# On the golden ratio and odd perfect numbers

Here is my question:

Is $$I(n^2) - 1 > 1/I(n^2)$$ true when $$I(n^2)=\sigma(n^2)/n^2$$ is the abundancy index of $$n^2$$ and $$q^k n^2$$ is an odd perfect number with special prime $$q$$ satisfying $$k>1$$?

My Attempt

If $$k>1$$, then since $$q$$ is the special prime, then $$q$$ satisfies $$q \equiv k \equiv 1 \pmod 4$$ and $$\gcd(q,n)=1$$. In particular, we know that $$q \geq 5$$ and $$k \geq 5$$.

We know that $$I(q^k) = \frac{\sigma(q^k)}{q^k} = \frac{q^{k+1} - 1}{q^k (q - 1)} < \frac{q^{k+1}}{q^k (q - 1)} = \frac{q}{q - 1} \leq \frac{5}{4}.$$

It follows that $$I(n^2) = \frac{2}{I(q^k)} > \frac{2(q - 1)}{q} \geq \frac{8}{5}.$$

Thus, $$I(n^2) - 1 > \frac{2(q - 1)}{q} - 1 = \frac{(2q - 2) - q}{q} = \frac{q - 2}{q} > \frac{q}{2(q - 1)} > \frac{1}{I(n^2)}$$ where the inequality $$\frac{q - 2}{q} > \frac{q}{2(q - 1)}$$ holds provided $$q > 3+\sqrt{5} \approx 5.23607$$.

However, the resulting inequality for $$I(n^2)$$ from $$I(n^2) - 1 > \frac{1}{I(n^2)}$$ together with the following upper bound for $$I(n^2)$$ (which holds when $$k>1$$) $$\frac{2q}{q+1} > I(n^2)$$ only yields $$\frac{2q}{q+1} > I(n^2) > \frac{\sqrt{5}+1}{2}$$ thereby giving $$q > \frac{1+\sqrt{5}}{3-\sqrt{5}} = 2+\sqrt{5} \approx 4.23607.$$

• Note that there is no discrepancy when $k=1$, as then we have $$I(n^2) - 1 \geq \frac{2}{3} > \frac{3}{5} \geq \frac{1}{I(n^2)}$$ yielding the lower bound $q > 2 + \sqrt{5} \approx 4.23607$ from $$\frac{2q}{q+1}=I(n^2) > \frac{\sqrt{5}+1}{2}.$$ – Jose Arnaldo Bebita-Dris Feb 23 at 10:18
• When $k>1$, we get $$I(n^2) - 1 > \frac{3}{5} \not\gt \frac{5}{8} > \frac{1}{I(n^2)}.$$ – Jose Arnaldo Bebita-Dris Feb 23 at 10:19
• Consider the case when $k > 1$. If $q > 5$, then since $q$ is a prime satisfying $q \equiv 1 \pmod 4$, it follows that $q \geq 13$, from which we obtain $$I(q^k) < \frac{q}{q - 1} \leq \frac{13}{12},$$ so that we get $$I(n^2) = \frac{2}{I(q^k)} > \frac{24}{13} = 1.\overline{846153},$$ which vastly improves on $$I(n^2) > \frac{\sqrt{5}+1}{2} \approx 1.61803.$$ – Jose Arnaldo Bebita-Dris Feb 23 at 10:57

Here is a partial answer for what happens when $$k=1$$.

We initially have $$I(n^2) - 1 = \frac{2q}{q+1} - 1 = \frac{q-1}{q+1} > \frac{q+1}{2q} = \frac{1}{I(n^2)},$$ where the inequality $$\frac{q-1}{q+1} > \frac{q+1}{2q}$$ holds when $$q > 2+\sqrt{5} \approx 4.23607.$$

In particular, the inequality $$I(n^2) - 1 > \frac{1}{I(n^2)}$$ holds when $$q \geq 5$$.

Unfortunately, the inequality $$I(n^2) - 1 > \frac{1}{I(n^2)}$$ implies that $$I(n^2) > \frac{\sqrt{5}+1}{2} \approx 1.61803$$, whereas $$k=1$$ implies that $$I(q^k)=I(q)=\frac{\sigma(q)}{q}=\frac{q+1}{q}=1+\frac{1}{q} \leq 1+\frac{1}{5}=\frac{6}{5}$$ (since $$q \geq 5$$), from which we obtain $$I(n^2)=\frac{2}{I(q^k)}=\frac{2}{I(q)}\geq \frac{5}{3} = 1.\overline{666}.$$

Here is a full answer to the original question.

Let $$q^k n^2$$ be an odd perfect number with special prime $$q$$. Furthermore, let $$I(x)=\sigma(x)/x$$ be the abundancy index of the positive integer $$x$$. (Note that $$\sigma(x)$$ is the sum of divisors of $$x$$.)

Since the inequality $$I(n^2) - 1 > \frac{1}{I(n^2)}$$ holds when $$q>5$$ or when $$k=1$$, it suffices to show that the inequality does not hold when $$q=5$$ and $$k>1$$.

First, note that the inequality $$I(n^2) - 1 > \frac{1}{I(n^2)}$$ implies that $$I(n^2) > \frac{\sqrt{5}+1}{2} \approx 1.61803.$$

Next, from the following answer to a related MSE question, we have:

Cohen and Sorli ruled out $$5^5$$ as a possible Eulerian component $$q^k$$ for an odd perfect number in page 4 of their paper titled On Odd Perfect Numbers and Even 3-Perfect Numbers.

Thus, under the assumption $$q=5$$ and $$k>1$$, we have that $$k \geq 9$$ (since $$k \equiv 1 \pmod 4$$), whereupon we obtain $$I(n^2) = \frac{2}{I(q^k)} \leq \frac{2}{I(5^9)} \leq \frac{1953125}{1220703} \approx 1.60000016384,$$ resulting in a contradiction.

Hence, the inequality $$I(n^2) - 1 > \frac{1}{I(n^2)}$$ does not hold when $$q=5$$ and $$k>1$$.

QED