# Any positive divisor of a deficient or perfect number other than the number itself is deficient

Prove that any positive divisor of a deficient or perfect number other than the number itself is deficient. A number is deficient if $\sigma(n)<2n$

Let the divisors of $N$ be $1, n_1, n_2, \ldots , N$

So $\sigma(N)=1+n_{1}+n_{2}+\dots +N$ $\le 2N$ (because perfect or deficient) $1/N+{n_1}/N+{n_2}/N+\dots +N/N \le 2$

The product of first and last divisors of $N$ is $N$. The product of $2$nd and $2$nd-to-last divisors of N is N. Etc.

So you get $$1/N+ \dots +1/{n_2}+1/{n_1}+1 \le 2$$

Got stuck here. Pls help.

let $q$ be a divisor of $n$ where $n$ is a deficient or perfect number. $\sigma(n) \leq 2n$ and $\exists r \in \mathbb{Z},r>1, qr=n$

Notice that if $d|q$, then $dr|qr=n$

Suppose on the contrary that $q$ is not deficient, then $\sigma(q)\geq 2q$,

$$\sigma(n)=\sum_{d|n}d>\sum_{d|q}(dr)=r\sum_{d|q}d=r\sigma(q)\geq 2qr=2n$$

• How do you know $\sum_{d|n}d>\sum_{d|q}(dr)$? Commented Oct 27, 2016 at 18:21
• The term on the right only contain part of the the factors of $n$, in particular, it doesn't contain $1$. Commented Oct 28, 2016 at 2:27
First note $$\sigma(ab)\geq a\sigma(b).$$ This is because $$\{ad\,|\,d>0\land d|b\}\subseteq\{d\,|\,d>0\land d|ab\}.$$(Note equality when $$a=1$$, otherwise proper inclusion).
Let $$n$$ be a whole number s.t. $$\sigma(n)\leq2n.$$ Suppose for contradiction that $$1 is not deficient (so $$\sigma(d)\geq2d$$). Then $$2n\geq\sigma(n)=\sigma(\frac{n}{d}\cdot d)>\frac{n}{d}\sigma(d)\geq\frac{n}{d}\cdot2d=2n,$$a contradiction. (Note we have strict equality above because we chose $$1).