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It is known from Subbarao, "On two congruences for primality" that $n>22$ is a prime iff $$n\sigma(n)\equiv 2\pmod{\varphi(n)},$$ where $\varphi(n)$ is Euler's function and $\sigma(n)$ is sum of divisors of $n$. See also Mathematics Stack Exchange

Can we switch $\varphi(n)$ and $\sigma(n)$:

Is $n>9$ a prime if $$n\varphi(n)\equiv 2\pmod{\sigma(n)}?$$ (This congruence holds for $n=8$ and $9$.) I can only prove that this congruence does not hold for prime powers.

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    $\begingroup$ 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. $\endgroup$
    – Chris Wuthrich
    Commented Nov 1, 2013 at 13:30
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    $\begingroup$ 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 $2.$ $\endgroup$
    – Alvin
    Commented Nov 1, 2013 at 14:48
  • $\begingroup$ :-) I should have been thinking before randomly trying to run through examples, indeed. Thanks Alvin. $\endgroup$
    – Chris Wuthrich
    Commented Nov 1, 2013 at 16:46
  • $\begingroup$ $n\varphi(n)\not\equiv0\pmod4$ only for some prime powers. It means that the most significant case is $\sigma(n)\equiv2\pmod4$. $\endgroup$
    – Alexey Ustinov
    Commented Nov 2, 2013 at 10:08
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    $\begingroup$ 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 $\endgroup$
    – duje
    Commented Nov 8, 2013 at 19:36

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