Section 19.4 Perfect Numbers
Subsection 19.4.1 A perfect definition and theorem
Definition 19.4.1.
When the ratio
Theorem 19.4.2.
If
Proof.
Euclid's proof (in the link) of this is worth looking at.
Theorem 19.4.3. Characterization of Even Perfect Numbers.
If
Proof.
First, assume that \(2^n-1\) is prime. Then the factors of \(N = 2^{n-1}\left(2^n-1\right)\) are coprime, so
The steps are because of multiplicativity and the formulas we had earlier (see Theorem 19.2.5) for \(\sigma\) of powers of two and primes. But then
so that the sum of divisors is exactly twice the original number.
Now for the converse, which is somewhat longer. Let us start with an even perfect number \(N\text{,}\) which is perforce divisible by some power of two.
Looking ahead, call this power the \((n-1)\)th power! Then our even perfect number may be written as \(N = 2^{n-1}q\text{,}\) where \(q\) is the (odd) quotient.
Let's divide the rest of the proof into several pieces. First, two facts.
-
We know that this number is perfect, so
\begin{equation*} \sigma\left(2^{n-1}q\right)=2\cdot 2^{n-1}q=2^nq \end{equation*} -
We also know how to compute \(\sigma\text{,}\) so
\begin{equation*} \sigma\left(2^{n-1}q\right)=\sigma\left(2^{n-1}\right)\sigma(q)=\left(2^n-1\right)\sigma(q) \end{equation*}
We can combine these observations to see that
Note that this means \(2^n-1\mid q\text{,}\) since \(q\) is the only odd part of the left-hand side (implicitly using some of Theorem 6.3.2). Let's write
Substituting, we have
Since \(m\) and \(q\) both divide \(q\text{,}\) by the definition of \(\sigma\) we have
Since these two divisors (\(q\) and \(m\)) alone add up to \(\sigma(q)\text{,}\) it must be true that \(q\) has exactly these two divisors, so it is prime. That means \(m=1\text{,}\) and \(q=2^n-1\text{,}\) and so the perfect number \(N\) equals \(2^{n-1}\left(2^n-1\right)\text{.}\) Great!
Subsection 19.4.2 Speculation and more terminology
There are many things people have claimed about numbers of this type. A Hellenistic Roman in the first century in Gerasaβ2β named Nichomachus claimed that thexxxxxxxxxx
(2^13-1)*2^12
Definition 19.4.4.
Recall that if
If
for some integer then we say that is -perfect.Or, if
then is abundant.If
we say is deficient.
Definition 19.4.5.
Here are some less well-known, but nonetheless interesting, terms.
A number is pseudoperfect if it is the sum of some of its divisors (other than itself).
A number
is superabundant if the ratio for is bigger than the value of the ratio for all smallerA number is weird if it is abundant but not pseudoperfect. (There is a famous paper of ErdΕs on this topic.)
Question 19.4.6.
Is a perfect number pseudoperfect?
Subsection 19.4.3 The abundancy index
It's time to give a name to the mysterious ratio at the core of this section.Definition 19.4.7.
The ratio
Question 19.4.8.
Rather than asking which integers can be gotten, which rational numbers can be gotten as
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def _(n=(20,[1..200])):
cols = ceil(n/10)
T = [cols*['$n$',r'$\sigma(n)/n$']]
list = [[i,(sigma(i)/i)] for i in range(1,n+1)]
list.extend((10-(len(list)%10))*['',''])
for k in range(10):
t = [item for j in range(cols) for item in list[k+10*j]]
T.append(t)
pretty_print(html(table(T,header_row = True, frame = True)))
Fact 19.4.9.
Proof.
We have that
Now note that for every \(d\mid n\text{,}\) the quotient is also an integer divisor \(d'\) of \(n\text{.}\) So
This is the same list as the original divisor list, so reordering gives
Fact 19.4.10.
Clearly all such numbers are in the interval
If
thenIf
in lowest terms, thenIf
is βcaughtβ between and (such that ) and is relatively prime to then is not an abundancy index.
Proof.
We skip the proof, but proving the first two facts is left as Exercise 19.6.22.
Subsection 19.4.4 Amicable Numbers
Another interesting idea of summing divisors is still of ancient provenance, though not quite as old as Euclid.Definition 19.4.11.
A pair of positive integers

Historical remark 19.4.13. Thabit ibn Qurra.
The ninth-century Arab doctor and mathematician Thabit ibn Qurra was probably responsible for a number of Arabic translations of Greek mathematics in his time in the βHouse of Wisdomβ of the Caliphs of Baghdad. Interestingly, he did not include a single example of either perfect numbers or amicable numbers, despite clearly being in control of effective information about them. He made important contributions to the question of the parallel postulate in geometry.
Algorithm 19.4.14. Get Amicable Numbers.
Here is one way to get amicable numbers.
Make a list of numbers of the form
andThen check if
and are all prime.If so, then
and are an amicable pair.
Proof.
Since only primes and powers of two are involved, it's easy to calculate \(\sigma\) in this case, so proving it is left as an exercise (see Exercise 19.6.21).
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def _(n=[2..20]):
pretty_print(html("We have $p_{%s}=%s$ and $p_{%s}=%s$"%(n-1,3*2^(n-1)-1,n,3*2^(n)-1)))
pretty_print(html("And $q_{%s}=%s$ as well."%(n,9*2^(2*n-1)-1)))
if is_prime(3*2^n-1) and is_prime(3*2^(n-1)-1) and is_prime(9*2^(2*n-1)-1):
pretty_print(html("Then the pair $%s$ and $%s$ is amicable!"%(2^n*(3*2^(n-1)-1)*(3*2^(n)-1), 2^n*(9*2^(2*n-1)-1))))
else:
pretty_print(html("Doesn't give an amicable pair"))
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sigma(1184),sigma(1210),1184+1210