1
$\begingroup$

Let $\sigma(x)$ be the sum of the divisors of $x$. A number $X$ is called perfect if $\sigma(X) = 2X$. Denote the abundancy index $\sigma(X)/X$ by $I(X)$.

If $N$ is odd and perfect, then $N$ can be written in the Eulerian form $N = {q^k}{n^2}$, where $q$ is prime with $q \equiv k \equiv 1 \pmod 4$ and $\gcd(q, n) = 1$.

Here is my question:

If $N = {q^k}{n^2}$ is an odd perfect number given in Eulerian form, does $\sigma(n^2)/q$ divide $2n^2 - \sigma(n^2)$?

Context (as requested by Hagen von Eitzen)

Since $N = {q^k}{n^2}$ is perfect, we have $$2{q^k}{n^2} = 2N = \sigma(N) = \sigma(q^k)\sigma(n^2).$$

Now write $\sigma(q^k)$ as $$\sigma(q^k) = q^k + \sigma(q^{k-1})$$ so that $${q^k}\left(2n^2 - \sigma(n^2)\right) = \sigma(q^{k-1})\sigma(n^2).$$ Since $\gcd\left(q,\sigma(q^{k-1})\right) = 1$, then $q \mid \sigma(n^2)$.

Consequently, we obtain $$I(q^{k-1}) = \frac{2n^2 - \sigma(n^2)}{\sigma(n^2)/q}.$$

Thus, if $k \geq 5$, $$1 < \frac{2n^2 - \sigma(n^2)}{\sigma(n^2)/q} = I(q^{k-1}) < \frac{5}{4}$$ so that $\sigma(n^2)/q \nmid \left(2n^2 - \sigma(n^2)\right)$.

By the contrapositive, $\sigma(n^2)/q \mid \left(2n^2 - \sigma(n^2)\right)$ implies the Descartes-Frenicle-Sorli conjecture that $k = 1$.

Observe that $k = 1$ implies that $$1 = I(1) = I(q^{k-1}) = \frac{2n^2 - \sigma(n^2)}{\sigma(n^2)/q}$$ so that $\sigma(n^2)/q \mid \left(2n^2 - \sigma(n^2)\right)$.

Thus, the biconditional $$k = 1 \Longleftrightarrow \sigma(n^2)/q \mid \left(2n^2 - \sigma(n^2)\right)$$ holds, as already pointed out by Hagen von Eitzen in his answer.

$\endgroup$
  • $\begingroup$ At least I can't think of a counterexample right now ... So how come you suspect this to hold? Can you give some context? $\endgroup$ – Hagen von Eitzen Feb 7 '16 at 15:48
  • $\begingroup$ Hold on, @HagenvonEitzen. =) $\endgroup$ – Jose Arnaldo Bebita-Dris Feb 7 '16 at 15:52
1
$\begingroup$

As $\sigma(n^2)$ is a multiple of $ \frac{\sigma(n^2)}{q}$, the desired claim is equivalent to $\frac{\sigma(n^2)}{q}\mid 2n^2$. We have $2q^kn^2=\sigma(N)=(1+q+q^2+\ldots+q^k)\sigma(n^2)$, hence $$ \frac{\sigma(n^2)}{q}=\frac{2n^2q^{k-1}}{1+q+q^2+\ldots+q^k}$$ and we ask whether $$\frac{1+q+q^2+\ldots+q^k}{q^{k-1}}=q+1+q^{-1}+\ldots+q^{1-k}$$ is an integer. Quite apparently, this is the case if and only $k=1$.

$\endgroup$
  • $\begingroup$ $\sigma(n^2)/q$ is always odd, so you can just say "the desired claim is equivalent to $\frac{\sigma(n^2)}{q} \mid n^2$". The rest of your answer remains valid. $\endgroup$ – Jose Arnaldo Bebita-Dris Feb 7 '16 at 16:12

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.