Holdener and Rachfal show that, if $N = q^k n^2$ is an odd perfect number with special prime $q$ (satisfying $q \equiv k \equiv 1 \pmod 4$ and $\gcd(q,n)=1$), then $q^{(k-1)/2} n^2$ is deficient-perfect.

Descartes, in the year 1638, knew that $22021 \cdot \left(3^2 \cdot 7^2 \cdot {11^2} \cdot {13^2}\right)$ would have been an odd perfect number if $22021$ were prime. (Notice that $3^2 \cdot 7^2 \cdot {11^2} \cdot {13^2}$ is the lone odd deficient-perfect number that we know of to this day.)

Here is my question:

Did Descartes and/or Frénicle consider deficient-perfect numbers in their attempts to solve the problem of existence of odd perfect numbers?

I dare to conjecture that the answer is YES, but I am not a math historian and I currently do not know which (authoritative) sources to check.


This isn't an answer, just remarks with myself ideas about even perfect numbers, that I think similar than the statement of previous authors. I am an amateur mathematician.

My belief is that the following conjeture holds (thus that it is very difficult to find a counterexample). I don't know if this conjecture is in the literature.

Conjecture. An integer $n\geq 1$ is an even perfect number if an only if $$\operatorname{rad}(n)=\frac{1}{\frac{1}{2}-2\frac{\varphi(n)}{\sigma(n)}}.\tag{1}$$

Here $\sigma(m)$ is the sum of divisors function, $\varphi(m)$ the Euler's totient function and $\operatorname{rad}(m)$ the radical of the integer $m\geq 1$ (see the Wikipedia Radical of an integer).

Thus with this answer you can make a comparison with the statement you cite from the authors, because it is easy/obvious to prove, by cases, that any even perfect number satisfies $(1)$. Thus it is obvious that if $n$ is an even perfect number then $$\frac{1}{\frac{1}{2}-2\frac{\varphi(n)}{\sigma(n)}}=\frac{1}{\frac{1}{2}-\frac{\varphi(n)}{n}}\tag{2}$$ is integer, and it is the integer $\operatorname{rad}(n)$. We see that $(2)$ can be written as $\frac{n}{\frac{1}{2}n-\varphi(n)},$ thus we've next easy/obvious fact.

Fact. If $n$ is an even perfect number then $$\frac{n}{\frac{1}{2}n-\varphi(n)}$$ is an integer.

That is the fact that I wanted to evoke if you want to make comparisons to yourself problem. I hope that you get an answer for your nice question. Good luck.

I tried to relate even perfect numbers to the Euler's totient, my guess is that even perfect numbers maybe are closely-related to the Euler's totient function, in fact here we've the following conjecture.

Conjecture. An integer $m\geq 1$ satisfies $$2\varphi(\varphi((m+1)(2m+1)))=(m+1)\varphi(m)\tag{3}$$ if and only if $(m+1)(2m+1)$ is an even perfect number.

Again I know how to prove $\Rightarrow$, but I can not to get the full proof or find a counterexample.

  • $\begingroup$ Yes, I hope help with my answer, if you don't know it. $\endgroup$ – user686930 Jul 5 at 19:08
  • $\begingroup$ Taking off from @user686930's answer, I want to check the following conjecture: If $N = q^k n^2$ is an odd perfect number, then $$\frac{2N}{N - 2\varphi(N)}$$ is an integer. Attempt at proof: Note that $$\frac{2N}{N - 2\varphi(N)} = \frac{\sigma(N)}{N - 2\varphi(N)} = \frac{\sigma(q^k)\sigma(n^2)}{q^k n^2 - 2\varphi(q^k)\varphi(n^2)} = \frac{\sigma(q^k)\sigma(n^2)}{q^k n^2 - 2\varphi(q^k)\bigg(n\varphi(n)\bigg)}.$$ In particular, the numerator is even and the denominator is odd. Hence, the quotient is not an integer. QED $\endgroup$ – Jose Arnaldo Bebita-Dris Jul 5 at 19:38
  • $\begingroup$ If I remember well (I take notes in my notebook but these are sparse) if it easy to prove that odd perfect numbers don't satisfy $(1)$ (dividing by $2$). On the other hand there is a typo in the last paragraph concerning the conjecture related to the equation $(3)$ since if I remember well I know how to prove $\Leftarrow$, and not $\Rightarrow$, thus I know how to prove that if $(m+1)\cdot(2m+1)$ is an even perfect number then the integer $m$ satisfies $$2\varphi(\varphi((m+1)\cdot(2m+1)))=(m+1)\varphi(m).$$ I add thus this comment to clarify my words and mistake @JoseArnaldoBebita-Dris $\endgroup$ – user686930 Jul 6 at 10:03
  • $\begingroup$ Kindly take a look at this other question, @user686930. $\endgroup$ – Jose Arnaldo Bebita-Dris Jul 6 at 21:12

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