Modulo Perfect

This problem is actually still an open question.

There haven't been a proof that there is no odd numbers .

"..a prolonged meditation on the subject has satisfied me that the existence of any one such [odd perfect number] — its escape, so to say, from the complex web of conditions which hem it in on all sides — would be little short of a miracle."

Should such a number exist, $x = 1$ is another solution.

And of course, proving that it gives a remainder of $4$ if there's no perfect odd number. All odd perfect numbers greater than $6$ is in the form of $2^{p-1} * (2^{p} - 1)$ for prime $p$

When consider modulo $6$ implies we consider modulo $2$ and $3$

Obviously modulo $2$ gives $0$ . And with $p$ as odd, modulo $3$ gives $(-1)^{\text{even}} \times ( (-1)^{\text{odd}} - 1 ) \equiv 1$ . Which gives $4$ modulo $6$ .