I am wondering if there is any perfect number that is of the form $\,n^3+1$ (i.e. one more than a perfect cube).
Since perfect number minus one (for example, $495$ and $8127$) $\equiv$ $0\pmod{9}$ and $7\pmod{8}$, are there any perfect numbers that is form of $\,n^3+1$ ? For example, $495$ isn't a perfect cube and $8127$ is not also a perfect cube.
There are certainly no even perfect numbers other than $\ 28\ $ that are of the form $\ n^3+1\ $. Every even perfect number must be of the form $\ 2^{p-1}\big(2^p-1\big)\ $ with $\ p\ge2\ $. For $\ p=2\ $, $\ 2^{p-1}\big(2^p-1\big)=6\ $ is not of the form $\ n^3+1\ $, while $\ 28=2^2\big(2^3-1\big)=3^3+1\ $ is of that form. If $\ p\ge4\ $, and $\ 2^{p-1}\big(2^p-1\big)\ $ were of that form we would have \begin{align} 2^{p-1}(2^p-1)&=n^3+1\\ &=(n+1)(n^2-n+1)\ , \end{align} and since $\ n\ $, and hence $\ n^2-n+1\ $ must then be odd, it would have to be the case that $\ 2^{p-1}\,\big|\,n+1\ $—that is $\ n=2^{p-1}k-1\ $ for some positive integer $\ k\ $. Thus, we would then have \begin{align} 2^p-1&=k\big(n^2-n+1\big)\\ &=k\big(\big(2^{p-1}k-1\big)^2-2^{p-1}k+1+1\big)\\ &=k\big(k^22^{2p-2}-2^pk+3-2^{p-1}k\big)\\ &=k\big(k2^{p-1}\big(k2^{p-1}-3\big)+3\big)\\ &>2^{p-1}\big(2^3-3)\\ &>2^p\ , \end{align} which is a contradiction. Therefore, $\ 2^{p-1}\big(2^p-1\big)\ $ is not of the form $\ n^3+1\ $ for any $\ p\ge4\ $,and hence $\ 28\ $ is the only even perfect number of that form.