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Let a Z. Using long division, we may write a = 6q+r for some integers q and r with r = {0, 1, 2, 3, 4, 5). The goal of this problem is to prove, under the assumption that ged(a, 6) = 1 (which we now officially adopt for the remainder of the problem), that 6a² - 1. (a) Explain why the condition gcd(a, 6) = 1 implies that neither 2 nor 3 divides a. (b) Write a = 6q+r with q, r € Z and r = {0, 1, 2, 3, 4, 5). Prove that we must in fact have r = 1 or r = 5. (Hint: use (a) to rule out all other possibilities for r.) (c) Use (b) to prove that 6 | a² - 1.

Let a Z. Using long division, we may write a = 6q+r for some integers q and r with r = {0, 1, 2, 3, 4, 5). The goal of this problem is to prove, under the assumption that ged(a, 6) = 1 (which we now officially adopt for the remainder of the problem), that 6a² - 1. (a) Explain why the condition gcd(a, 6) = 1 implies that neither 2 nor 3 divides a. (b) Write a = 6q+r with q, r € Z and r = {0, 1, 2, 3, 4, 5). Prove that we must in fact have r = 1 or r = 5. (Hint: use (a) to rule out all other possibilities for r.) (c) Use (b) to prove that 6 | a² - 1.

Advanced Engineering Mathematics
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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Let a Z. Using long division, we may write a = 6q+r for some integers q and r with r = {0, 1, 2, 3, 4, 5}. The goal of this problem is to prove, under the assumption that ged(a, 6) = 1 (which we now officially adopt for the remainder of the problem), that 6 | a² - 1. = (a) Explain why the condition ged(a, 6) = 1 implies that neither 2 nor 3 divides a. (b) Write a 6q+r with q, r € Z and r = {0, 1, 2, 3, 4, 5). Prove that we must in fact have r = 1 or r = 5. (Hint: use (a) to rule out all other possibilities for r.) (c) Use (b) to prove that 6 | a² - 1.
Transcribed Image Text:Let a Z. Using long division, we may write a = 6q+r for some integers q and r with r = {0, 1, 2, 3, 4, 5}. The goal of this problem is to prove, under the assumption that ged(a, 6) = 1 (which we now officially adopt for the remainder of the problem), that 6 | a² - 1. = (a) Explain why the condition ged(a, 6) = 1 implies that neither 2 nor 3 divides a. (b) Write a 6q+r with q, r € Z and r = {0, 1, 2, 3, 4, 5). Prove that we must in fact have r = 1 or r = 5. (Hint: use (a) to rule out all other possibilities for r.) (c) Use (b) to prove that 6 | a² - 1.
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