# My Talk on Aliquot Parts

Copy of talk given to The Math Club of The University of Dayton by Harry J. Smith in December 1953.

### Aliquot Parts

Aliquot parts are the divisors of a number, for example the aliquot parts of 12 are 1, 2, 3, 4, 6, 12. The main theorem necessary for solving problems related to aliquot parts is that every integer greater than one can be represented uniquely as the product of prime factors. A number N > 1 can therefore be written

### `N = P1^A1 P2^A2 ... Pr^Ar (1)`

where the Pi's are the various different prime factors and Ai the number of times Pi occurs in the prime factorization.

For any divisor Dj of N we have

### `N = Dj Ej (2)`

where Ej is also a divisor of N. When multiplied together, the prime factorization of Dj and Ej must give that of N so that

### `Dj = P1^B1 P2^B2 ... Pr^Br (3)`

where the exponent Bi do not exceed the corresponding Ai in (1). Since the second factor in (2) must contain the remaining factors, it becomes

### `Ej = P1^(A1-B1) P2^(A2-B2) ... Pr^(Ar-Br)`

In the expression (3), for a divisor, the exponent B1 can take on (A1+1) values, namely 0, 1, 2, ..., A1. Similarly B2 can take on the (A2+1) values 0, 1, 2, ..., A2; and so on. Since any choice of B1 can be combined with any choice of B2 and these when chosen can be combined with any choice of B3 and so on, and since for every set of values for Bi there is a unique divisor determined, we have

Theorem 1. The number of divisors of a number N in the form (1) is

Example:
The number

has

### ` V(60) = (2+1)(1+1)(1+1) = 12`

divisors. They are

### `1, 2, 3, 4, 5, 6, 10, 12, 15, 20, 30, 60 (5)`

Now let us consider the product of all the divisors of N. In (2) let Dj run through all V(N) divisors of N. The corresponding Ej will then also run through these divisors in some order, so that the product of all Dj is the same as the product of all Ej's. This we can write

### `PI(Dj) = PI(Ej)`

where the symbol PI() denotes the product. We can now form the product of all V(N) equations (2) and obtain

### `N^V(N) = PI(Dj) PI(Ej) = (PI(Dj))^2`

This yields the desired result:

Theorem 2. The product of all divisors of a number N is

### `PI(Dj) = N^(V(N)/2) (6)`

Example:
The product of all divisors of 60 is

### `PI(Dj) = 60^6 = 46,656,000,000`

as we could check by multiplying together the divisors (5).

The determination of the sum sig(N) of the divisors of a number N is not quite as simple. If N has more than one prime factor, it can be written as the product of two relatively prime factors

### `N = A B`

Since the prime factors of A and B are different, it is concluded that any divisor Dj of N must be of the form

### `Dl = Ak Bk (7)`

where Ak is a divisor of A and Bk is a divisor of B. We can denote the divisors of A and B, respectively, by

### `1, A1, A2, ..., A 1, B1, B2, ..., B`

so that the sums of their divisors are

### `sig(B) = 1 + B1 + B2 + ... + B`

In (7) let us take all divisors of N with the same Ak. Their sum is

### `Ak (1 + B1 + B2 + ... + B) = Ak sig(B)`

Next, sum this for all possible Ak and obtain the total sum of all the divisors of N

### `sig(B) + A1 sig(B) + A2 sig(B) + ... + A sig(B) = sig(A) sig(B)`

Thus when A and B are relatively prime

### `sig(N) = sig(A B) = sig(A) sig(B) (8)`

We can split A and B further into relatively prime factors and apply the same rule (8) again.

### `sig(N) = sig(A B C ... G) = sig(A) sig(B) sig(C) ... sig(G)`

This may be continued until the factors become the powers of the various prime divisors of N. So that when N is in the form (1)

### `sig(N) = sig(P1^A1) sig(P2^A2) ... sig(Pr^Ar) (9)`

For a prime power P^A the divisors are

so that

### `sig(P^A) = 1 + P + P^2 + ... + P^A`

This is a geometric series. It can be summed by multiplying the series by P

### `P sig(P^A) = P + P^2 + ... + P^A + P^(A+1)`

and subtracting the original series

so that

### `sig(P^A) = (P^(A+1) - 1) / (P-1) (10)`

Apply this result to every factor in (9). We have

Theorem 3. The sum of divisors of a number N in the form (1) is

### `sig(N) = ((P1^(A1+1) - 1) / (P1-1)) * ... * ((Pr^(Ar+1) - 1) / (Pr-1)) (11)`

Every prime power P^A in the factorization (1) contributes a factor (10) to the expression (11) for the sum of the divisors of N. It is useful to observe the following simple cases, which occur commonly. When a prime appears only once, i.e. A = 1, we have

When P^A = 2^A

### `sig(2^A) = (2^(A+1) - 1) / (2-1) = 2^(A+1) - 1`

Example:
The sum of the divisors of

is

### `sig(60) = (2^3 - 1)(3+1)(5+1) = 7 4 6 = 168`

as can be verified by summing the divisors (5).

A perfect number can be defined as a number equal to the sum of its proper divisors, that is divisors other than itself. We can denote the sum of such divisors as

### `sig0(N) = sig(N) - N (12)`

and then condition for a perfect number becomes

### ` sig(N) = 2 N (14)`

By means of either of these conditions we can check whether a number is perfect. For instance,

### `sig(28) = sig(2^2 7) = (2^3 - 1)(7+1) = 7 8 = 56`

Therefore 6 and 28 are perfect numbers.

Only one type of perfect number is known:

Theorem 4. A number of the form

### `N = 2^(P-1) * (2^P - 1) (15)`

is perfect when 2^P - 1 is prime.

This is proven by applying theorem 3.

### `sig(N) = (2^P - 1) (2^P) = 2 N`

which gives the condition for a perfect number.

At this date there are only 17 known perfect numbers. They correspond to

### `P = 2,3,5,7,13,17,19,31,61,89,107,127,521,607,1279,2203,2281.`

The corresponding perfect numbers are

### ```N2 = 2 (2^2 - 1) = 6 N3 = 2^2 (2^3 - 1) = 28 N5 = 2^4 (2^5 - 1) = 496 N7 = 2^6 (2^7 - 1) = 8128 N13 = 2^12 (2^13 - 1) = 33550336 . . . ```

All of these perfect numbers are even. Leonhard Euler (1707-1783) succeeded in proving the following theorem, which is the most general result known for perfect numbers:

Theorem 5. Every even perfect number is of the type (15).

To prove this theorem we write the general even perfect number N in the form

### `N = (2^(P-1)) Q (16)`

where Q is some odd number. Since the two factors in (16) are relatively prime, we can use (8) and obtain

### `sig(N) = sig(2^(P-1)) sig(Q) = (2^P - 1) sig(Q)`

joining this with the condition (14) for perfect numbers

### ` sig(N) = (2^P - 1) sig(Q) = 2 N = (2^P) Q`

In this relation let us use the sum sig0(Q) as defined in (12) instead of sig(Q). We obtain

or

or finally

### `Q = (2^P - 1) sig0(Q) (17)`

This condition permits us to draw some strong conclusions. It implies first of all that sig0(Q) is a proper divisor of Q. On the other hand sig0 is the sum of all proper divisors, including itself, so that there can be no other proper divisor besides sig0(Q). But a number Q with a single proper divisor sig0(Q) must be prime and sig0(Q) = 1. From (17) we conclude finally that

### `Q = 2^P - 1`

and Q is prime. Thus the even perfect number (16) is of the form (15) with Q prime.

The major unsolved problem of perfect numbers is whether or not an odd perfect number exists.

END OF TALK. ----------------------