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)

For any divisor Dj of N we have

N = Dj Ej (2)

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

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

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

V(N) = (A1+1)(A2+1)(A3+1) ... (Ar+1) (4)

The number

60 = 2^2 3 5

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

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

PI(Dj) = PI(Ej)

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

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

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

The product of all divisors of 60 is

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

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

Dl = Ak Bk (7)

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

sig(A) = 1 + A1 + A2 + ... + A

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

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

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

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

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

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

1, P, P^2, P^3, ..., P^A

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

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

P sig(P^A) - sig(P^A) = P^(A+1) - 1

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

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)

sig(P) = (P^2 - 1) / (P-1) = P+1

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

The sum of the divisors of

60 = 2^2 3 5

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

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)

sig0(N) = N (13)

sig(N) = 2 N (14)

sig(6) = sig(2 3) = (2+1)(3+1) = 3 4 = 12

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

Only one type of perfect number is known:

Theorem 4. A number of the form

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

This is proven by applying theorem 3.

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

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.

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 . . .

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)

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

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

(2^P - 1) [sig0(Q) + Q] = (2^P) Q

(2^P - 1) sig0(Q) + (2^P) Q - Q = (2^P) Q

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

Q = 2^P - 1

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

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

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