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

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

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

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

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

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

Since

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

we can subtract 1 from both sides and arrive at

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)