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(Note: This question has been cross-posted to MO.)

Is it true that every odd perfect number $N = q^k n^2$ can be written in the form $$N = \frac{r\sigma(r)}{2r - \sigma(r)}?$$

(Here, $q$ is a prime satisfying $q \equiv k \equiv 1 \pmod 4$ and $\gcd(q,n)=1$.)

In particular, setting $$r = ({q^{(k-1)/2}}n)^2$$ appears to work.

The proof is contained in this paper.

MY ATTEMPT

Setting $$r = ({q^{(k-1)/2}}n)^2$$ in $$\frac{r\sigma(r)}{2r-\sigma(r)}=\frac{q^{k-1}n^2\sigma(q^{k-1}n^2)}{2q^{k-1}n^2 - \sigma(q^{k-1}n^2)},$$ we shall show that this last quantity is equal to $q^k n^2 = N$.

We assume (?) that equality indeed holds and work our way backwards. (That is, we show that the steps are reversible.) $$\frac{q^{k-1}n^2\sigma(q^{k-1}n^2)}{2q^{k-1}n^2 - \sigma(q^{k-1}n^2)}={q^k}{n^2}.$$ Cancelling ${q^{k-1}}{n^2}$ from both sides of the last equation, we get $$\frac{\sigma(q^{k-1}n^2)}{2q^{k-1}n^2 - \sigma(q^{k-1}n^2)}=q.$$ $$\sigma(q^{k-1}n^2) = 2q^k n^2 - q\sigma(q^{k-1}n^2)$$ $$(q+1)\sigma(q^{k-1})\sigma(n^2) = 2q^k n^2.$$

Alas, here is where I get stuck. I do not know how to rewrite $$(q+1)\sigma(q^{k-1})$$ as $$\sigma(q^k).$$

quid
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Jose Arnaldo Bebita Dris
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  • Hmmm, those last two quantities are not equal, so either the proposed $r$ is inaccurate or there is another calculation error elsewhere (possibly even in the paper). To see this, note that $\sigma(q^k)=1+q\sigma(q^{k-1})$ from the simple expression $\sigma(q^k) = 1 + q + q^2 + \cdots + q^k$. – Erick Wong Jan 02 '18 at 18:40
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    @ErickWong, indeed those last two quantities are only equal when $k=1$. – Jose Arnaldo Bebita Dris Jan 03 '18 at 07:37

1 Answers1

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As pointed out by Pace Nielsen via MO, the value of $r$ should be $$r = q^{(k-1)/2} n^2$$ as considered in the paper.

Setting $$r = q^{(k-1)/2} n^2$$ in $$\frac{r\sigma(r)}{2r-\sigma(r)}=\frac{q^{(k-1)/2}n^2\sigma(q^{(k-1)/2}n^2)}{2q^{(k-1)/2}n^2 - \sigma(q^{(k-1)/2}n^2)},$$ we shall show that this last quantity is equal to $q^k n^2 = N$.

We assume (?) that equality indeed holds and work our way backwards. (That is, we show that the steps are reversible.) $$\frac{q^{(k-1)/2}n^2\sigma(q^{(k-1)/2}n^2)}{2q^{(k-1)/2}n^2 - \sigma(q^{(k-1)/2}n^2)}={q^k}{n^2}.$$ Cancelling ${q^{(k-1)/2}}{n^2}$ from both sides of the last equation, we get $$\frac{\sigma(q^{(k-1)/2}n^2)}{2q^{(k-1)/2}n^2 - \sigma(q^{(k-1)/2}n^2)}=q^{(k+1)/2}.$$ $$\sigma(q^{(k-1)/2}n^2) = 2q^k n^2 - q^{(k+1)/2}\sigma(q^{(k-1)/2}n^2)$$ $$(q^{(k+1)/2}+1)\sigma(q^{(k-1)/2})\sigma(n^2) = 2q^k n^2.$$

Alas, here is where I get stuck. I do not know how to rewrite $$(q^{(k+1)/2}+1)\sigma(q^{(k-1)/2})$$ as $$\sigma(q^k).$$

Update (March 1, 2018)

But, of course! $$\sigma(q^k)=\frac{q^{k+1}-1}{q-1}=\frac{\bigg(q^{(k+1)/2}+1\bigg)\bigg(q^{(k+1)/2}-1\bigg)}{q-1}=\bigg(q^{(k+1)/2}+1\bigg)\sigma(q^{(k-1)/2}),$$ where we have used the fact that $k+1 \equiv 2 \pmod 4$ (i.e., $k+1$ is even).

Jose Arnaldo Bebita Dris
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