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In an answer to an earlier question, it is shown that $$D(2^p - 1)D(2^{p-1}) = 2s(2^p - 1)s(2^{p-1}) = 2^p - 2,$$ if $2^{p-1}(2^p - 1)$ is an even perfect number, $D(x) = 2x - \sigma(x)$ is the deficiency of $x$, $s(x) = \sigma(x) - x$ is the sum of the aliquot divisors of $x$, and $\sigma(x)$ is the sum of divisors of $x \in \mathbb{N}$, the set of positive integers. (Lastly, denote the abundancy index of $x$ by $I(x) = \sigma(x)/x$.)

Here is my question in this post:

What is the common (and simplified) value for $D(q^k)D(n^2) = 2s(q^k)s(n^2)$ when $q^k n^2$ is an odd perfect number?

MOTIVATION

Since $\gcd(q^k, \sigma(q^k)) = 1$, we have $$\frac{\sigma(n^2)}{q^k} = \frac{2n^2}{\sigma(q^k)} = \gcd(n^2, \sigma(n^2))$$ $$\frac{D(n^2)}{s(q^k)} = \frac{2s(n^2)}{D(q^k)} = \gcd(n^2, \sigma(n^2))$$ $$D(q^k)D(n^2) = 2s(q^k)s(n^2) = D(q^k)s(q^k)\gcd(n^2, \sigma(n^2)).$$ Then, we have from this earlier question the following expression for $\gcd(n^2, \sigma(n^2)$ in terms of $q$, $k$, $n^2$, and $\sigma(n^2)$: $$\gcd(n^2,\sigma(n^2)) = \frac{I(q^k)}{s(q^k)}\sigma(n^2) - \frac{1}{I(q^k)s(q^k)}(2n^2).$$

Plugging in this value for $\gcd(n^2, \sigma(n^2))$ into our equation for $$D(q^k)D(n^2) = 2s(q^k)s(n^2) = D(q^k)s(q^k)\gcd(n^2, \sigma(n^2)),$$ we obtain $$D(q^k)D(n^2) = 2s(q^k)s(n^2) = D(q^k)s(q^k)\Bigg(\frac{I(q^k)}{s(q^k)}\sigma(n^2) - \frac{1}{I(q^k)s(q^k)}(2n^2)\Bigg) = D(q^k)\Bigg(I(q^k)\sigma(n^2) - \frac{1}{I(q^k)}(2n^2)\Bigg).$$

Note that "if" $I(q^k) = -1$ then the equation would be trivial and the common value would just be $D(q^k)D(n^2)$. But, of course, we know that $$1 < I(q^k) < \frac{5}{4}.$$

Alas, this is where I get stuck.

Jose Arnaldo Bebita Dris
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  • I do realize that the equation $I(q^k)I(n^2) = 2$ gives both $$I(q^k) = \frac{2}{I(n^2)} = \frac{2n^2}{\sigma(n^2)}$$ and $$\frac{2}{I(q^k)} = I(n^2) = \frac{\sigma(n^2)}{n^2}$$ so that $$I(q^k)\sigma(n^2) - \frac{1}{I(q^k)}(2n^2) = \frac{2n^2}{\sigma(n^2)}\cdot{\sigma(n^2)} - \frac{\sigma(n^2)}{n^2}\cdot{n^2} = 2n^2 - \sigma(n^2) = D(n^2),$$ as required. One of the points of this question is to ponder why, seemingly, $I(q^k) = -1$ appears to be possible. – Jose Arnaldo Bebita Dris Oct 02 '18 at 11:44

1 Answers1

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Assuming that $q$ is prime and using that $$\sigma(n^2)=\frac{2q^kn^2}{\sigma(q^k)}\qquad\text{and}\qquad \sigma(q^k)=\frac{q^{k+1}-1}{q-1}$$ we have $$\begin{align}D(q^k)D(n^2) &= 2s(q^k)s(n^2) \\\\&=2(\sigma(q^k)-q^k)(\sigma(n^2)-n^2) \\\\&=2(\sigma(q^k)-q^k)\left(\frac{2q^kn^2}{\sigma(q^k)}-n^2\right) \\\\&=2n^2\left(3q^k-\sigma(q^k)-\frac{2q^{2k}}{\sigma(q^k)}\right) \\\\&=2n^2\left(3q^k-\frac{q^{k+1}-1}{q-1}-\frac{2q^{2k}(q-1)}{q^{k+1}-1}\right) \\\\&=2n^2\cdot\frac{3q^k(q-1)(q^{k+1}-1)-(q^{k+1}-1)^2-2q^{2k}(q-1)^2}{(q-1)(q^{k+1}-1)} \\\\&=\frac{2n^2(q^k-1)(q^{k+1}-2q^k+1)}{(q-1)(q^{k+1}-1)}\end{align}$$

mathlove
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