A number $n \in \mathbb{N}$ is said to be [*superperfect*](http://mathworld.wolfram.com/SuperperfectNumber.html) if
$$\sigma(\sigma(n)) = 2n.$$


A number $m \in \mathbb{N}$ is said to be [*almost perfect*](http://mathworld.wolfram.com/AlmostPerfectNumber.html) if $$\sigma(m) = 2m - 1.$$


Here is my question:



> 
> Is it possible to have an even superperfect number and an odd superperfect number whose product is an almost perfect number?
> 
> 
> 


By the results in a paper (by Antalan and Tagle) mentioned in this [preprint](http://arxiv.org/pdf/1602.04248v3.pdf), an even almost perfect number which is not a power of two must take the form $M = {2^r}{b^2}$ where $r \geq 1$ and $b$ is an odd composite. Consequently, by the same preprint, we know that, if there exists an even almost perfect number which is not a power of two, then it must be a product of an even [solitary number](http://mathworld.wolfram.com/SolitaryNumber.html) and an odd solitary number. I wonder if a similar scenario holds for the product of two superperfect numbers of different parity.



Is it possible to have an even superperfect number and an odd superperfect number whose product is an almost perfect number?
Guy's Unsolved Problems in Number Theory states that it is a result of Suryanarayana and Kanold that the even superperfect numbers are precisely the numbers $2^{n-1}$ where $2^n-1$ is a Mersenne prime. So let $r=n-1$. Since even almost perfect numbers which have an odd factor greater than one have the form $2^rb^2$ (with the stated conditions on $b$) by the result of Antalan and Tagle you mention, then suppose $2^r$ is the even superperfect number (so that $2^{r+1}-1$ is the Mersenne prime) and $b^2$ is the odd superperfect number. Then since as mentioned in the proof of Lemma 2.3 of [1],
$$
\sigma(b^2)=\frac{2^{r+1}b^2-1}{2^{r+1}-1},
$$
and
$$
\sigma\left(\frac{2^{r+1}b^2-1}{2^{r+1}-1}\right)=2b^2
$$
by the assumed superperfect property of $b^2$, then $\frac{2^{r+1}b^2-1}{2^{r+1}-1}$ must be of the form $q^km^2$ where $q\equiv k\equiv 1 \bmod 4$ and $gcd(q,m)=1$. Since $2^{r+1}-1$ is a Mersenne prime, either $r=1$ or $r$ is even. If $r$ is even, then any prime $p$ dividing $\frac{2^{r+1}b^2-1}{2^{r+1}-1}$ satisfies 
$$
\left(\frac{2}{p}\right) = 1
$$
as $2(2^{r/2}b)^2\equiv 1 \bmod p$, hence $p\equiv \pm 1 \bmod 8$. So when $r$ is even, the $q$ previously mentioned is $1\bmod 8$.


Letting $I(n)=\sigma(n)/n$, we also have
$$
I(\sigma(b^2)) = \frac{2^{r+1}-1}{2^r}+\frac{1}{2^r\sigma(b^2)}.
$$


In general, if one would ask the more general question of whether $n$ could both be a product of an even superperfect number and an odd superperfect number, and also a [generalized quasi-perfect number](https://mathoverflow.net/questions/141445/generalized-quasi-perfect-numbers), that is, for fixed nonzero integers $a\_1, a\_2$, $n$ satisfies $\sigma(n)=a\_1n+a\_2$, suppose $n=2^rb^2$ where $b^2$ is odd. Then for the case $r>1$, if $2^{r+1}-1$ is a Mersenne prime, a prime $p$ dividing $a\_12^rb^2+a\_2$ satisfies
$$
\left(\frac{-a\_2a\_1^{-1}}{p}\right)=1
$$
and this would be satisfied by $p$ in certain residue classes mod some modulus $D(a\_1,a\_2)$, and there are no solutions to the above general question for those $r$ mod the period of the powers of $2$ mod $D(a\_1,a\_2)$, which do not take the Mersenne prime $2^{r+1}-1$ to one of those residue classes.


[1] Antalan, Dris. "Some New Results on Even Almost Perfect Numbers Which Are Not Powers of Two", arXiv 1602.04248.


