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A number $N$ is said to be perfect when $\sigma(N)=2N$, where $\sigma$ is the classical sum-of-divisors function.

An odd perfect number $N = {q^k}{n^2}$ is said to be given in Eulerian form if $q$ is prime with $q \equiv k \equiv 1 \pmod 4$ and $\gcd(q,n)=1$.

Here is my question:

(1) If $N={q^k}{n^2}$ is an odd perfect number given in Eulerian form, is it true that $$\sigma(q^{k-1}) \mid \left(2n^2 - \sigma(n^2)\right)?$$ (2) If the answer to question (1) is YES, is it then true that $$\gcd\left(n^2,\sigma(n^2)\right) = \frac{2n^2 - \sigma(n^2)}{\sigma(q^{k-1})}?$$

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    $\begingroup$ Since there are no odd perfect numbers (surely true, but unproved), then of course your two statements are true. So perhaps you need to ask not whether these are true, but whether these can be proved without first proving that there are no odd perfect numbers. $\endgroup$ – GEdgar Oct 10 '18 at 13:08
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    $\begingroup$ @GEdgar, or put in another way, assuming the existence of at least one odd perfect number, can it be proved that the two statements are true? This has been shown to be so by mathlove's answer below. $\endgroup$ – Jose Arnaldo Bebita-Dris Oct 10 '18 at 13:11
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True for both (1) and (2) : $$\begin{align}\frac{2n^2 - \sigma(n^2)}{\sigma(q^{k-1})}&=\frac{2n^2-\frac{2q^kn^2}{\sigma(q^k)}}{\frac{q^k-1}{q-1}}=\frac{2n^2-\frac{2q^kn^2(q-1)}{q^{k+1}-1}}{\frac{q^k-1}{q-1}} \\\\&=\frac{2n^2(q-1)(q^{k+1}-1)-2q^kn^2(q-1)^2}{(q^k-1)(q^{k+1}-1)}=\frac{2n^2(q-1)(q^k-1)}{(q^k-1)(q^{k+1}-1)} \\\\&=\frac{2n^2}{\sigma(q^k)}=\gcd(n^2,\sigma(n^2))\end{align}$$

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