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Let $N = q^k n^2$ be an odd perfect number with Euler prime $q$. (That is, $\gcd(q,n)=1$ and $q \equiv k \equiv 1 \pmod 4$.) Let $\sigma(x)$ denote the sum of the divisors of $x \in \mathbb{N}$.

Define $$D(n^2) := 2n^2 - \sigma(n^2)$$ to be the deficiency of the non-Euler part $n^2$.

CLAIM

$\gcd(n^2, D(n^2)) \neq 1$.

MY ATTEMPT

From this preprint, we have the relationships

$$\gcd(n^2, \sigma(n^2)) = \dfrac{D(n^2)}{\sigma(q^{k-1})} = \dfrac{\sigma(n^2)}{q^k}.$$

We also have the lower bound

$$\dfrac{\sigma(n^2)}{q^k} \geq 3.$$

Since $\gcd(a,b) = \gcd(a, ax+by)$ for $x, y \in \mathbb{Z}$, we have

$$\gcd(n^2, \sigma(n^2)) = \gcd(n^2, 2n^2 - \sigma(n^2)) = \gcd(n^2, D(n^2)) \geq 3.$$

Here is my question:

QUESTION

Is this proof correct?

Added February 13 2017

Note that, for the Descartes spoof $d = 198585576189 = KM$ (where $K$ is a square and $M$ is the quasi-Euler prime), then $$D(K) = D({3^2}\cdot{7^2}\cdot{{11}^2}\cdot{{13}^2}) = 819 = {3^2}\cdot{7}\cdot{13}$$ which divides $K = {3^2}\cdot{7^2}\cdot{{11}^2}\cdot{{13}^2}$. (In other words, $K$ is an odd deficient-perfect number.)

In particular, note that $$\gcd(K, D(K)) = {3^2}\cdot{7}\cdot{13} = D(K) \neq 1.$$

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  • $\begingroup$ Probably more suitable for mathoverflow.com (if I remember correctly, according to the rules of this website). $\endgroup$ – barak manos Feb 13 '17 at 11:04
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    $\begingroup$ @barakmanos, thank you for your comment. However, when I checked MO, there is no proof-verification tag. Hence, I have posted my question here instead. $\endgroup$ – Jose Arnaldo Bebita-Dris Feb 13 '17 at 11:11
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    $\begingroup$ OK, at first glance, I thought you meant to prove an unproven conjecture. $\endgroup$ – barak manos Feb 13 '17 at 11:16
  • $\begingroup$ @barakmanos I think, it is meant. $\endgroup$ – Dietrich Burde Mar 5 '17 at 19:25
  • $\begingroup$ Your proof looks correct to me except that the claim "$\gcd(a,b) = \gcd(a, ax+by)$ for $x, y \in \mathbb{Z}$" is not true in general. Take $a=2,b=3,x=2,y=2$. Your proof is correct because $\gcd(a,b)=\gcd(a,2a-b)$ is true. $\endgroup$ – mathlove Oct 11 '18 at 11:43
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Is this proof correct?

Your proof looks correct to me except that the claim "$\gcd(a,b) = \gcd(a, ax+by)$ for $x, y \in \mathbb{Z}$" is not true in general. Take $a=2,b=3,x=2,y=2$. Your proof is correct because $\gcd(a,b)=\gcd(a,2a-b)$ is true.

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