Let $p$ be an odd prime number which large enough. I am interested in the study of the sums of reciprocals in the field $\mathbb{F}_p$.
In particular, I have the following question:
which primes $p$ do satisfy
\begin{equation*}
\sum_{j\equiv
1\pmod{5}}^{\frac{p-1}{2}}j^{-1}\not\equiv\sum_{j\equiv
3\pmod{5}}^{\frac{p-1}{2}}j^{-1}\pmod{p}\quad ?
\end{equation*}
Thanks for any comments or any helpful references.
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$\begingroup$ Have you done any numerical experiments? $\endgroup$– Gerry MyersonCommented Jan 27, 2018 at 10:21
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$\begingroup$ According to my calculations, the equality fails e.g. for $p=17$ and $p=1867$. $\endgroup$– მამუკა ჯიბლაძეCommented Jan 27, 2018 at 12:59
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$\begingroup$ This is true for almost all primes up to $10^{14}$ using pariGP. $\endgroup$– Zakariae.BCommented Jan 27, 2018 at 13:07
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2$\begingroup$ Are there any other exclusions? Can you list them, if so? $\endgroup$– მამუკა ჯიბლაძეCommented Jan 27, 2018 at 13:10
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3$\begingroup$ A reasonable guess might be that the residues of the the left and right hand sides modulo $p$ act like independent, uniformly-distributed random variables. If so, we should expect that there are infinitely many $p$ for which the two sides are congruent, and that the number of $p\leq x$ for which there is congruence grows like $\log\log x$. But questions like this are hard, and I expect it will be difficult to prove that there are infinitely many $p$ for which congruence holds, or even that there are infinitely many $p$ for which congruence does not hold. $\endgroup$– Julian RosenCommented Jan 27, 2018 at 16:19
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