If $q^k n^2$ is an odd perfect number with Euler prime $q$, is it possible to have $n\sigma(n)=q\sigma(q^k)$?

QUESTION

If $q^k n^2$ is an odd perfect number with Euler prime $q$, is it possible to have $n\sigma(n)=q\sigma(q^k)$?

BACKGROUND

If $\sigma(N)=2N$ (where $\sigma(N)$ is the sum of the divisors of $N$), then $N$ is said to be perfect. (Denote the abundancy index of $x \in \mathbb{N}$ by $I(x)=\sigma(x)/x$.)

Euler proved that every odd perfect number has the form $N=q^k n^2$, where $q$ is prime (called the Euler prime) satisfying $q \equiv k \equiv 1 \pmod 4$ and $\gcd(q,n)=1$.

MY ATTEMPT TO ANSWER THE QUESTION

Let $N=q^k n^2$ be an odd perfect number with Euler prime $q$, and assume to the contrary that $n\sigma(n)=q\sigma(q^k)$. Since $\gcd(q,n)=1$, this implies that $$\frac{\sigma(n)}{q}=\frac{\sigma(q^k)}{n} \in \mathbb{N}.$$

Since $\sigma(q^k) \equiv k+1 \equiv 2 \pmod 4$ and $n$ is odd, then $\sigma(q^k) \neq n$. This implies that $$2 \leq \frac{\sigma(n)}{q}=\frac{\sigma(q^k)}{n}.$$

But $qn$ and $q^k n$ are both deficient. This means that $$\frac{\sigma(q)}{n}\cdot{2} \leq \frac{\sigma(q)}{n}\cdot\frac{\sigma(n)}{q} = I(qn) < 2 \implies \frac{\sigma(q)}{n} < 1 \implies q < n,$$ and $$\frac{\sigma(n)}{q^k}\cdot{2} \leq \frac{\sigma(n)}{q^k}\cdot\frac{\sigma(q^k)}{n} = I({q^k}n) < 2 \implies \frac{\sigma(n)}{q^k} < 1 \implies n < q^k.$$

Together, $q < n$ and $n < q^k$ imply that $k > 1$. Additionally, the equation $$n\sigma(n) = q\sigma(q^k)$$ is equivalent to $$\frac{\sigma(n)}{\sigma(q^k)}=\frac{q}{n},$$ which together with $q < n$, implies that $$\sigma(n) < \sigma(q^k).$$

Consequently, we have $$q < \sigma(q) < n < \sigma(n) < q^k < \sigma(q^k)$$ so that $$\frac{\sigma(q)}{n} < 1 < \frac{\sigma(n)}{q}$$ and $$\frac{\sigma(n)}{q^k} < 1 < \frac{\sigma(q^k)}{n},$$ whereupon I do not obtain any contradictions.

• Since we have the inequality $I(q^k) < I(n)$, then if $n\sigma(n)=q\sigma(q^k)$, we know that $$\frac{n}{q}=\frac{\sigma(q^k)}{\sigma(n)}<\frac{q^k}{n}.$$ This implies that $n^2 < q^{k+1}$. If one could prove the Dris conjecture that $q^k < n$, then we obtain the contradiction $$q^{2k} < n^2 < q^{k+1} \implies 2k < k + 1 \implies k < 1.$$ – Arnie Bebita-Dris Jun 23 '17 at 10:20

This isn't a answer, only the calculations that I can get (by the nature of the problem concerning odd perfect numbers, one can do deductions with the purpose to prove a more elaborated statement or well with the purpose to finish a proof by contradiction of the statement) . I hope that if there are mistakes some user can tell me.

I prefer the notation $N=q^{4\lambda+1}m^2$, where $q^{4\lambda+1}$ is the Euler's factor associated to the odd perfect number. Since the sum of divisors function $\sigma(n)$ is multiplicative and $N$ is perfect we've the factorization $$\sigma(q^{4\lambda+1})\sigma(m^2)=2q^{4\lambda+1}m^2.\tag{1}$$ On the other hand we've your condition $$m\sigma(m)=q\sigma(q^{4\lambda+1}).\tag{2}$$ Thus multiplying by $\sigma(m^2)$ your condition $(2)$ with the purpose to combine with $(1)$, we get from $$m\sigma(m)\sigma(m^2)=q\sigma(N)=2q^{4\lambda+2}m^2,$$ and thus this

Claim 1. Under previous assumptions $$\sigma(m)\sigma(m^2)=2q^{4\lambda+2}m.\tag{3}$$

Now we search a more elaborated statement combining previous claim with the fact that $$\sigma(m)\cdot\sigma(m^2)\geq (1+m)\cdot(1+m^2+(\sigma(m)-1)),$$ to get $$m^2+\sigma(m)\leq 2q^{4\lambda+2}\frac{m}{m+1},$$ and thus $m^2+\sigma(m)< 2q^{4\lambda+2}$, since $m/(m+1)<1$. From this, we can write

Claim 2. Under previous assumptions $$\sigma(m)<2q^{4\lambda+2}-m^2.$$

• user243301, how do you prove $$\sigma(m)\cdot\sigma(m^2) \geq (1+m)\cdot(1 + m^2 + (\sigma(m) - 1))?$$ I understand that $\sigma(m) > m + 1$ (since $m$ is composite) but I do not comprehend how you (seemingly) get $$\sigma(m^2) \geq 1 + m^2 + (\sigma(m) - 1).$$ – Arnie Bebita-Dris Apr 4 '17 at 14:29
• Additionally, from your Claim 1, I could prove that $\sigma(m) \equiv 2 \pmod 4$. – Arnie Bebita-Dris Apr 4 '17 at 14:44
• user243301, now that you have mentioned it, I just realized that you are right. =) – Arnie Bebita-Dris Apr 5 '17 at 0:19
• Since $\sigma(m) \equiv 2 \pmod 4$ and $m$ is odd, $m$ then takes the form $m = p^{\alpha} {\mu}^2$, where $\gcd(p, \mu)=1$ and $p \equiv \alpha \equiv 1 \pmod 4$. (See here for a proof of this fact.) – Arnie Bebita-Dris Apr 25 '17 at 17:17
• Many thanks for your reference and attention. Good luck @JoseArnaldoBebitaDris – user243301 Apr 25 '17 at 17:30

(What follows are just partial results for the present problem.)

Building from the results in the answer to this MSE question, I have $$\frac{\sigma(n)}{q}=\frac{\sigma(q^k)}{n}=x \in \mathbb{N}.$$

If the immediately preceding equation is true, then $$xn = \sigma(q^k)$$ $$\sigma(n) = xq$$ $$\sigma\left(\frac{1}{x}\sigma(q^k)\right)=xq.$$ There are no solutions to this last equation for $k>1$ and $k \equiv 1 \pmod 4$, for particular values of $x$. We have $\sigma(X) \geq X$, where the inequality is strict for $X>1$; hence, $$xq=\sigma\left(\frac{1}{x}\sigma(q^k)\right) \geq \frac{1}{x}\sigma(q^k) \geq {\frac{1}{x}}{q^k},$$ so we would need $$q^{k-1} \leq x^2.$$

But for $k>1$ and $k \equiv 1 \pmod 4$, we have $k \geq 5$, so that $$q^{k-1} \geq q^4 \geq 5^4.$$ Hence the original system of equations will only have a solution when $$x^2 \geq 5^4 \implies x \geq 25.$$ Consequently, we obtain $$25 \leq \frac{\sigma(q^k)}{n} = \frac{\sigma(n)}{q}.$$ Multiplying both sides of the inequality and equation by $\sigma(n)/{q^k}$, we get $${25}\cdot\frac{\sigma(n)}{q^k} \leq \frac{\sigma(n)}{q}\cdot\frac{\sigma(n)}{q^k} = \frac{\sigma(q^k)}{n}\cdot\frac{\sigma(n)}{q^k} = I({q^k}n) < 2.$$ Multiplying both sides of the inequality and equation by $\sigma(q)/n$, we get $${25}\cdot\frac{\sigma(q)}{n} \leq \frac{\sigma(q)}{n}\cdot\frac{\sigma(q^k)}{n} = \frac{\sigma(q)}{n}\cdot\frac{\sigma(n)}{q} = I(qn) < 2.$$

Therefore, we have $\sigma(n)/{q^k} < 2/{25}$ and $\sigma(q)/n < 2/{25}$, from which it follows that $$\frac{\sigma(n)}{q^k} < \frac{2}{25} < 25 \leq \frac{\sigma(q^k)}{n}$$ and $$\frac{\sigma(q)}{n} < \frac{2}{25} < 25 \leq \frac{\sigma(n)}{q},$$ whereupon I still do not get any contradictions.

This is only a partial answer and serves to add perspective to the problem.

Let $N=q^k n^2$ be an odd perfect number with Euler prime $q$.

Since it is not possible to have both $$\frac{\sigma(n)}{q} < 1 < \frac{\sigma(q)}{n}$$ and $$\frac{\sigma(q^k)}{n} < 1 < \frac{\sigma(n)}{q^k},$$ then we know that $$\lnot\bigg(\left(\frac{\sigma(n)}{q} < 1 < \frac{\sigma(q)}{n}\right) \land \left(\frac{\sigma(q^k)}{n} < 1 < \frac{\sigma(n)}{q^k}\right)\bigg)$$ must be true.

Therefore, it must be the case that $\bigg(\frac{\sigma(q)}{n} < 1\bigg) \lor \bigg(\frac{\sigma(n)}{q} \geq 1\bigg) \lor \bigg(\frac{\sigma(n)}{q^k} \leq 1\bigg) \lor \bigg(\frac{\sigma(q^k)}{n} \geq 1\bigg).$ (A)

Towards the end of the original post, it is shown that the problem at hand is equivalent to proving that not all disjuncts in (A) are true.