Does there exists an integer satisfying $\sigma(n)=10$?

Question

Define $\sigma(n)$: Sum of all positive divisors of $n$.

Now,does there exists a number $n\in\mathbb{Z}$ such that $\sigma(n)=10$. If not, then what is the reason?

I think that the statement is not true because $10$ cannot be represented as

$$\prod\limits_{i=1}^r(1+p_i+{p_i}^2+\cdots +{p_i}^{\alpha_i})$$

What is the general condition for $n\in\mathbb{Z}$ such that $\sigma(n)=k$ for a given fixed $k\in \mathbb N$?

Answer 1

Method 1: $n|n$ and $1|n$ so $n < n+1 \le \sigma n$ so if $\sigma(n) =10$ then $n < 10$.

Just test $\sigma(1...9)$ and see if any of them equal $10$.

Method 2:

If $n = \prod p_i^{k_i}$ then $\sigma(n) = \prod_{i; p_i|n} (\sum_{m=0}^{k_i} p_i^{k_i})= \prod_{i; p_i|n}\frac {p_i^{k_i +1} - 1}{p_i - 1}$.

So we need $\prod_{i; p_i|n} (\sum_{m=0}^{k_i} p_i^{k_i})=10$

So we can have $10 = 10$ so $10 = 1 + p+ p^2 + .... + p^k$ and $n = p^k$ for some prime $p$. That is easily verified as impossible for $p = 2,3,5,7$ and if $p > 10$ that is clearly impossible as $1+p + p^2 + ... + p^k > p > 10$.

Or we can have $10 = 2\times 5$ and so $2 = 1+p + p^2 + ... +p^k$ and $5 = 1+q + q^2 + ..... + q^j$ for some primes $p, q$. That's clearly impossible. $1+p \ge 1 + 2> 2$ so that's impossible.

Method 3:

(baby version of method 2).

If we consider the sums that add to $10$ (such as $1 + 9$ and $2+8$ and $1+1+1+2+5$ etc...) and we consider that the terms must all be distinct and as $1|n$ the must include $1$ the sums are

$1+ 9$

$1+ 2+7$

$1+2 + 3 +4$

$1 + 3+ 6$

$1 + 4 + 5$.

We can rule out $1 + 9, 1+3+6, 1+4+5$ and they include composite numbers without including all their prime factors. We can rule out $1 + 2 + 7$ and $1+2 + 3 + 4$ as it includes two relatively prime factors without including the multiple of the relatively prime factors.

...

In general to have $\sigma(n) = k$ we need $k = \prod \frac {p_i^{m_i} -1}{p_i -1}$ for some collection of primes $p_i$.

And if you prime factorize $k$ and compare to possible factorizations to see if any are of the form above, by trial and error, you may be able to do it but... I think it's mostly guess work.

Answer 2

1. In the case $k=10$, it doesn't exist. Because $\sigma(n)>n$, all you need to do is check every number below 10. 9 doesn't work, 8 doesn't work, nor does 7, 6, 5, 4, 3, 2, or 1. So one doesn't exist. Unless you count negative numbers, but it's clear the results will be similar.

2. For the general case I don't know of any general condition that fully encapsulates it, but I imagine there are many algorithms that can be used to narrow it down, given that you only need to check at most $k$ numbers.