Show that, there do not exist two consecutive perfect numbers. Show that, there do not exist two consecutive perfect numbers
I know that it is unknown whether there exist any odd perfect number(s) or not. 
Also, I know that all the even perfect numbers are determined by Euler's theorem. 
Which states that:

If $N$ is an even perfect number, then N can be written in the form $N =
 2^{n−1} (2^n − 1)$, where $2^n − 1$ is prime

But, still, I am unable to crack the above-mentioned problem asked in some olympiad, I guess.
Any progress is appreciated. Please help with the ideas. Thanks in advance.
 A: Suppose that $E$ and $P$ are consecutive perfect numbers with $E$ even. Since $5$ and $7$ are not perfect, $E\ne 6$. Then from Euclid's formula, $E$ is $4$ (mod $12$).
First suppose that $P>E$ and then $P$ is $2$ (mod $3$). Let $x$ be any divisor of $P$ and consider $y=\frac{P}{x}$. Then $xy$ is $2$ (mod $3$) and so $x+y$ is $0$ (mod $3$). Thus the divisors of $P$ can be paired so the sum of each pair is divisible by $3$. Hence $\sigma (P)$ is divisible by $3$ and so $P$ is not perfect.
Now suppose $P<E$ and then $P$ is $3$ (mod $4$). The same argument proves that $\sigma (P)$ is divisible by $4$ and so, again, $P$ is not perfect.
A: Your question/problem is covered exactly in the following M. Sc. thesis, completed in 2008, at De La Salle University - Manila.
You may refer to pages 53 to 55.
Basically, the following facts about perfect numbers are used:

*

*A number $M$ (odd or even) satisfying $M \equiv 2 \pmod 3$ cannot be perfect.

*An odd perfect number $N$ must satisfy $N \equiv 1 \pmod 4$.

The proof proceeds by first showing that if $O$ is odd and perfect, then $O - 1$ and $O + 1$ cannot be a(n) (even) perfect number.  Next, supposing that $E$ is even and perfect, we show that $E - 1$ and $E + 1$ cannot be a(n) (odd) perfect number.
This completes the proof.
