Perfect Numbers

A perfect number is a positive integer that equals the sum of its proper divisors, that is, positive divisors excluding the number itself. For example, \(6\) is a perfect number because the proper divisors of \(6\) are \(1,2,\) and \(3,\) and \(6=1+2+3.\)

The sum of all positive divisors of a number \(n\) is denoted by \(\sigma(n)\). A perfect number is therefore a positive integer \( n \) such that \( \sigma(n) = 2n.\)

Perfect numbers were of great interest to ancient mathematicians including the Greeks, who attributed mystical significance to the property. It is perhaps somewhat surprising that many elementary questions about perfect numbers are still open. Most notably, it is not known whether there are infinitely many perfect numbers, and it is not known whether there are any odd perfect numbers.

Euclid proved in the Elements that if \( 2^p-1 \) is prime, then \( 2^{p-1}(2^p-1) \) is perfect.

The first four perfect numbers are

\[\begin{align} 6 &= 2^{2-1}(2^2-1) \\ 28 &= 2^{3-1}(2^3-1) \\ 496 &= 2^{5-1}(2^5-1) \\ 8128 &= 2^{7-1}(2^7-1). \end{align}\]

More than 2000 years later, Euler was the first to give a proof that every even perfect number was of this form. This is known as the Euclid-Euler theorem. Euler's proof is quite elementary:

A positive integer \( n\) is an even perfect number if and only if \( n = 2^{p-1}(2^p-1)\) for some positive prime \(p \) such that \( 2^p-1\) is prime. \((\)Here \( 2^p-1 \) is known as a Mersenne prime.\()\)

For the "if" (Euclid) statement, let \( P \) be the Mersenne prime \( 2^p-1 \), and write down the divisors of \( n \) explicitly: they are \( 1,2,4,\ldots,2^{p-1} \) and \( P,2P,\ldots,2^{p-1}P \). The sum of these divisors, by the geometric series formula, is

\[(2^p-1)+(2^p-1)P = (2^p-1)(P+1) = (2^p-1)2^p = 2n.\]

So the sum of the proper divisors is \(n\).

For the "only if" (Euler) statement, some facts about the sum of divisors function, denoted \(\sigma(n) \), will be relevant. In particular, \( \sigma \) is multiplicative: \( \sigma(a)\sigma(b) =\sigma(ab) \) if \(a,b \) are relatively prime.

Now suppose \( n \) is an even perfect number. Then write \( n = 2^{a-1} b,\) where \( a\ge 2 \) and \( b \) is odd. Since \(n\) is perfect, \(\sigma(n) = 2n\). Then

\[2n = \sigma(n) = \sigma(2^{a-1})\sigma(b) = (2^a-1)\sigma(b).\]

So \( (2^a-1)|2n\), hence \( (2^a-1)|n \), hence \( n = 2^{a-1}(2^a-1)c \) for some integer \( c \), and expanding the above equation again yields

\[\begin{align} 2^a(2^a-1)c &= (2^a-1)\sigma\big((2^a-1)c\big) \\ 2^ac &= \sigma\big((2^a-1)c\big). \end{align}\]

But note that the right side is at least \( c + (2^a-1)c \) since these are two divisors of \( (2^a-1)c \), and those two divisors already add up to \( 2^ac \). So the conclusion is that equality holds and these are the only two divisors of \( (2^a-1)c \). This can only happen if \( c = 1 \) and \( 2^a-1 \) is prime. This completes the proof. \(_\square \)

\((\)Exercise: Where did the proof use the fact that \( a \ge 2?) \)

The following is a list of some interesting properties of perfect numbers. Unlike many lists in the literature, this list is confined to nontrivial properties: for instance, the fact that every even perfect number is a triangular number is a trivial consequence of the formula

\[ N = 2^{p-1}(2^p-1) = \frac{(2^p-1)2^p}2. \]

1. Every even perfect number ends in the digits \(6\) or \(28\) (when written in decimal).
2. The iterative digital sum of any even perfect number except \(6\) is \(1\).
3. The only square-free perfect number is \(6\).
4. A perfect number is an Ore harmonic number; that is, the harmonic mean of its divisors is an integer. (Not every Ore harmonic number is perfect, e.g. 140.)
5. If an odd perfect number exists, it has more than \(300\) digits, at least \(75\) prime factors, at least \(9\) distinct prime factors, and at least one prime factor of at least \(20\) digits.

Proofs: Here are hints for 1-4:

1. This follows from a mod-10 and mod-100 analysis. If \( p \equiv 1 \) mod \( 4\) or \( p =2 \), the corresponding perfect number ends in \( 6 \), and if \( p \equiv 3 \) mod \( 4 \), it ends in \( 28 \) \((\)hint: show that it is congruent to \( 3 \) mod \( 25). \)

2. Show that every even perfect number except \( 6 \) is \( 1 \) mod \( 9 \). Summing the digits and iterating preserves the congruence class mod \( 9 \).

3. For even perfect numbers this is clear from Euclid-Euler; for odd perfect numbers, two odd prime factors would lead to a factor of \( 4 \) in \( \sigma(n) \), but \( 2n \) isn't divisible by \( 4\).

4. The harmonic mean of the divisors of a perfect number is \( \frac{d(n)}2 \) (why?), where \( d(n) \) is the number of divisors. Now \( d(n) \) is even unless \( n \) is a square, but a perfect square cannot be a perfect number because the sum of its divisors is odd (why?).

5. This represents the current frontier of mathematical research on odd perfect numbers. The methods used to establish these results seem exceedingly unlikely to lead to a proof that odd perfect numbers do not exist since they provide lower bounds only. Perhaps the existence of an odd perfect number seems unlikely given 5, but sometimes interesting phenomena only happen for large enough integers.