## 6.5.10

### OPNs Part 2 - Circa 2010

In my last blog post on odd perfect numbers (OPNs), I asked the following question:

Does the implication "m < (p^k) --> k = 1" imply that the Euler prime p is the largest prime in the canonical factorization of the OPN N?

I answer this today with a resounding "YES". Here is the proof:

[m < (p^k) --> k = 1] --> m < p (upon substituting k = 1 in p^k)

Since an OPN must contain at least 9 distinct prime factors (proof by Pace Nielsen), it follows that m must contain at least 8 distinct prime factors. Furthermore, we know that the exponent of each prime p_i in m must be at least 1. (Note that N = (p^k)*(m^2) is an OPN, by assumption.)

Thus: p > m >= (p_1)(p_2)(p_3)...(p_r) (where r = omega(N) - 1 >= 8)

Obviously:

(p_1)(p_2)(p_3)...(p_r) > p_i (for all i = 1, 2, 3, ..., r)

Consequently:

p > m >= (p_1)(p_2)(p_3)...(p_r) > p_i (for all i = 1, 2, 3, ..., r)

In other words, the Euler prime p is the largest prime in the canonical factorization of the OPN N, under the given assumption.

QED

N.B Note that this still does not contradict anything, as it is previously only known that the Euler prime p cannot be the smallest prime in the canonical factorization of an OPN.

For the proof:
Let N = (p^k)*(m^2) be an OPN. Suppose p is the smallest prime in the factorization of N.

sigma(N) = sigma(p^k)*sigma(m^2) = 2N = 2*(p^k)*(m^2)

By the factor chain approach, we know that a divisor of
sigma(p^k) (not necessarily prime) must also be a divisor of 2N.

But sigma(p^k) = 1 + p + (p^2) + (p^3) + ... + [p^(k-1)] + (p^k). (where k is congruent to 1 modulo 4)

sigma(p^k) has k + 1 addends, where k + 1 is divisible by 2 but not by 4. The important thing is that k + 1, the number of terms in sigma(p^k), is even. Thus, we may group the terms in two's, as follows:

sigma(p^k) = 1 + p + (p^2) + (p^3) + ... + [p^(k-1)] + (p^k)

= [1 + p] + [(p^2) + (p^3)] + ... + {[p^(k-1)] + (p^k)}

= (1 + p) * {1 + p^2 + ... + [p^(k-1)]}

In other words, 1 + p divides sigma(p^k).

Since p is a prime which is congruent to 1 modulo 4, p is odd. Thus, 1 + p is even. Consequently:

(1/2)*(1 + p) divides sigma(p^k).

Hence, we have shown that (1 + p) is a divisor of sigma(p^k) (and thus, also of 2N), and
(1/2)*(1 + p) is also a divisor of sigma(p^k) (and consequently, also of N).

But 1 < p. The number (1/2)*(1 + p) is the arithmetic mean of 1 and p, and so:

1 < (1/2)*(1 + p) < p

We now have a divisor of the OPN N which is smaller than the Euler prime p. This contradicts our assumption that p is the smallest prime.

End of proof.