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For n = 4k+3. to any. residue O. Prove that 4 σ (+k+3) for each positive integer k. m = 3 (mod 4), it's · M=.3. modulo 4. it's divisors can be congruent. k = 1 f n = 4 (1) + 3 7: divisors of 7 = 1,7 of = σ (7) = 1+7 = 8. 8 is divisible by 4. k=2 divisors n= n = 4 (²) +3. = 11. .11 = 1,.11. O₁ (11) = 1. +11=12 = 12. is divisible by 4.. Hence, 4/0 (4K+3) is true for each positive integer k. CS 掃描全能王 創建

For n = 4k+3. to any. residue O. Prove that 4 σ (+k+3) for each positive integer k. m = 3 (mod 4), it's · M=.3. modulo 4. it's divisors can be congruent. k = 1 f n = 4 (1) + 3 7: divisors of 7 = 1,7 of = σ (7) = 1+7 = 8. 8 is divisible by 4. k=2 divisors n= n = 4 (²) +3. = 11. .11 = 1,.11. O₁ (11) = 1. +11=12 = 12. is divisible by 4.. Hence, 4/0 (4K+3) is true for each positive integer k. CS 掃描全能王 創建

Algebra & Trigonometry with Analytic Geometry
Algebra & Trigonometry with Analytic Geometry
13th Edition
ISBN:9781133382119
Author:Swokowski
Publisher:Swokowski
Chapter10: Sequences, Series, And Probability
Section10.4: Mathematical Induction
Problem 26E
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For n = 4k+3. to any. residue O. Prove that 4 σ (+k+3) for each positive integer k. m = 3 (mod 4), it's · M=.3. modulo 4. it's divisors can be congruent. k = 1 f n = 4 (1) + 3 7: divisors of 7 = 1,7 of = σ (7) = 1+7 = 8. 8 is divisible by 4. k=2 divisors n= n = 4 (²) +3. = 11. .11 = 1,.11. O₁ (11) = 1. +11=12 = 12. is divisible by 4.. Hence, 4/0 (4K+3) is true for each positive integer k. CS 掃描全能王 創建
Transcribed Image Text:For n = 4k+3. to any. residue O. Prove that 4 σ (+k+3) for each positive integer k. m = 3 (mod 4), it's · M=.3. modulo 4. it's divisors can be congruent. k = 1 f n = 4 (1) + 3 7: divisors of 7 = 1,7 of = σ (7) = 1+7 = 8. 8 is divisible by 4. k=2 divisors n= n = 4 (²) +3. = 11. .11 = 1,.11. O₁ (11) = 1. +11=12 = 12. is divisible by 4.. Hence, 4/0 (4K+3) is true for each positive integer k. CS 掃描全能王 創建
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