Let $n\ge 2$ be a natural number, $\sigma(n)$ the sum of its divisors.
- Can $\sigma(n)-n$ be a PROPER divisor of $n$ ?
If $\sigma(n)-n=n$ , $n$ is a perfect number.
If $\sigma(n)-n=1$ , $n$ is a prime number.
I checked the numbers upto $n=10^9$ and did not find a number $n$ with the desired property. My conjecture is that there is no such number. Can anyone prove or disprove this conjecture ?
There is no such $n$ such that $\sigma(n)-n$ is a proper divisor of $n$, except for prime numbers, for which it is trivial (as you stated). Assume $n$ is not prime, and let $m$ be its biggest divisor (not necessarily a prime). Then $n=m\cdot p$ for some prime $p$. If $m=p$, then $n=p^2$ and $\sigma(n)-n=p+1\nmid p^2$. If $m\neq p$, then $\sigma(n)-n> m+p> m$, so $\sigma(n)-n$ is larger than any proper divisor of $n$, and can't be any of them.