Timeline for Is the asymptotic density of positive integers $n$ satisfying $\gcd(n, \sigma(n^2))=\gcd(n^2, \sigma(n^2))$ equal to zero?
Current License: CC BY-SA 4.0
5 events
| when toggle format | what | by | license | comment | |
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| Sep 18, 2020 at 8:27 | vote | accept | Jose Arnaldo Bebita | ||
| Aug 19, 2020 at 2:03 | comment | added | Aaron Meyerowitz | OK, I fixed it. Though I hadn't used it. The idea was, roughly $1-\frac{1}{p}$ | |
| Aug 19, 2020 at 1:58 | history | edited | Aaron Meyerowitz | CC BY-SA 4.0 |
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| Aug 19, 2020 at 1:41 | comment | added | Jose Arnaldo Bebita | Thank you for your answer, @AaronMeyerowitz. Assuming $$1 - \frac{p-1}{p^2} < 1 - \frac{1}{p-1}$$ is correct, then we obtain $$\frac{1}{p-1} < \frac{p-1}{p^2}$$ which implies that $$p^2 < (p-1)^2 = p^2 - 2p + 1$$ resulting in the contradiction $$p < \frac{1}{2}.$$ Hence, I am led to conclude that there must be a typo in your upper bound for $$1 - \frac{p-1}{p^2}.$$ | |
| Aug 18, 2020 at 22:30 | history | answered | Aaron Meyerowitz | CC BY-SA 4.0 |