Sum of factors be n-1

libojie

Let $1=d_1<d_2<...<d_k=n$ be all positive factors of positive integer $n$ .
If $d_1+d_2+...+d_{k-1}=n-1$ , prove: there exists a positive integer $m$ , such that $n=2^m$ .

**Posted:** Fri Oct 23, 2009 10:07 am

libojie

No one answer? Please try this problem, thanks.

**Posted:** Sat Oct 31, 2009 8:19 am

wauwau · P versus NP Offline Joined: 04 Aug 2006 Posts: 34

Just an idea.....

Let $\sigma(n)$ the ordinary sum of divisors function. Then your problem reads
If $\sigma(n) = 2n - 1$ (1) then $n = 2^\alpha$
$\sigma(n)$ is multiplicative
Be $n = 2^kr$ with odd r

$\sigma(n) = (2^{k + 1} - 1)\sigma(r)$

(1) $(2^{k + 1} - 1)\sigma(r) = 2^{k + 1}r - 1$

$2^{k + 1}(\sigma(r) - r) = \sigma(r) - 1$ (2)

If $\sigma(r) = r$ gives r=1 and (2) is true that is the desired result.

So $\sigma(r) > r$ with $t = 2^{k + 1}$

(2) writes as: $(t - 1)\sigma(r) = rt - 1$

$r = p^ks$ with $gcd(s,p) = 1$ , $\sigma(r) = (1 + p + p^2 + .. + p^k)\sigma(s)$
we have

$(t - 1)p^k\sigma(s) < (t - 1))(1 + p + p^2 + .. + p^k)\sigma(s) = (t - 1)\sigma(r) = p^kst - 1$

only integers involved we have

$(t - 1)p^k\sigma(s) \le p^kst$

$(t - 1)\sigma(s) \le st$

and now I am Lost...... to be continued...

**Posted:** Thu Nov 05, 2009 9:44 am

sdkudrgn88 · Riemann Hypothesis Offline Joined: 06 Apr 2009 Posts: 300

This is actually an open question.

A related question seems to be whether there exist any numbers whose proper factors, including 1, add up to exactly 1 more than the number itself.

**Posted:** Thu Nov 05, 2009 5:54 pm