when \(n\) is a product of the prime powers \(p_i^{e_i}\text{.}\)
An odd perfect number cannot be a prime power. This is easy; using the computation for \(k=1\) would require \(2=\frac{\sigma(n)}{n}<\frac{p}{p-1}\text{.}\) Even for \(p=2\text{,}\)\(2< p/(p-1)\) isn’t possible; since we are looking for an odd perfect number, it definitely won’t be possible!
An odd perfect number cannot be a product of exactly two prime powers. Use the same idea, but now with the biggest possible values for odd primes.
An odd perfect number cannot be a product of exactly three prime powers unless the first two are \(3^e\) and \(5^f\text{.}\) This proof is slightly longer.
Suppose that \(3\) is not the smallest prime involved. Then the biggest that
and this fraction is still less than \(2\text{.}\)
Suppose that \(5\) is not the second-smallest prime involved (assuming \(3\) is the smallest). We again get a contradiction.
This proof is from [E.2.8, Section 3.3A], which has even more details – including a full elementary proof that an odd perfect number must have four different prime factors!
Subsection19.5.2The abundancy index and odd perfect numbers
What is particularly interesting about this is that we can connect odd perfect numbers to a non-integer abundancy index in a surprising way! The connection below is due to P. Weiner in [E.7.14].
We begin with a useful lemma, which answers questions very closely related to Exercises 19.6.11 and 19.6.12.
Lemma19.5.3.
If \(n\) and \(\sigma(n)\) are both odd, then \(n\) is a perfect square.
Proof.
If \(n\) is odd, it is a product of odd prime powers. Let’s look at \(\sigma\) as applied to each piece, thanks to multiplicativity.
If \(\sigma(n)\) is odd, then each factor \(1+p+p^2+\cdots +p^e\) is odd. Such a factor of \(\sigma(n)\) is a sum of odd numbers, which is only odd if there is an odd number of them.
Since there are \(e+1\) summands, \(e\) must be even for every primes \(p\) dividing \(n\text{.,}\) which finishes proving the lemma.
Theorem19.5.4.
If \(\frac{5}{3}\) is the abundancy index of \(N\text{,}\) then \(5N\) is an odd perfect number.
Proof.
Assume this works for some \(N\text{.}\) Then \(3\sigma(N)=5N\text{.}\)
Let’s look at divisors. First, \(3\mid N\text{.}\) So if \(N\) is even, then \(6\mid N\text{,}\) so by Fact 19.4.10,
Subsection19.5.3Even more about odd perfect numbers, if they exist
Naturally, all of this is somewhat elementary; there are many more criteria. They keep on getting more complicated, so I can’t list them all, but here is a selection, including information from a big computer-assisted search 12
www.lirmm.fr/~ochem/opn/
13
There was another search at oddperfect.org but they seem to have let their domain lapse, so it is unclear whether it is still a going concern. (Search web.archive.org for the status in 2019.)
some desired numbers to help compute up to \(10^{2100}\text{.}\))
Have at least 101 prime factors (not necessarily distinct).
Have at least 10 distinct prime factors. (This is new and relies on heavy computation by Pace Nielsen in Odd perfect numbers, Diophantine equations, and upper bounds in Mathematics of Computation 16
Have a largest prime factor at least \(10^8\text{.}\)
Have a second largest prime exceeding \(10000\text{.}\)
Have the sum of the reciprocals of the prime divisors of the number between about\(0.6\) and \(0.7\text{.}\)
Have the sum of the reciprocals of odd perfect numbers be finite (since the sum of the reciprocals of all perfect numbers is finite!). In fact, the sum of the reciprocals of odd perfects must be less than \(2\times 10^{-150}\) (see [E.7.6]), and that of all perfects is less than about \(0.0205\text{.}\)
Obey the rule that if \(n\) is an odd perfect number, then \(n\equiv 1\text{ mod }12\) or \(n\equiv 9\text{ mod }36\text{.}\)
As an appropriate way to finish up this at times overwhelming overview, since Euler finished the characterization of even perfect numbers, let us present his own criterion for odd perfects! (See also the linked article 18
An odd perfect number must be of the form \(p^e m^2\text{,}\) where \(m\) is odd, \(p\) is prime, and \(p\)and\(e\) are both \(\equiv 1\text{ (mod }4)\text{.}\)