Divisor of Perfect Number is Deficient

Theorem
Let $n$ be a perfect number.

Let $n = r s$ where $r$ and $s$ are positive integers such that $r > 1$ and $s > 1$.

Then $r$ and $s$ are both deficient.

Proof
, consider $r$.

We have by definition of $\sigma$ function and perfect number that:
 * $\sigma \left({n}\right) = 2 n$

Each of the divisors of $r$ can be multiplied by $s$, and these numbers will all be divisors of $n$.

Thus all the divisors of $r$ can be expressed in the form $\dfrac k s$ whose sum will be no greater than $\dfrac {2 n} s$.

Thus:
 * $\sigma \left({r}\right) \le \dfrac {2 n} s = 2 r$

But note that $1$ is a divisors of $n$ which is not a multiple of $s$.

So the sum of divisors of $r$ is at most $\dfrac {2 n} s - 1 = 2 r - 1$.

That is:
 * $\sigma \left({r}\right) < 2 r$

Hence the result by definition of abundant.