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6.2.12

OPN Research - Feb 06, 2012 - Part 2

Let N = {q^k}{n^2} be an odd perfect number with Euler
factor q^k.

Suppose that sigma(q) < n.

Then we have q + 1 < n, which implies that:

q + 2 <= n

Therefore:

q <= n - 2   

Consequently:

q < n

Result #1: If sigma(q) < n, then q <= n - 2.

Now suppose that n < sigma(q).

This implies that:

n < q + 1

n - 1 < q

Therefore:

n <= q

Consequently:

n < q       since gcd(q,n) = 1.

Result #2: If n < sigma(q), then n < q. 

Hence, we have the following equivalence result:

Lemma: sigma(q)/n > 1 if and only if n < q.


Note that, in case sigma(q)/n > 1, then

sigma(q^k)/n >= sigma(q)/n > 1.