Lecture 15 - Perfect Numbers


In today's class, we discussed perfect numbers: those numbers n so that $\sigma(n) = 2n$. We gave a characterization of even perfect numbers and talked about properties that odd perfect numbers have — even though we don't know whether odd perfect numbers actually exist. We then briefly mentioned amicable pairs.

Numeric Perfection

Last class period we saw that $\sigma(6) = 12 = 2\cdot 6$ and $\sigma(28) = 56 = 2\cdot 28$. We said that these numbers n have the additional property that the sum of their proper divisors (i.e., those divisors smaller than n) is n. We used $\sigma'(n)$ to denote the sum of the proper divisors of n. Hence, for example, we have

\begin{align} \sigma'(6) = 1+2+3 = 6. \end{align}

This leads us to the main topic of the day

Definition: An integer n is called perfect if $\sigma(n) = 2n$. Equivalently, a number n is perfect if $\sigma'(n) = n$.

Example: Some perfect numbers

In addition to 6 and 28, here are the next perfect numbers: $496, 8128, 33550336$. We noticed in class that these numbers have particular factorizations

\begin{align} \begin{split} 6 &= 2^1\cdot 3\\ 28 &= 2^2\cdot 7\\ 496 &= 2^4\cdot 31\\ 8128 &= 2^6 \cdot 127\\ 33550336 &= 2^{12}\cdot 8191.\\ \end{split} \end{align}

We noticed a few things about these numbers in class. First, these numbers all seemed to factor as some power of two times some other odd prime. This odd prime number also seemed to be one less than a power of 2, and indeed the power of 2 which appeared in the factorization of n seemed to determine the power of 2 which was just ahead of the odd prime. In summary, we said that it seemed that the perfect numbers we were seeing took the form

\begin{equation} 2^{p-1}(2^p-1), \end{equation}

where $2^p-1$ was prime (and hence p itself was prime; see Exercise 26 in Chapter 1 of Strayer). It turns out we're not the first people to recognize this.$\square$

Theorem: If $n = 2^{p-1}(2^p-1)$ where $2^p-1$ is prime, then n is perfect.

Proof: To see that this is true, we just compute $\sigma(n)$:

\begin{align} \sigma(n) = \sigma(2^{p-1}(2^p-1)=\sigma(2^{p-1})\sigma(2^p-1), \end{align}

where the last equality follows because $(2^{p-1},2^p-1) = 1$. Now we can compute the value of $\sigma$ on each of the components individually; for the former we can use our formula for $\sigma$ on a prime power, and for the latter we can use the fact that $2^p-1$ is prime to conclude that $\sigma(2^p-1) = 2^p-1+1$:

\begin{align} \sigma(2^{p-1})\sigma(2^p-1) = \frac{2^p-1}{2-1} (2^p-1+1) = 2^p(2^p-1). \end{align}

Notice that this final expression is just $2\cdot (2^{p-1}(2^p-1)) = 2n$, and hence we've proven that n is perfect.$\square$

Hence this theorem says whenever we find a prime of the form $2^p-1$, we can use this to produce a new even perfect number. Since it was recently shown that $2^{43112609}-1$ is a prime number, this means that

\begin{equation} 2^{43112608}(2^{43112609}-1) \end{equation}

is a perfect number. This number has around 26 million digits!

Is This It?

Now that we've shown that numbers of a particular form are perfect, it is natural to ask if there are numbers not of this form which are perfect. A partial answer to this question was provided by Euler.

Theorem: If n is even and perfect, then $n = 2^{p-1}(2^p-1)$, where $2^p-1$ is prime.

Notice in particular this means that we have a correspondence between even perfect numbers and Mersenne primes: finding one gives rise to another.

Proof: Suppose that n is an even perfect number, and factor n as $n=2^a b$; we assume that this factorization leaves $2\nmid b$. Our goal will be to show that $b = 2^{a+1}-1$, and that b is prime.

Now let's compute $\sigma(n)$ in two ways. First, we'll use the factorization of n to determine a factorization of $\sigma(n)$:

\begin{align} \sigma(n) = \sigma(2^a b) = \sigma(2^a)\sigma(b) = (2^{a+1}-1)\sigma(b). \end{align}

On the other hand, we know that n is perfect, so that $\sigma(n) = 2n$. Hence we have

\begin{align} \sigma(n) = 2n = 2(2^ab) = 2^{a+1}b. \end{align}

Putting these two equations together leaves

\begin{align} 2^{a+1}b = (2^{a+1}-1)\sigma(b). \end{align}

Our goal will be to use this equation to tell us that b has the desired form.

To do this, notice that $2^{a+1}$ divides the left hand side, so hence we also have $2^{a+1} \mid (2^{a+1}-1)\sigma(b)$. Since $2^{a+1}-1$ is relatively prime to $2^{a+1}$, this means that we must have $2^{a+1} \mid \sigma(b)$. This means there is an integer c so that

\begin{align} 2^{a+1}c = \sigma(b). \end{align}

Notice that if we substitute this expression back in for $\sigma(b)$ in Equation (9) and cancel out the $2^{a+1}$ term common to both sides, then we get the equation

\begin{equation} b = (2^{a+1}-1)c. \end{equation}

We now claim that — in fact — $c=1$. To do this, suppose that $c>1$. Then Equation (11) tells us that c is a divisor of b which is distinct from 1 and b, and hence we get

\begin{align} \sigma(b) = \sum_{d\mid b} d \geq 1+c+b. \end{align}

On the other hand, we know from Equation (10) that

\begin{align} \sigma(b) = 2^{a+1}c = 2^{a+1}c-c+c = (2^{a+1}-1)c+c = b + c, \end{align}

where the last equality comes from Equation (11). Together these equations give an obvious contradiction, and hence we conclude that $c=1$.

Now that we know $c=1$, we can substitute this back into Equations (10) and (11) to find

\begin{align} \begin{split} b &= 2^{a+1}-1\\ \sigma(b) &= 2^{a+1}. \end{split} \end{align}

Hence we have $\sigma(b) = b+1$, so that we know b is prime.

Believe it or not, we've now established everything we set out to prove. $\square$

Odd Perfects

Since we have such a nice description of even perfect numbers, it's natural to wonder if we've captured all the perfect numbers out there. In other words, we'd like to know if there are odd perfect numbers. It turns out that no one yet knows whether odd perfect numbers exist. There's been a lot of work done to find these numbers, and a surprising amount about odd perfect numbers is known. That is to say, if an odd perfect number exists, it has to satisfy a number of properties that one might not expect a priori. For instance, it is known that if n is an odd perfect number, then

  • $n = p^a m^2$ where p is an odd prime that is congruent to 1 modulo 4, a is an integer which is congruent to 1 modulo 4, and m is an integer relatively prime to p
  • n has at least 8 distinct prime factors (at least 75 if you count with multiplicity)
  • $n \geq 10^{300}$

Amicable Pairs

Another related topic which you'll explore briefly in your homework is that of amicable pairs. We start with the following

Definition: A pair of integers n and m are called an amicable pair if

$\displaystyle \sigma'(n) = m \quad \mbox{ and } \quad \sigma'(m) = n$

or, equivalently, if

$\displaystyle \sigma(n) = \sigma(m) = n+m$.

As with perfect numbers, there are quite a bit of things known about amicable pairs, but also plenty about amicable pairs which isn't known. You'll see in your homework that 220 and 284 are an amicable pair. There were two other pairs that were known up until about the 1730s, and then Euler was able to find about 59 more pairs. Here are a few open questions about amicable pairs:

  • All known amicable pairs to date are either both even or both odd. It is unknown whether there exist amicable pairs for which one number is even and the other number is odd.
  • All known amicable pairs have a nontrivial common factor. It isn't known whether there exists an amicable pair where the numbers of the pair are relatively prime to each other.
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