With $\sigma (n)$ we denote the sum of natural divisors of the natural number $n$. Prove that, if $n$ is the product of different prime numbers of the form $2^k-1$ for $k \in \mathbb{N}$($Mersenne's$ prime numbers) , than $\sigma (n)=2^m$, for some $m \in \mathbb{N}$. Is the inverse statement true?
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Tags: algebra, polynomial, function, inequalities, number theory, prime numbers, number theory unsolved
22.05.2010 23:22
ridgers wrote: With $\sigma (n)$ we denote the sum of natural divisors of the natural number $n$. Prove that, if $n$ is the product of different prime numbers of the form $2^k-1$ for $k \in \mathbb{N}$($Mersenne's$ prime numbers) , than $\sigma (n)=2^m$, for some $m \in \mathbb{N}$. Is the inverse statement true? Assume $n=\prod_{i=1}^{k}{(2^{\alpha_i}-1)}$. We have $\sigma(n)=\prod_{i=1}^{k}{\sigma(2^{\alpha_i}-1)=\prod_{i=1}^{k}{2^{\alpha_i}}=2^{\sum_{i=1}^{k}{\alpha_i}}}$.
02.09.2010 18:29
02.09.2010 19:55
15.06.2016 06:08
I think the inverse statement is not correct. Consider number \(7\). \(\sigma(7)=1+7=8=2^3\) but \(7\) itself is a prime, which can't be a product of any two primes.