Timeline for $n\varphi(n)\equiv 2\pmod{\sigma(n)}$ as a primality test
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| Jul 29, 2019 at 20:49 | comment | added | user142929 | (3/3) $ n\sigma(\operatorname{rad}(n))\equiv 2\text{ mod }\varphi(n)$ then $n$ is a prime number. I hope don't disturb feel free to study these or other congruences of your invention. | |
| Jul 29, 2019 at 20:49 | comment | added | user142929 | (2/3) and that aren't prime numbers are $1,25$ and $169$. C) It seems, as a variant of Wolstenholme's theorem a congruence that seems to have the same behaviour is $\binom{2\operatorname{rad}(n)-1}{\operatorname{rad}(n)-1}\equiv 1\text{ mod }n^3$, for integers $n\geq 5$. D) It seems that the integers satisfying (a congruence similar than Wall–Sun–Sun primes, the quotient is different) $F_{\operatorname{rad}(n)-(\frac{\operatorname{rad}(n)}{5})}\equiv 0\text{ mod }n$ are square-free. E) It seems that if an integer $n>22$ satisfies (the congruence inspired by Subbarao congruence) | |
| Jul 29, 2019 at 20:49 | comment | added | user142929 | (1/3) Hi I tried to change some factors of the congruence by $\operatorname{rad}(n)$, the so-called radical of an integer. Then A) It seems that $\operatorname{rad}(n)\varphi(n)\equiv2\text{ mod }\sigma(n)$, if and only if $n$ is prime. With the same trick one can to get other congruences (I don't know if this "method" improve the mathematical content of the genuine problem/congruence). From the Wikipedia's article dedicated to Table of congruences I wrote the following congruences. B) It seems that the only integers that satisfy $(\operatorname{rad}(n)-1)!\equiv-1\text{ mod }n$ | |
| Apr 13, 2017 at 12:19 | history | edited | CommunityBot |
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| Nov 10, 2013 at 5:45 | comment | added | Alexey Ustinov | Thank you, Andrej. I think this is the best possible answer for today. | |
| Nov 8, 2013 at 19:36 | comment | added | duje | the congrunce was studied recently in A. Dujella and F. Luca, On a variation of a congruence of Subbarao, J. Aust. Math. Soc. 93 (2012), 85-90. web.math.pmf.unizg.hr/~duje/pdf/DL2and3.pdf | |
| Nov 6, 2013 at 20:58 | history | edited | user9072 |
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| Nov 2, 2013 at 10:08 | comment | added | Alexey Ustinov | $n\varphi(n)\not\equiv0\pmod4$ only for some prime powers. It means that the most significant case is $\sigma(n)\equiv2\pmod4$. | |
| Nov 1, 2013 at 16:46 | comment | added | Chris Wuthrich | :-) I should have been thinking before randomly trying to run through examples, indeed. Thanks Alvin. | |
| Nov 1, 2013 at 14:48 | comment | added | Alvin | well, for $n=pq$ the result immediately follows from the fact that $\sigma(n)=(1+p)(1+q)=0(\mod 4)$ and $n\phi(n)=0(\mod 4)$ unless $p$ or $q$ is $2.$ | |
| Nov 1, 2013 at 13:30 | comment | added | Chris Wuthrich | It could be true. At least I did not find any $N=p\, q$ with $p<10^5$ and $q<10^6$ where it fails. But then, it is not likely to fail and it could just be one of these $\log\log(x)$-symptomes. | |
| Nov 1, 2013 at 3:50 | history | edited | Alexey Ustinov | CC BY-SA 3.0 |
added 140 characters in body; edited tags
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| Nov 1, 2013 at 3:35 | history | asked | Alexey Ustinov | CC BY-SA 3.0 |