An even integer is perfect if and only if it can be written as $2^{p1}(2^p1)$, where both $p$ and $2^p1$ are prime.
We show that:
If $n = 2^{p1}(2^p1)$ and $p$ and $2^p1$ are prime then $n$ is perfect.
\begin{align}
\sigma(n) &= \sigma(2^{p1}(2^p1)) \\
&= \sigma(2^{p1}) \sigma(2^p1) \ \\
& = ( 2^p1 ) 2^p \\
& = 2^p ( 2^p1 ) \\
& = 2 (2^{p1}(2^p1)) \\
& = 2n
\end{align}
Since $\sigma(n)=2n$ we conclude that $n$ is perfect.
We show that:
If $n$ is even and perfect then it can be represented as $n=2^{p1}(2^p1)$ where $p$ and $2^p1$ are prime.
Since $n$ is even we assume $n=2^{k1}m$ where $(2^{k1}, m) = 1$.
(1) We calculate the divisorsum of $n$ as follows:
\begin{align}
\sigma(n) &= \sigma(2^{k1}m)\\
& = \sigma(2^{k1})\sigma(m)\\
& = \frac{2^k1}{21}\sigma(n).
\end{align}
(2) We assume $n$ is perfect thus:
\begin{align}
\sigma(n) &= 2n\\
& = 2 (2^{k1} m)\\
& = 2^k m
\end{align}
Now (1) and (2) gives:
\begin{align}
\frac{2^k1}{21}\sigma(m) & = 2^k m \Leftrightarrow \\
\sigma(m) & = \frac{2^k m}{2^k1} \\
& = \frac{((2^k1)+1)m}{2^k1} \\
& = m + \frac{m}{2^k1}
\end{align}
Since $\sigma(m)$ is the sum of all divisors $\frac{m}{2^k1}$ must be $1$. So $m=2^k1$ with divisors $1$ and $m$ and is thus prime.
We conclude that $n=2^{k1}(2^k1)$ with $2^k1$ and $k$ prime.
QED
Friday, February 25, 2011
Perfect numbers
An example of a note in my NT Wiki. In M381 only the first part of the proof is given which was known as early as Euclid. Euler was the first who gave, a not so very clear proof of the reverse. Many followed Euler with subsequent improvements of the proof. I have used a proof of Dickson published in 1911.  Perfect numbers are still actively researched. It is for example still unknown if odd perfect numbers exist. If they exist however they are surely very large.
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