# Talk:Quasiperfect number

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I see no reason, in principle, why a qusiperfect number cannot be an even square or twice a square, which also have odd σ(n)'s. I would therefore include this in the request for citation. Septentrionalis 18:48, 2 October 2006 (UTC)

I can give a reason: Assume a quasiperfect number n σ(n) = 2n + 1 Now, n = 2^e k for some e and odd k, and ${\displaystyle \sigma (2^{e})=\,\!2^{e+1}-1}$

${\displaystyle (2^{e+1}\,\!-1)\sigma (k)=2^{e+1}k+1}$

Then, σ(k) is even unless k = l^2 for some l

${\displaystyle (2^{e+1}-1)\sigma (l^{2})=2^{e+1}l^{2}+\,\!1}$

Since σ(l^2) is an integer, we can write:

${\displaystyle 2^{e+1}l^{2}+1\,\!\equiv 0\ ({\mbox{mod}}\ 2^{e+1}-1)}$

Since

${\displaystyle 2^{e+1}\equiv 1\ ({\mbox{mod}}\ 2^{e+1}-1)}$

we can subtract 1 from both sides and arrive at

${\displaystyle l^{2}\equiv -1\ ({\mbox{mod}}\ 2^{e+1}-1)}$

Since l is an integer, we conclude that -1 is a quadratic residue of 2^(e+1) - 1. A familiar theorem from elementary number theory states that -1 can only be a quadratic residue of an integer r if r is of the form 4p + 1. Therefore, 2^(e+1) - 1 must be of the form 4p + 1 for some p; however, this is only true if e+1 = 1 and therefore e = 0, and n is an odd perfect square.

I can't remember exactly where I found that. 69.163.197.224 01:44, 9 November 2006 (UTC)

I'd wondered whether any such numbers exist. I was considering four cases of positive integers that equal the sum of their divisors:
• all divisors (the only such number is 1)
• all except 1 (which turn out to be the primes)
• all except itself (perfect numbers)
• all except 1 and itself
After I'd thought a bit about it and written a program to see if it finds any, I came here and found this name for them. By this point I had a simple proof that any odd quasiperfect must be square, but hadn't got anywhere near discovering that they can't be even. I wonder if there's a simpler proof of the latter. Meanwhile, I guess I'll have to have a look at that quadratic residue theorem. — Smjg (talk) 00:08, 15 September 2011 (UTC)

Refer to the above. If ${\displaystyle (2^{e+1}\,\!-1)\sigma (k)=2^{e+1}k+1}$ then ${\displaystyle \sigma (k)}$ is odd, not even. SophieAthena (talk) 22:47, 5 January 2013 (UTC)