Are all nonunitary perfect numbers in the form 4k where k is an even perfect number

Question

It appears that all numbers $n$ such that the sum of the nonunitary divisors of $n$ equals itself are in the form $4k$ where $k$ is an even perfect number. For example, the smallest nonunitary perfect number is $24$, because the sum of the nonunitary divisors of 24 (divisors $d$ such that $d$ and $n/d$ share a common divisor) is $2 + 4 + 6 + 12 = 24$, which is $4$ times the first ordinary perfect number. My question is are there any counterexamples, or is there a way to prove that all nonunitary perfect numbers are in this form?

Answer 1

Not a complete answer, but here is a proof that every one of them of the form $2^k p^a$ where $p$ is an odd prime is of the form in question. Note that all even perfect numbers are of the form $2^{q-1}(2^q -1)$ where $2^q-1$ is prime. So our goal is to show that if $n=2^k p^a$, and $n$ is nonunitary perfect, then $a=1$ and $p=2^{k-1}-1$.

Following Ligh and Wall - Functions of non-unitary divisors(pdf), we will write $\sigma^{\#}(n)$ to be the sum of the the nonunitary divisors of $n$, and use from their paper the following formula:

$$\newcommand\div{\mid}\newcommand\ddiv{\parallel}\sigma^{\#}(n) = \left(\prod_{p\ddiv n} (p+1)\right) \left(\prod_{p^e\ddiv n, e>1} \frac{p^e -1}{p-1} - \prod_{p^e\ddiv n, e>1} \left(p^e +1\right)\right).\tag{1}\label{463487_1}$$

First, let's show that if $n=2^k p^a$ and $n$ is nonunitary perfect, and $a \geq 2$, then $a = 2$.

Assume for now that $a \geq 2$. Then using Equation \eqref{463487_1}, we have:

$$(2^{k+1}-1)\left(\frac{p^a -1}{p-1}\right) - (p^a +1)(2^k+1) = p^a 2^k,$$ which is equivalent to

$$2^k = \frac{2p^{a+1} +p -p^a -2}{2p^a - p -1}. $$

Thus, $$2p^a - p -1\div 2p^{a+1} +p -p^a -2.\tag{2}\label{463487_2}$$

Now, Equation \eqref{463487_2} implies that $$2p^a - p -1\div 2p^{a+1} +p -p^a -2 - p(2p^a - p -1) = p^2 - p^a +2p +1.\tag{3}\label{463487_3}$$

Now, let us consider the sign of $p^2 - p^a +2p +1$. If this quantity is zero or positive, then we have $p^a \leq p^2 +2p +1$ which implies that $a=2$. So we may assume that $p^2 - p^a +2p +1 <0$. Thus, Equation \eqref{463487_3} implies that we have

$$2p^a - p -1\div p^a -p^2 -2p -1\tag{4}\label{463487_4}$$ and the right-hand side of \eqref{463487_4} is positive. Thus, $$2p^a -p -1 \leq p^a -p^2 -2p -1,$$

which implies that $p^a \leq -p^2 +p$, which is nonsense. So one possibility led to $a=2$ and the other led to a contradiction so we must have $a=2$.

We will now show that $a=2$ leads to a contradiction. Equation \eqref{463487_1} for $a=2$ becomes

$$\sigma^{\#}(n) = \left(p^2 +p+1\right)\left(2^{k+1}-1\right) -\left(p^2 +1\right)\left(2^k+1\right).\tag{5}\label{463487_5}$$

Equation \eqref{463487_5} is equivalent to a quadratic in $p$ with discriminant $9(2^{k+2}-5)$. So the discriminant is 3 (mod 4), and thus cannot be a perfect square. So there are no integer solutions to Equation \eqref{463487_5}.

Thus, the only situation we care about is when $a=1$.

In this situation, Equation \eqref{463487_1} yields

$$2(p+1)(2^{k+1}-1) = 2^kp,$$ which is just a linear equation in $p$, and solving for $p$ gives $p=2^{k-1}-1$ and so we are done.