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Let $\sigma(x)$ denote the sum of divisors of the positive integer $x$. A number $r$ is called deficient-perfect if $(2r - \sigma(r)) \mid r$.

Here is my question:

Does $\gcd(r, \sigma(r)) = 2r - \sigma(r)$ always hold when $r$ is deficient-perfect?

MY ATTEMPT

Since $r$ is deficient-perfect, then $(2r - \sigma(r)) \mid r$. Also, by this answer, then we have $(2r - \sigma(r)) \mid \sigma(r)$. So, $2r - \sigma(r)$ is a common divisor between $r$ and $\sigma(r)$. How do I show that it is always the greatest common divisor? This is where I get stuck.

Jose Arnaldo Bebita Dris
  • 8,471
  • Any common divisor of $r,s$ that is a linear combination $, m r + n s,$ is necessarily the greatest common divisor - see the linked dupe. – Bill Dubuque Aug 11 '19 at 13:15

1 Answers1

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Does this work?

Suppose that $d$ is a common divisor between $r$ and $\sigma(r)$. Then $d \mid r$ and $d \mid \sigma(r)$, from which it follows that $$d \mid (2r - \sigma(r)).$$

Thus $2r - \sigma(r)$ is the greatest common divisor between $r$ and $\sigma(r)$.

That is, $$\gcd(r,\sigma(r))=2r-\sigma(r).$$

QED (?)

Jose Arnaldo Bebita Dris
  • 8,471