[Theorem 6.4] the product of all n's divisors amount to n^(T(n)/2). [Proof] Suppose d is a divisor of n. Therefore, n = dd', where d' is another divisor of n. Do this for all divisors of n, and we have T(n)/2 pairs where the product of each pair is n. Q.E.D.

[Theorem 6.4] the product of all n's divisors amount to n^(T(n)/2). [Proof] Suppose d is a divisor of n. Therefore, n = dd', where d' is another divisor of n. Do this for all divisors of n, and we have T(n)/2 pairs where the product of each pair is n. Q.E.D.

[Definition 6.2] If for natural number n, S(n) = 2n (that is, the aliquot sum of n is equal to itself), n is called a perfect number. If S(n) < 2n, it is a deficient number, and if S(n) > 2n it is an abundant number.

[Theorem 6.3] T(n) and S(n) are multiplicative functions. The proof is trivial from Theorem 5.2.

[Theorem 6.5] Given that p=2^n-1 is prime, then (2^(n-1))p is a perfect number. Conversely, all even perfect numbers must be in this form. [Proof] The first half of this question can be proven with Theorem 6.2. For the second half, if m is an even perfect number, m=pa, where a is an odd number and p>=2. Apply theorem 6.3, and we have S(m) = S(a)*2^(n-1). Since m is perfect, we have S(m) = 2m, therefore S(a)*2^(n-1)=pa. Divide both sides by 2p-1 and we have S(a) = a + a/p. Since S(a) is an integer, p must be a divisor of S(a). Since a is also a divisor of a, a + a/p must be the sum of all two of a's divisors. Therefore, a is prime and a = p, Q.E.D.

[Theorem 6.5] Given that p=2^n-1 is prime, then (2^(n-1))p is a perfect number. Conversely, all even perfect numbers must be in this form. [Proof] The first half of this question can be proven with Theorem 6.2. For the second half, if m is an even perfect number, m=pa, where a is an odd number and p>=2. Apply theorem 6.3, and we have S(m) = S(a)*2^(n-1). Since m is perfect, we have S(m) = 2m, therefore S(a)*2^(n-1)=pa. Divide both sides by 2p-1 and we have S(a) = a + a/p. Since S(a) is an integer, p must be a divisor of S(a). Since a is also a divisor of a, a + a/p must be the sum of all two of a's divisors. Therefore, a is prime and a = p, Q.E.D.

The perfect numbers, starting from the lowest, are 6, 28, 496, 8128, etc. We still haven't found an odd perfect number yet.

496 is really close to 495, so is 495 special too?

Actually, it is. I'll introduce why on the next page.