Let $N = p^k m^2$ be an odd perfect number with special prime $p$ satisfying $p \equiv k \equiv 1 \pmod 4$ and $\gcd(p,m)=1$.
It is known that
$$D(p^k)D(m^2)=2s(p^k)s(m^2) \tag{0}$$
where $D(x)=2x-\sigma(x)$ is the deficiency of the positive integer $x$, $s(x)=\sigma(x)-x$ is the aliquot sum of $x$, and $\sigma(x)=\sigma_1(x)$ is the classical sum of divisors of $x$. Denote the abundancy index of $x$ by $I(x)=\sigma(x)/x$.
Since $p \equiv k \equiv 1 \pmod 4$ and $p$ is prime, we obtain the lower bound $p^k \geq 5$, whereupon we have the lower bound
$$\sigma(p^k) \geq p^k + 1.$$
This means that we have the upper bound
$$D(p^k) = 2p^k – \sigma(p^k) \leq p^k – 1.$$
Noting that $s(p^k) = (p^k – 1)/(p – 1)$, this means that we obtain
$$\frac{D(p^k)}{p – 1} \leq s(p^k),$$
or equivalently,
$$\frac{D(p^k)}{s(p^k)} \leq p – 1. \tag{1}$$
Note that equality cannot occur in Equation $(1)$, because $\gcd(D(p^k),s(p^k))=1$.
Hence, we have the inequality
$$\frac{D(p^k)}{s(p^k)} < p – 1. \tag{2}$$
But note that we have (from Equation $(0)$)
$$\frac{D(p^k)}{s(p^k)} = \frac{2s(m^2)}{D(m^2)}. \tag{3}$$
From Inequality $(2)$ and Equation $(3)$, we get
$$\frac{2s(m^2)}{D(m^2)} < p – 1$$
$$\frac{I(m^2) – 1}{2 – I(m^2)} < \frac{p – 1}{2}. \tag{4}$$
Now suppose to the contrary that $k = 1$. Since $N$ is perfect, and because the abundancy index function is multiplicative, we derive
$$I(p^k)I(m^2) = I(p^k m^2) = I(N) = 2$$
$$I(m^2) = \frac{2}{I(p^k)}$$
By assumption, $k = 1$ so that
$$I(m^2) = \frac{2}{I(p)} = \frac{2p}{p + 1}. \tag{5}$$
Substituting $(5)$ in $(4)$, we obtain
$$\frac{p – 1}{2} = \dfrac{\dfrac{2p}{p + 1} – 1}{2 – \dfrac{2p}{p + 1}} < \frac{p – 1}{2},$$
which is a contradiction.
We therefore conclude that $k \neq 1$.
Here is my:
QUESTION: Does the proof hold water (particularly in going from $(1)$ to $(2)$)? I have some doubts since $D(p^k) = p – 1$ and $s(p^k) = 1$ when $k = 1$, whereupon $\gcd(D(p^k),s(p^k)) = 1$ still holds.
Best Answer
It is true that $\gcd(D(p^k),s(p^k))=1$ holds unconditionally. However, it is wrong to say that equality cannot occur in $(1)$. In fact, equality occurs in $(1)$ whenever $k=1$. If we assume that $k=1$ then $(1)$ becomes $p-1=p-1$ and equality is achieved. This is because $s(p^k)=1$ if $k=1$ and $D(p^k)=p-1$ if $k=1$.