# If $q^k n^2$ is an odd perfect number with Euler prime $q$, can $q=17$ hold?

Note: This question is an offshoot of this earlier MSE post.

If $N$ is odd and $\sigma(N)=2N$ where $\sigma=\sigma_{1}$ is the classical sum-of-divisors function, then $N$ is said to be an odd perfect number. Euler proved that $N$ must necessarily have the form $N=q^k n^2$ where $q$ is prime (called the Euler prime) with $q \equiv k \equiv 1 \pmod 4$ and $\gcd(q,n)=1$. Descartes, Frenicle, and subsequently Sorli (page 89) conjectured that $k=1$ always holds.

Hereinafter, denote the abundancy index $I$ of $n^2$ as $$I(n^2)=\dfrac{\sigma(n^2)}{n^2}.$$

Here is my question:

If $q^k n^2$ is an odd perfect number with Euler prime $q$, can $q=17$ hold?

My Attempt

Suppose $q=17$. Note that $q \equiv 1 \pmod 4$.

Since $N$ is perfect, we have $$\sigma(q^k)\sigma(n^2)=\sigma(N)=2N=2q^k n^2.$$ Because $\gcd(q^k,\sigma(q^k))=1$, we obtain $$\sigma(q^k) \mid 2n^2$$ so that $(q+1)/2 = \sigma(q)/2 \mid \sigma(q^k)/2 \mid n^2 \mid N$. In particular, $9 = (q+1)/2 \mid N$. This implies that $\sigma(9)=\sigma(3^2)=13 \mid N$, if $3^2 \mid\mid N$. Otherwise, $$I(n^2) \geq I(3^4) = \dfrac{\sigma(81)}{81} = \dfrac{121}{81} \approx 1.49383.$$

If $3^2 \mid\mid N$, then since $q \neq 13$, we have ${13}^2 \mid n^2 \mid N$. Proceeding similarly as before, if ${13}^2 \mid\mid N$, then $\sigma(169)=\sigma({13}^2)=183={3}\cdot{61} \mid N$, violating $3^2 \mid\mid N$. Hence, we obtain $$3^2 \mid\mid N \implies I(n^2) \geq I(3^2)\cdot{I({13}^4)} = \left(\dfrac{\sigma(9)}{9}\right)\cdot\left(\dfrac{\sigma(28561)}{28561}\right) = \dfrac{30941}{19773} \approx 1.56481.$$

If we assume that the Descartes-Frenicle-Sorli conjecture is false, then $k \geq 5$ (since $k \equiv 1 \pmod 4$). This will imply that $$I(q^k)=I({17}^k) \geq I({17}^5) = \dfrac{\sigma(1419857)}{1419857} = \dfrac{1508598}{1419857} \approx 1.0625.$$

Consequently, I obtain (when $k \neq 1$) $$2=I(q^k)I(n^2) \geq {I({17}^5)}\cdot{I(3^2)}\cdot{I({13}^4)} = \dfrac{398953254}{239955833} \approx 1.6626.$$

Since $q \mid q^k$, then $(q+1)/q = I(q) \leq I(q^k)$, so that we have the bound $$I(n^2) = \frac{2}{I(q^k)} \leq \frac{2q}{q+1}.$$
In particular, when $q = 17$, we have $$I(n^2) \leq \frac{17}{9} = 1.\overline{888}.$$