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Question 4. We define o(n) as the sum of all the divisors of n, i.e., o(n) = Σd. d|n For example o(6) = 1+2+3+ 6 = 12. A perfect number is a number n such that σ(n) = 2n. The numbers 6 and 28 are perfect, as 1+2+3=6 and 1+2+4+7+ 14 = 28. Show that, if M₁ = 2ª – 1 is prime, then 2ª-¹ (2ª − 1) is perfect.

Question 4. We define o(n) as the sum of all the divisors of n, i.e., o(n) = Σd. d|n For example o(6) = 1+2+3+ 6 = 12. A perfect number is a number n such that σ(n) = 2n. The numbers 6 and 28 are perfect, as 1+2+3=6 and 1+2+4+7+ 14 = 28. Show that, if M₁ = 2ª – 1 is prime, then 2ª-¹ (2ª − 1) is perfect.

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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n, i.e., Question 4. We define o() as the sum of all the divisors of o(n) =d. u|p For example o (6) = 1+2+ 3+ 6 = 12. A perfect number is a number n such that o(n) 6 and 1+2 +4+7+14 = 2n. The numbers 6 and 28 are perfect, as 1+2+3 = = 28. Show that, if M, = 2ª – 1 is prime, then 24-1(2ª – 1) is perfect.
Transcribed Image Text:n, i.e., Question 4. We define o() as the sum of all the divisors of o(n) =d. u|p For example o (6) = 1+2+ 3+ 6 = 12. A perfect number is a number n such that o(n) 6 and 1+2 +4+7+14 = 2n. The numbers 6 and 28 are perfect, as 1+2+3 = = 28. Show that, if M, = 2ª – 1 is prime, then 24-1(2ª – 1) is perfect.
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