# If $N = qn^2$ is an odd perfect number with $\gcd(q,n)=1$, is it possible to have $q + 1 = \sigma(n)$?

The title says it all.

Question

If $$N = qn^2$$ is an odd perfect number with Euler prime $$q$$ and $$\gcd(q,n)=1$$, is it possible to have $$q + 1 = \sigma(n)$$?

Heuristic

From the Descartes spoof, with quasi-Euler prime $$q_1$$:

$$n_1 = 3003 < \sigma(n_1) = 5376 < q_1 = 22021$$

So it appears that it might be possible to prove that $$q + 1 \neq \sigma(n)$$.

Some Essential Estimates

Acquaah and Konyagin recently obtained the estimate $$q < n\sqrt{3}$$. We will use this estimate to obtain an upper bound for $$\sigma(q)/n$$.

Ochem and Rao recently obtained the lower bound $$N > {10}^{1500}$$ for the magnitude of an odd perfect number. Using this bound, together with the inequality $$n < q$$, gives $$I(q) < 1 + {10}^{-500}.$$

Motivation

We wish to prove the following proposition:

If $$N = {q^k}{n^2}$$ is an odd perfect number with Euler prime $$q$$, then $$3 \nmid N$$ implies that $$q < n$$.

If $$q + 1 \neq \sigma(n)$$, then it follows that $$I(q) + I(n) \neq \frac{\sigma(q)}{n} + \frac{\sigma(n)}{q}$$ from which we obtain $$I(q) + I(n) < \frac{\sigma(q)}{n} + \frac{\sigma(n)}{q},$$ since the reverse inequality $$\frac{\sigma(q)}{n} + \frac{\sigma(n)}{q} < I(q) + I(n)$$ will violate the inequality $$I(q) < \sqrt[3]{2} < I(n)$$ (see this paper). (Edit February 8 2016: Assuming $$\frac{\sigma(q)}{n} + \frac{\sigma(n)}{q} < I(q) + I(n)$$ is true, then $$\left(q < n\right) \land \left(\sigma(n) < \sigma(q)\right)$$ is false. However, I am currently unable to rule out $$\left(n < q\right) \land \left(\sigma(q) < \sigma(n)\right).$$ This particular case remains open.)

But the inequality $$I(q) + I(n) < \frac{\sigma(q)}{n} + \frac{\sigma(n)}{q}$$ implies that the biconditional $$q < n \Longleftrightarrow \sigma(q) < \sigma(n)$$ holds.

This biconditional is then a key ingredient in the proof of the proposition mentioned earlier.

My method is able to rule out $$\sigma(q) = q + 1 = \sigma(n)$$ if $$3 \nmid n$$, since we obtain $$2.799 \approx 1 + 2^{\frac{\log(6/5)}{\log(31/25)}} \leftarrow \frac{q + 1}{q} + \left(\frac{2q}{q + 1}\right)^{\frac{\log(I(5))}{\log(I(5^2))}} \leq I(q) + \left(I(n^2)\right)^{\frac{\log(I(u))}{\log(I(u^2))}}$$ $$< \frac{\sigma(q)}{q} + \frac{\sigma(n)}{n} = \frac{\sigma(q)}{n} + \frac{\sigma(n)}{q} < \sqrt{3}\left(1 + {10}^{-500}\right) + \left(1 + {10}^{-500}\right) \approx 2.732,$$ (where the smallest prime factor $$u$$ of $$N$$ satisfies $$u \geq 5$$), whence we arrive at a contradiction.

Further Considerations

If $$\sigma(q) = q + 1 = \sigma(n)$$ and $$3 \mid n$$, then the same method does not force a contradiction, because we then have $$2.7199 \approx 1 + 2^{\frac{\log(4/3)}{\log(13/9)}} \leftarrow \frac{q + 1}{q} + \left(\frac{2q}{q + 1}\right)^{\frac{\log(I(3))}{\log(I(3^2))}} \leq I(q) + \left(I(n^2)\right)^{\frac{\log(I(u))}{\log(I(u^2))}}$$ $$< \frac{\sigma(q)}{q} + \frac{\sigma(n)}{n} = \frac{\sigma(q)}{n} + \frac{\sigma(n)}{q} < \sqrt{3}\left(1 + {10}^{-500}\right) + \left(1 + {10}^{-500}\right) \approx 2.732,$$ where $$u$$ is the smallest prime factor of $$N$$.

If $$q + 1 = \sigma(n)$$, then $$\sigma(n) \equiv 2 \pmod 4$$, so that $$n$$ takes the form $$n = {p^r}{m^2}$$ where $$p$$ is a prime with $$p \equiv r \equiv 1 \pmod 4$$ and $$\gcd(p, m) = 1$$. If $$3 \mid n$$, then $$p \neq 3$$, so that $$3 \mid m$$.

So I have $$\sigma(n) = \sigma(p^r)\sigma(m^2)$$ where $$\sigma(p^r) \equiv r + 1 \equiv 2 \pmod 4$$.

Since $$\gcd(q, q+1) = 1$$, then $$\gcd(q, \sigma(n)) = 1$$, so that $$\gcd(q, \sigma(p^r)\sigma(m^2)) = 1.$$ Thus, $$q \nmid \sigma(p^r)$$ and $$q \nmid \sigma(m^2)$$.

However, note that $$q \mid \sigma(n^2) = \sigma(p^{2r})\sigma(m^4).$$

Alas here is where I get stuck.

• What is an "Euler prime"? What is a "quasi-Euler prime"? What is $I(n)$?
– Seva
Commented Feb 4, 2016 at 20:06
• @Seva, the Euler prime of an odd perfect number is the unique prime that occurs to an odd exponent (conjectured to be $1$ by Descartes, Frenicle, and most recently by Sorli). Commented Feb 4, 2016 at 20:09
• The quasi-Euler prime of the lone spoof odd perfect number that we know of (i.e., $D = 198585576189$), is the factor $q_1 = 22021$ that has odd exponent and that makes $D$ "perfect", if we pretend that $q_1 = {{19}^2}\cdot{61}$ is "prime". Commented Feb 4, 2016 at 20:13
• The abundancy index of $n$, denoted by $I(n)$, is defined as the ratio $$I(n) = \frac{\sigma(n)}{n}.$$ Commented Feb 4, 2016 at 20:14
• Arnie, have you read this: arxiv.org/abs/1602.01591 ? Commented Feb 5, 2016 at 14:33

Suppose that $N=qn^2$ is a perfect number with $q$ a prime and $\sigma(n)=q+1$. If $q \mid n$, then $\sigma(n)\geq q+1$ with equality iff $n=q$. However, if $n=q$, then $N=q^3$, so $N$ is not perfect.
Hence $q \nmid n$. This implies that $$2N=\sigma(N)=\sigma(qn^2)=\sigma(q)\sigma(n^2)=(q+1)\sigma(n^2)$$ This implies that $\frac{q+1}{2} \mid N$.
• Thank you for your answer, wythagoras. $q \nmid n$ follows from the fact that $\gcd(q, n) = 1$, while your second result $$\frac{q+1}{2} \mid N$$ could be strengthened to $$\frac{q+1}{2} \mid n$$ if $\frac{q+1}{2}$ happens to be prime. Commented Feb 6, 2016 at 7:02
• Additionally, the divisibility constraint $\frac{q + 1}{2} \mid N$ is true even when $\sigma(n) \neq q + 1$. This is because $$\sigma(q) \mid \sigma(q^k) \mid \sigma(N) = 2N.$$ Commented Feb 6, 2016 at 16:45
• @ArnieDris $\sigma(q) \nmid \sigma(q^k)$. For example $\sigma(3)=4$ while $\sigma(9)=13$. Commented Feb 6, 2016 at 20:25
• For an odd perfect number $N = {q^k}{n^2}$, we have $q \equiv k \equiv 1 \pmod 4$. Commented Feb 7, 2016 at 9:45