Let $\sigma(x)$ be the sum of divisors of the positive integer $x$. A number $l$ is called perfect if $\sigma(l)=2l$.

Let $n$ be an odd perfect number given in the so-called Eulerian form $n = p^k m^2$ where $p$ is the special/Euler prime satisfying $p \equiv k \equiv 1 \pmod 4$ and $\gcd(p,m)=1$.


Has anybody seen

Enrique Santos L's "Proof that no odd perfect number exists"?

From Eq. (6) in that paper, we have $$p^x a = \sigma(a) \frac{\sigma(p^x)}{2},$$ where it is implied that $a = m^2$ and $x = k$ (to use our notation).

Then in the section Separation in two equations in that paper, Enrique claims that $\sigma(a)$ has to be coprime to $a$, which I know to be false since $$\gcd(m^2,\sigma(m^2))=\frac{\sigma(m^2)}{p^k}=\frac{m^2}{\sigma(p^k)/2} \geq 3,$$ a result of Dris from 2012.


Can the rest of the "proof argument" be salvaged? Is it possible to mend Enrique's "proof argument" to hopefully produce some partial results on odd perfect numbers?


1 Answer 1


Eulerian form is not used in the paper "Proof that no odd perfect number exists", as it is not used the assumption that $n$ has to be odd. Then it is not correct the claim that "it is implied that $a = m^2$", which would be valid only under the odd $n$ assumption. That is why your result cannot be applied also. In fact, $σ(a)$ is co-prime to $a$ in every known perfect number.

The cited paper has an important flaw, anyway, that is the affirmation that "it should be obvious" that the following relation

$$ \begin{align} \ p^x = \prod_i {σ(q_i^{s_i + r_i}) \over q_i^{r_i}} = {\prod_i ( 1 + q_i + ... + q_i^{s_i + r_i} ) \over \prod_i q_i^{r_i}} \end{align} $$

can not be integer "unless the denominator is 1". It is not obvious at all, and it has not been proved later also.

  • $\begingroup$ Thank you for your response, @enrique. Just so we are on the same page: You are saying that in the equation $$p^x a = \sigma(a) \frac{\sigma(p^x)}{2}$$ $a$ is not $m^2$, therefore $x$ is not odd, hence $x$ is even, and consequently $$\frac{\sigma(p^x)}{2}$$ is not an integer? $\endgroup$ Commented Apr 20, 2020 at 14:21
  • $\begingroup$ I think $\gcd(a, \sigma(a)) = 1$ only works when $a = 2^{q-1}$, where $2^{q-1} (2^q - 1)$ is an even perfect number. So logically, you are correct in claiming that $\gcd(a, \sigma(a)) = 1$ in every known perfect number (as no odd perfect number has been discovered). Nonetheless, because of the Eulerian form, and the fact that the special/Euler prime is the lone prime that has an odd exponent in the prime factorization of an odd perfect number, then my stated result implies that $$\gcd(a, \sigma(a)) = \gcd(m^2, \sigma(m^2)) \geq 3.$$ Hence, my contention that... (continued) $\endgroup$ Commented Apr 20, 2020 at 14:29
  • $\begingroup$ $a$ is not coprime to $\sigma(a)$ for odd perfect numbers, and thus my rebuttal to your claim that $\gcd(a,\sigma(a))=1$ holds in all cases. $\endgroup$ Commented Apr 20, 2020 at 14:31
  • $\begingroup$ Note that your definition of $p$ and $x$ is such that $$\frac{\sigma(p^x)}{2}$$ must be an integer. In which case, $x$ must be odd. If we are considering the case of odd perfect numbers, $p$ must then be the special/Euler prime. I hope that you now see my point, @enrique. $\endgroup$ Commented Apr 20, 2020 at 14:37
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    $\begingroup$ I think I see your point now, @enrique. However, I do believe that you will never be able to prove that $\gcd(a, \sigma(a)) = 1$ unconditionally, since as I have already explained at length above, $\gcd(a, \sigma(a)) \geq 3$ for odd perfect numbers. $\endgroup$ Commented Apr 26, 2020 at 4:38

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