JOHN VOIGHT

Abstract. This serves as an elementary introduction to the history and theory

surrounding even perfect numbers .

One would be hard put to find a set of whole numbers with a

more fascinating history and more elegant properties surrounded

by greater depths of mystery—and more totally useless—than the

perfect numbers.

—Martin Gardner [2]

The number 6 is unique in that 6 = 1+2+3, where 1, 2, and 3 are all of the
proper

divisors of 6. The number 28 also shares this property, for 28 = 1 + 2 + 4 + 7 +
14.

These “perfect” numbers have seen a great deal of mathematical study —indeed,

many of the basic theorems of number theory stem from the investigation of the

Greeks into the problem of perfect and Pythagorean numbers [16]. Moreover, it

was while investigating these numbers that Fermat discovered the (little)
theorem

that bears his name and which forms the basis of a substantial part of the
theory of

numbers. Though it is rooted in ancient times , remarkably this subject remains
very

much alive today, harboring perhaps the “oldest unfinished project of
mathematics”

[17].

This paper surveys the history and elementary results concerning perfect
numbers.

## 1. Early History

The almost mystical regard for perfect numbers is as old as the mathematics

concerning them. The Pythagoreans equated the perfect number 6 to marriage,

health, and beauty on account of the integrity and agreement of its parts.

Around 100 c.e., Nicomachus noted that perfect numbers strike a harmony

between the extremes of excess and deficiency (as when the sum of a number ’s

divisors is too large or small), and fall in the “suitable” order: 6, 28, 496,
and

8128 are the only perfect numbers in the intervals between 1, 10, 100, 1000,
10000,

and they end alternately in 6 and 8. Near the end of the twelfth century, Rabbi

Josef b. Jehuda Ankin suggested that the careful study of perfect numbers was an

essential part of healing the soul. Erycius Puteanus in 1640 quotes work
assigning

the perfect number 6 to Venus, formed from the triad (male, odd) and the dyad

(female, even). Hrotsvit, a Benedictine in the Abbey of Gandersheim of Saxony

and perhaps the earliest female German poet, listed the first four perfect
numbers

in her play Sapientia as early as the tenth century:

JOHN VOIGHT

We should not leave unmentioned the principal numbers . . . those

which are called “perfect numbers.” These have parts which are neither

larger nor smaller than the number itself, such as the number

six, whose parts, three, two, and one, add up to exactly the same

sum as the number itself. For the same reason twenty-eight, four

hundred ninety-six, and eight thousand one hundred twenty-eight

are called perfect numbers [2].

Saint Augustine (among others, including the early Hebrews) considered 6 to

be a truly perfect number—God fashioned the Earth in precisely this many days

(rather than at once) to signify the perfection of His work. Indeed, as recorded
by

Alcuin of York (who lived from 732 to 804 c.e.), the second origin was
imperfect, as

it arose from the deficient number 8 > 1+2+4, this number counting the 8 souls
in

Noah’s ark (Noah, his three sons, and their four wives, in Genesis, chapter 7)
from

which sprung the entire human race [8]. Philo Judeus, in the first century c.e.,

called 6 the most productive of all numbers , being the smallest perfect number.

Throughout the centuries that followed, various mathematicians carefully
studied

perfect numbers (see the continued extensive history given by Dickson [6] and

also by Picutti [14]). Up to the time of Descartes and Fermat, a sizeable pool
of

important results—as well as much misinformation—had been collected.

## 2. Elementary Results

We now turn to some definitions and some elementary results.

**Definition 1. **The sum of divisors is the function
,
where d runs

over the positive divisors of n including 1 and n itself.

For example,

**Definition 2.** The number N is said to be perfect if
.
When

2N, we say N is deficient; when
,
we say N is abundant.

The definition of perfect is equivalent to saying that the sum of the proper
(or

aliquot) divisors of N is equal to N (we just do not add N itself to the sum).
While

this may seem more natural, the central reason for using the function
is that it

possesses some very special properties. First:

**Proposition 3.**
whenever gcd(m, n) = 1.

Proof. If d is a divisor of mn, then by unique factorization we can represent
d

uniquely as the product of a divisor of m and a divisor of n (since they have no

common factors ). That is, every term of
(the sum of all divisors of mn)

occurs exactly once in the sum
(the product of all divisors of m and n).

The converse is also true: every such product is a divisor of mn, so the sums
must

be the same.

This suffices to establish the proposition. It might be easier also just to
appeal

to the very special way in which is defined as a divisor sum (for the details
consult

[12]).

Hence
is completely determined when its value is known for every prime-power

argument. (Such functions are, of course, the multiplicative functions.) This
yields

the very useful:

**Theorem 4.** If
is a factorization of N into primes,

then

Proof. The only divisors of

Recall the geometric series has the closed form

which implies that
.
Since is multiplicative,

as claimed.

To give us just a little bit practice and feel for this formula for the sum
of divisors,

we prove the following two short lemmas. If
it is clear that
,
but

an even stronger result holds:

**Lemma 5. **If
,
then

where equality holds only if η = N.

Proof. Note that if d | N then kd = N for some k, so k = (N/d) | N. This
argument

works just as well in reverse, so we have that d | N if and only if (N/d) | N,
which

implies that

If η is a proper divisor of N, we have

Otherwise, equality holds.

For free, we also have:

Corollary 6. If N is perfect,

Proof. We just showed d | N is equivalent to (N/d) | N, hence

Cancelling N proves the corollary.

We assume henceforth that the reader is familiar with congruence arithmetic

and the fundamentals of group theory (consult [12] or [7] for some of the
basics),

of which we will use often use various facts without proof.

## 3. Even Perfect Numbers

The first individual to really categorize the perfect numbers was the Greek
mathematician

Euclid. He noticed that the first four perfect numbers are of a very

specific form:

Notice, though, that the numbers 90 =
= 8 · 15 and 2016 =

= 32 · 63 are missing from this sequence. As Euclid pointed out,

this is because 15 = 3 · 5 and 63 =
are both composite, whereas the numbers

3, 7, 31, 127 are all prime.

As it appears in Book IX, proposition 36 of his Elements, Euclid writes: “If

as many numbers as we please beginning from an unit be set out continuously in

double proportion, until the sum of all becomes prime, and if the sum multiplied

into the last make some number, the product will be perfect.”

We state this observation in a slightly more compact form:

Theorem 7 (Euclid). If
is prime, then N =
is perfect.

Proof. Clearly the only prime divisors of N are 2^{n} − 1 and 2. Since 2^{n} − 1
occurs

as a single prime, we have simply that
and thus

So N is perfect.

The task of finding perfect numbers, then, is intimately linked with finding

primes of the form 2^{n} −1. Such numbers are referred to as Mersenne
primes, after

the seventeenth century monk Marin Mersenne, a colleague of Descartes, Fermat,

and Pascal. He is credited with investigating these unique primes as early as
1644.

Mersenne knew that 2^{n} − 1 is prime for n = 2, 3, 5, 11, 13, 17, and
19—and, more

brilliantly, conjectured the cases n = 31, 67, 127, 257. It took nearly two
hundred

years to test these numbers.

There is one important criterion that hones in on the primality of Mersenne

numbers:

**Proposition 8** (Cataldi-Fermat). If 2^{n} − 1 is prime, then n
itself is prime.

Proof. Note from before, x^{n} − 1 = (x − 1)(x^{n-1} + · · ·
+ x + 1). Suppose we can

write n = rs, where r, s > 1. Then

so that (2^{r} − 1) | (2^{n} − 1) which is prime, a
contradiction.

Note that the converse is not true—the number 2^{11} − 1 = 2047 = 23
· 89, for

instance.

PERFECT NUMBERS

Must all perfect numbers be of Euclid’s type? Leonard Euler, in a posthumous

paper, proved that every even perfect number is of this type. Many ingenious
proofs

of this fact exist.

Theorem 9 (Euler). If N is an even perfect number, then N can be written in
the

form N = 2^{n-1}(2^{n} − 1), where 2^{n} − 1 is prime.

Proof. The first proof is Euler’s own [6]. Let N = 2^{n-1}m be
perfect, where m is

odd; since 2 does not divide m, it is relatively prime to 2^{n-1}, and

N is perfect so and with the
above, 2^{n}m =

Let
We then have
since 2^{n} does not divide

2^{n} − 1, it must divide s (because m is an integer), so that m = (2^{n}
− 1)q for some

q = s/2^{n}. If q = 1, we have a number of Euclid’s type, for then m = 2^{n}
− 1 and

Since
is the sum of all of the divisors

of m, m = 2^{n} − 1 must be a prime, and

If q > 1, we retotal the sum of the divisors of m = (2^{n} − 1)q. The
factors of m

then include 1, q, 2^{n} − 1, and m itself, so that

But this implies

an impossibility, for we have previously established equality:

1)s implies that m/s =

Proof 2. A second, simpler proof is given by Dickson [5]. From

we note

Since both
and m are integers, so too must d = m/(2^{n} − 1) be an integer.

Thus (2^{n} − 1) | m and consequently d itself divides m.

But
= m+d is the sum of all of the divisors of m. How

can this be? Certainly 1 divides m, so are forced to conclude that d = 1; if
this were

not the case, then we would have
a contradiction. Therefore,

and in partiular, m has no other positive divisors other than one and

itself, so
is prime.

Proof 3. A third proof is given by McDaniel [10]. Since
every

prime divisor of
must also divide m (for it is odd and cannot divide
).
So

suppose
divides
with p prime.

By Lemma 5,

Hence,

JOHN VOIGHT

This is only satisfied when the fraction on the right is zero, so that p = 2^{n}
− 1,

α = 1 and m = p, hence N is of Euclid’s type.

Proof 4. A similar proof is provided by Cohen [4]. Again we start with 2^{n}m
=

where 2^{n} − 1 | m. We then write

From the previous lemma,

To sustain equality throughout this expression, we must have 2^{n} −
1 = m and

is prime.

Proof 5. A fifth proof is given by Carmichael [3]. Let

Then

Write
Rearranging and dropping terms on the right, we have

Note that
Immediately, then,

some p_{i} must divide d, for the right-hand side has been completely factored into

primes and d is odd. If p_{i} is an aliquot divisor, then p_{i} < d and 1+1/p_{i} will
exceed

the left side of the inequality. Hence p_{i} = d. But with that, if k > 2, the
inequality

is again exceeded, so k = 2. Likewise,
and we have a number of Euclid’s

type.

Proof 6. As if that weren’t enough, a sixth and final proof is given by
Knopfmacher

[9]. Very familiar now, we have

Since
cannot divide 2, it divides N, and we may write

It follows that

PERFECT NUMBERS

and therefore

so

Since
for all i, expanding the product on the right will yield each of

the terms on the left. But there are the only two, which means that
for all

values of i except one; so then k = 2, and
The above equation then yields

simply

so
and N is of Euclid’s type as claimed.

Even perfect numbers have a number of nice little properties. We say that a

number is triangular if it can be arranged in triangular fashion, that is, one
atop

two atop three (and so on) arranged in a triangular lattice. While perfect
numbers

can be written in a geometric expression `a la Euclid, they may also be derived
from

arithemetic progressions:

6 = 1 + 2 + 3, 28 = 1 + 2 + 3 + 4 + 5 + 6 + 7, 496 = 1 + 2 + 3 + · · · + 31.

We state this result as follows:

**Proposition 10.** If N is an even perfect number, then N is triangular.

Proof. We have m triangular if
for

some k. But note that

We can also write any perfect number as the sum of cubes:

**Proposition 11 **([1]). If
is perfect then

Proof. Recall the formula

a fact which can be proved by induction. Let
we have then that

By substituting for m we get the desired result.

Perfect numbers have very unique representations. As first noted by

we have 6 = 110_{2}, 28 = 11100_{2}, 496 = 111110000_{2}, and 8128
= 1111111000000_{2}

written in binary; in general:

**Proposition 12.** If
is perfect and N is written in base 2, then

it has 2n − 1 digits, the first n of which are unity and the last n − 1 are
zero.

JOHN VOIGHT

Proof. This follows immediately from how numbers are written in binary, by
recalling

that

We also have the historically interesting fact:

**Proposition 13.** Every even perfect number ends in either 6 or 8.

Proof. Every prime greater than 2 is of the form 4m + 1 or 4m + 3. In the
first

case,

since by induction it is clear that
(mod 10) for all m. Similarly, in the

second case,

Finally, if n = 2, N = 6, and so we have captured all the possibilities.

Even perfect numbers exhibit another amazing property, first proven in 1844:

Proposition 14 (Wantzel). The iterative sum of the digits of an even perfect

number (other than 6) is one.

Proof. Let
The first claim we establish is that
(mod 9).

Now
(mod 3), for then 3 | n so n is not prime (N ≠ 6) and N
is not perfect.

Hence
or 2 modulo 3. Suppose we have the first case, and n = 3k + 1 for

some k. Now n is odd, so k is even and we could have written n = 6k + 1. Hence

In the second case, n = 3k + 2 for some k, so as n is not even, k is odd and

n = 6k + 5. Thus,

The work is now done, for the iterative sum of the digits of a number is
merely

its congruence class mod 9. For if
where the a_{i} are digits, then the

sum of digits is

By iterating this sum we get the desired result.

Historically, there are a number of problems that arise in
the investigation of

even perfect numbers. For instance, Fermat declared that if n is a prime, then

is divisble by 2n. Of course, this
immediately follows from the fact that

modulo n if n is prime by Fermat’s little
theorem.

Fermat also showed that if n is prime, then
is divisible by no prime other

than those of the form 2kn + 1. If q | for
some prime

modulo q. Since we have a power of 2 equal to 1, the identity, the order of 2
divides

n; and since n is prime, we have the order just equal to n. But also n | (q −1),
the

order of the multiplicative group. q is odd so q −1 is even and we have even
better

that 2n | (q − 1), or 2nk = q − 1, and the result follows.

Furthermore, Philiatrus discovered two primes n < 50 where
is not

prime: is divisible by 47 and
is divisble by 83. Lucas proved the

following: if p = 8m + 7 is a prime, then is
not. It can be shown that if

PERFECT NUMBERS

or 7 modulo 8 then
− 1 is divisible by the same factors as

In the second case, we have (p − 1)/2 = (8m+ 7 − 1)/2 = 4m+ 3 so
− 1 by

Fermat’s little theorem is divisible by p = 8m + 7.

## 4. Mersenne Primes

From the previous section, we know that finding even
perfect numbers reduces

simply to finding primes of the form where n
is prime. There are 37 known

Mersenne primes, summarized in the following table [2]:

**Index Even Perfect Number Number of Digits Date**

With each one of these primes, there is a story that goes
with its discovery. For

instance, Peter Barlow in his Theory of Numbers published in 1811, wrote about

the eighth perfect number: “It is the greatest that will ever be discovered,
for, as

they are merely curious without being useful, it is not likely that any person
will

attempt to find one beyond it.”

Quite to the contrary, the latest Mersenne prime was
discovered on January 27,

1998 by Roland Clarkson, a 19-year-old student at California State University–

Dominguez Hills. Roland used a 200 MHz Pentium computer part-time for 46

days to prove the number prime—he is just one of over 4000 volunteers world-wide

participating in the Great Internet Mersenne Prime Search (GIMPS). Indeed, Scott

Kurowski, the co-author of software Clarkson used to find this enormous prime,
has

offered a cash prize to the discoverer of the 38th Mersenne prime. The prize
pool

is $1.00 for every 1000 digits in the new prime or US $1,000.00, whichever is
larger.

He says, “If you have a desktop computer, we’ve got something for it to do
between

keystrokes and mouse clicks, and many more machines are needed for the next,

even larger, prime.”

JOHN VOIGHT

The Mersenne primes appear to be regularly
distributed —“ballpark” estimates

(meaning within thousands of orders of magnitude) can be given [15], but
ultimately

these numbers must be checked out one by one. We close with one interesting
result:

Conjecture 15 (Cunningham). If (mod 4),

and 2p + 1 is a prime, then is prime.