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 ???
