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__

**Note that this still does not contradict anything, as it is previously only known that the Euler prime p**

__N.B__

*cannot***in the canonical factorization of an OPN.**

*be the smallest prime*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.__
## 1 comment:

My bad. Here is the addendum:

The implication "m < (p^k) --> k = 1" implies that the Euler prime p is the largest prime in the canonical factorization of the OPN N = (p^k)*(m^2) (assuming at least one such number exists), provided that the inequality m < (p^k) is true.

My past professors on logic would kill me :-(:

The last assertion follows from a direct application of the Modus Ponens rule. That is:

If P, then Q.

P is true.

Therefore, Q is true.

To my past professors, "sorry po". @waaaaah@

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