ii) to find the solutions to the congruences x² = 23 mod 751 x² = 15 mod 797
Expert Solution
steps

Step by step

Solved in 3 steps with 1 images

Blurred answer