Exercise 4.2.16 (Perfect numbers)

Question

We say (following Euclid) that $m$ is a {\bf perfect number} if $\sigma(m) = 2m$, that is, if $m$ is the sum of all its positive divisors other that itself. If $2^n - 1$ is a prime $p$, prove that $2^{n-1}p$ is a perfect number. Use this result to find three perfect numbers.

Richard Ganaye · Accepted Answer

\begin{proof} 
Suppose that $p = 2^n - 1$ is prime, and put $n = 2^{n-1} p = 2^{n-1}(2^n - 1)$. Since $\sigma$ is multiplicative, by Theorem 4.5,
\begin{align*}
\sigma(n) &= \sigma(2^{n-1}) \sigma(p)\
&=(2^n -1) (p + 1)\
&=(2^n -1)2^n\
&= 2n.
\end{align*}
So $n  = 2^{n-1}(2^n - 1)$ is a perfect number if $2^n - 1$ is prime.

\begin{verbatim}
sage: [2^(n-1) * (2^n -1) for n in range(50) if is_prime(2^n - 1)]
[6, 28, 496, 8128, 33550336, 8589869056, 137438691328, 2305843008139952128]
\end{verbatim}
So $6, 28, 496, 8128, 33550336, 8589869056, 137438691328, 2305843008139952128$ are perfect numbers.
\end{proof}
The last found Mersenne's number (October 2024) is $M_{136279841}$, with more than 40,000,000 digits. So
$$n = 2^{136279840}(2^{136279841} - 1)$$
is a (big) perfect number.

\bigskip
See the sequences A000043 and A000396 in the OEIS:
\begin{verbatim}
https://oeis.org/A000043
https://oeis.org/A000396
\end{verbatim}

Exercise 4.2.16 (Perfect numbers)

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Exercise 4.2.16 (Perfect numbers)

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