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The following condition is equivalent to Sorli's conjecture: (a) $\sigma(n^2)/q \mid \left(2n^2 - \sigma(n^2)\right) \Longleftrightarrow \sigma(n^2)/q \mid n^2$ (b) MSE question: If $N={q^k}{n^2}$ is an odd perfect number given in Eulerian form, does $\sigma(n^2)/q$ divide $2n^2 - \sigma(n^2)$?


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One condition that I could think of is $$\frac{\sigma(x)}{\sigma(y)} < \frac{x}{y} < 1.$$ Another one is $$\left(x \mid y\right) \land \left(x \neq y\right).$$ The case of greatest interest is of course when $\gcd(x, y) = 1$.


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(a) $\sigma(N/q^k)/q^k$ is a square. (b) K. A. Broughan, D. Delbourgo, and Q. Zhou, Improving the Chen and Chen result for odd perfect numbers, Integers 13 (2013), A39


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(a) $n < q$ (b) J. A. B. Dris, The abundancy index of divisors of odd perfect numbers, JIS, Vol. 15 (2012), Article 12.4.4, Published electronically on September 3 2012 (see Lemma 15 in page 6)


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(a) $rad(N) > \sqrt{N}$ (b) P. Ochem and M. Rao, Another remark on the radical of an odd perfect number, Fibonacci Quart. 52 (2014), no. 3, 215–217



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