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Consider $n \in \mathbb{N}$.

Define the aliquot sum function $s(n)$ to be the sum of the proper divisors of $n$ (the divisors not including $n$ itself). Call $n$ deficient if $s(n) < n$, abundant if $s(n) > n$ and perfect if $s(n) = n$.

My initial question is: Can $n$ be divisible by the sum of its proper factors?

In other words, do there exist $n, k \in \mathbb{N}$ such that $n = s(n) \cdot k$?

It is clear that $n$ perfect already satisfies this property (set $k = 1$), and that $n$ prime does too (set $k = n$). So I will call these the trivial solutions and am only concerned with finding composite, deficient solutions.

No even number will satisfy this property either as 1 and $\frac{n}{2}$ will be among its proper factors and $n$ is never divisible by a number greater than $\frac{n}{2}$, apart from itself.

So, does there exist a composite, deficient, odd number that is divisible by the sum of its proper factors?

We can also see that if $s(n)$ is even, then it cannot be a factor of $n$ odd. So $s(n)$ must be odd, which means $n$ has an odd number of odd proper factors (including 1) and therefore has an odd number of proper factors (as $n$ only has odd factors anyway).

I’ve not been able to make any real progress on this problem other than to check the first 50,000 or so integers and find no solutions.

What am I missing? Something obvious, most likely. I think it can probably be proven easily that there exist no such numbers.

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Basically the same argument as for $n$ even works in general for composite $n$. We have that $n$ is divisible by $1$, and the largest proper factor of $n$ is included in the sum $s(n)$. So $s(n)$ must be larger than the largest proper factor of $n$.

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  • $\begingroup$ I already knew I was stupid. But I didn’t know I was this stupid. Thank you. $\endgroup$ – 雨が好きな人 Apr 16 at 15:32
  • $\begingroup$ I definitely need more sleep. $\endgroup$ – 雨が好きな人 Apr 16 at 15:37

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