An even integer is perfect if and only if it can be written as $2^{p-1}(2^p-1)$, where both $p$ and $2^p-1$ are prime.

We show that:

If $n = 2^{p-1}(2^p-1)$ and $p$ and $2^p-1$ are prime then $n$ is perfect.

\begin{align}

\sigma(n) &= \sigma(2^{p-1}(2^p-1)) \\

&= \sigma(2^{p-1}) \sigma(2^p-1) \ \\

& = ( 2^p-1 ) 2^p \\

& = 2^p ( 2^p-1 ) \\

& = 2 (2^{p-1}(2^p-1)) \\

& = 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 $n=2^{p-1}(2^p-1)$ where $p$ and $2^p-1$ are prime.

Since $n$ is even we assume $n=2^{k-1}m$ where $(2^{k-1}, m) = 1$.

(1) We calculate the divisor-sum of $n$ as follows:

\begin{align}

\sigma(n) &= \sigma(2^{k-1}m)\\

& = \sigma(2^{k-1})\sigma(m)\\

& = \frac{2^k-1}{2-1}\sigma(n).

\end{align}

(2) We assume $n$ is perfect thus:

\begin{align}

\sigma(n) &= 2n\\

& = 2 (2^{k-1} m)\\

& = 2^k m

\end{align}

Now (1) and (2) gives:

\begin{align}

\frac{2^k-1}{2-1}\sigma(m) & = 2^k m \Leftrightarrow \\

\sigma(m) & = \frac{2^k m}{2^k-1} \\

& = \frac{((2^k-1)+1)m}{2^k-1} \\

& = m + \frac{m}{2^k-1}

\end{align}

Since $\sigma(m)$ is the sum of all divisors $\frac{m}{2^k-1}$ must be $1$. So $m=2^k-1$ with divisors $1$ and $m$ and is thus prime.

We conclude that $n=2^{k-1}(2^k-1)$ with $2^k-1$ and $k$ prime.

QED

Stephen Hawking RIP

3 days ago

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