In a letter to Dirichlet, Gottold Eisenstein stated the congruence: $$ q(u) \equiv u - \frac{u^2}{2}+ \frac{u^3}{3} - \frac{u^4}{4} + \cdots + \frac{u^{p-2}}{p-2}-\frac{u^{p-1}}{p-1} \pmod{p} $$ where $p$ in an odd prime number and $$ q(u) = \frac{(1+u)^p -1 - u^p}{p}. $$ Putting $u=1$ and using $$ 1+\frac{1}{2}+\frac{1}{3}+ \cdots + \frac{1}{p-2}+\frac{1}{p-1} \equiv 0 \pmod{p} $$ we get $$ 1+\frac{1}{3}+\frac{1}{5}+ \cdots + \frac{1}{p-4}+\frac{1}{x} \equiv \frac{q(1)}{2} \pmod{p} $$ for $x=p-2.$

Question: There is a $t\quad \quad$ ``odd" solution of the following congruence: $$ 1+\frac{1}{3}+\frac{1}{5}+ \cdots + \frac{1}{p-4}+ \cdots + \frac{1}{t} \equiv q(1) \pmod{p} $$ i.e.,

$$ t \in \{p-2,p+2,p+4, \ldots\}. $$

I just got $-2$ points on this post but I do not see why ? There not seem to exists negative votes ???

  • $\begingroup$ Of course, such a $t$ exists provided $q(1)+\frac1{p-3}\not \equiv 0 (mod p)$ by the previous identity. Do you want $t$ as a function of $p$? $\endgroup$ – J.C. Ottem Jan 6 '11 at 19:15
  • $\begingroup$ @J.C.: thanks for comment. From the comment I detected a technical error in the formula. Hope new edit is OK and clearer. $\endgroup$ – Luis H Gallardo Jan 6 '11 at 21:38
  • $\begingroup$ It's still unclear. As far as I remember, the sum of reciprocals of odds $<p$ is treated in my joint paper arxiv.org/abs/1004.4337 (one of the lemmas). $\endgroup$ – Wadim Zudilin Jan 6 '11 at 22:07
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    $\begingroup$ 17 doesn't have a solution. $\endgroup$ – Dror Speiser Jan 6 '11 at 22:10
  • $\begingroup$ @Wadim: Nice, thanks, you think to page 4, formula (21) of your paper ?. There are also very nice formulas for the fermat quotient , etc in an old paper of Emma Lemmer in Annals of Math 339, 2, 350--360, (1938). $\endgroup$ – Luis H Gallardo Jan 6 '11 at 22:27

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