This question is an offshoot of this earlier MSE post.

Citing Banks, et. al.: "Let us call an integer $n$ a Descartes number if $n$ is odd, and if $n = km$ for two integers $k, m > 1$ such that $\sigma(k)(m + 1) = 2n.$"

From the same paper, we have the divisibility constraints $$2k - \sigma(k) \mid k$$ and $$2k - \sigma(k) \mid \sigma(k).$$

Following the answer to this MSE post, is it possible to prove that $$\gcd(k,\sigma(k)) = 2k - \sigma(k)?$$

Here is my attempt:

$$\sigma(k)(m + 1) = 2n = 2km \Longleftrightarrow \sigma(k) = m(2k - \sigma(k))$$ $$\Longleftrightarrow 2k = \frac{(m+1)\sigma(k)}{m}=(m + 1)(2k - \sigma(k))$$

Can we now conclude that $\gcd(k, \sigma(k)) = 2k - \sigma(k)$?


For $k$ deficient, by the Euclidean algorithm, $$\gcd(\sigma(k),k)=\gcd(\sigma(k)-k,k)=\gcd(\sigma(k)-k,2k-\sigma(k)),$$ so you're asking if $2k-\sigma(k)\mid \sigma(k)-k$, but $$\dfrac{\sigma(k)-k}{2k-\sigma(k)}+1=\dfrac k{2k-\sigma(k)},$$ and that the latter is an integer is given. QED


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