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Thank for my previous post. Also, thank you so much for this site (m.s.e)

1) If odd perfect numbers there, those numbers can be expressible $12k + 1$ or $324k + 81$ or $468k + 117$. If yes, please discuss, how far I am correct.

2) If $K$ = $(4^n - 2^n)$/2 is perfect, when $k = 1^3 + 3^3 + ...$

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You might want to add how you came up with these results ... –  Hagen von Eitzen Feb 3 '13 at 8:56
    
@HagenvonEitzen! sinece even perfect numbers cannot wirtien in 12k + 1 and so on,.. –  Jiha Feb 3 '13 at 9:19
    
Well, even perfect numbers surely cannot be written as $12k+1$. What was your reasoning about odd perfect numbers? –  Hagen von Eitzen Feb 3 '13 at 9:24
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2 Answers

up vote 4 down vote accepted

1) is stated at Wikipedia, and attributed to Roberts, T (2008). "On the Form of an Odd Perfect Number". Australian Mathematical Gazette 35 (4): 244. Here is a direct link.

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For part 2):

The formula $$\sum_{k=1}^nk^3=\frac{n^2(n+1)^2}{4}$$ is well-known and leads to $$ 1^3+3^3+\ldots+(2m-1)^3=\sum_{k=1}^{2m}k^3-\sum_{k=1}^{m}(2k)^3\\=\frac{(2m)^2(2m+1)^2}{4}-8\cdot\frac{m^2(m+1)^2}{4}\\=m^2(2m^2-1).$$ Since it is well-known that even perfect numbers $N$ are of the form $N=2^{p-1}(2^p-1)$ with $p=2n+1$ an odd prime and $2^p-1$ a Mersenne prime, letting $m=2^n$, you find that indeed $$N=\frac{4^n-2^n}{2}=m^2(2m^2-1)=1^3+3^3+\ldots+(2m-1)^3.$$

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! Thank you so much for your kind help. –  Jiha Feb 5 '13 at 4:09
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