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        riddles >> medium >> Sigma(n) and r(n)
        
(Message started by: ThudanBlunder on Mar 15^th, 2009, 6:36pm)

Title: Sigma(n) and r(n)
Post by ThudanBlunder on Mar 15^th, 2009, 6:36pm

Let sigma(n) denote the sum of the positive integer n
and
Let r(n) be the number of such divisors.

Prove sigma(n) is prime implies r(n) is also prime.

Title: Re: Sigma(n) and r(n)
Post by Immanuel_Bonfils on Mar 15^th, 2009, 6:49pm

Please, what is sum of the positive integer and the number of such divisors?

Title: Re: Sigma(n) and r(n)
Post by ThudanBlunder on Mar 15^th, 2009, 7:14pm

on 03/15/09 at 18:49:41, Immanuel_Bonfils wrote:

Please, what is sum of the positive integer and the number of such divisors?

Sorry,
let sigma(n) denote the sum of the positive divisors of the positive integer n
and
let r(n) be the number of such divisors.

Title: Re: Sigma(n) and r(n)
Post by Eigenray on Mar 18^th, 2009, 7:50am

[hideb]Since sigma is multiplicative, and sigma(n)>1 for n>1, it's clear that if sigma(n) is prime, then n=p^k is a prime power, in which case r(n) = k+1. Now if k+1 = ab, with a,b > 1, then
sigma(n) = 1+p+p²+...+p^ab
= (1+p+..+p^a-1)(1+p^a+...+p^a(b-1))
would be composite, a contradiction.[/hideb]