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 $N =
2^{nā1} (2^n ā 1)$, where $2^n ā 1$ 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.
Best Answer
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.