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.

  • $\begingroup$ Wikipedia lists many necessary conditions for an odd perfect number. Maybe, this helps to rule out numbers of the desired form. $\endgroup$
    – Peter
    Jan 25, 2022 at 11:34

2 Answers 2


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.


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.

  • $\begingroup$ Care to explain the quick downvote? $\endgroup$ Jan 25, 2022 at 11:38
  • 1
    $\begingroup$ Surely the second fact ($E-1$ and $E+1$ not perfect) is implied by the first ($O-1$ and $O+1$ not perfect)? $\endgroup$
    – TonyK
    Jan 25, 2022 at 11:44
  • $\begingroup$ No, @TonyK, the implication does not follow. I presented the proof in that way so as to ensure that an even perfect number and an odd perfect number cannot differ by $1$. $\endgroup$ Jan 25, 2022 at 11:48
  • $\begingroup$ @TonyK: You may refer to the papers by Luca, and Holdener, et. al for more details. The part of my thesis that I quoted in this answer is basically an exposition of their results on this topic. $\endgroup$ Jan 25, 2022 at 11:50
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    $\begingroup$ I don't need to refer to any papers to see that the first of those facts implies the second. $\endgroup$
    – TonyK
    Jan 25, 2022 at 12:12

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