Exercise 3.15

Question

\def\imp{\Rightarrow}
\def\gcro{\mbox{[\hspace{-.15em}[}}
\def\dcro{\mbox{]\hspace{-.15em}]}}

ewcommand{\D}{\mathrm{d}}

ewcommand{\Q}{\mathbb{Q}}

ewcommand{\Z}{\mathbb{Z}}

ewcommand{\N}{\mathbb{N}}

ewcommand{\R}{\mathbb{R}}

ewcommand{\C}{\mathbb{C}}

ewcommand{\F}{\mathbb{F}}

ewcommand{e}{\,\mathrm{Re}\,}

ewcommand{\ord}{\mathrm{ord}}

ewcommand{
}{\mathrm{N}}

ewcommand{\legendre}[2]{\genfrac{(}{)}{}{}{#1}{#2}}

For any prime $p$ show that the numerator of $1+ \frac{1}{2} + \frac{1}{3} + \ldots + \frac{1}{p-1}$ is divisible by $p$.

Richard Ganaye · Accepted Answer

\def\imp{\Rightarrow}
\def\gcro{\mbox{[\hspace{-.15em}[}}
\def\dcro{\mbox{]\hspace{-.15em}]}}

ewcommand{\D}{\mathrm{d}}

ewcommand{\Q}{\mathbb{Q}}

ewcommand{\Z}{\mathbb{Z}}

ewcommand{\N}{\mathbb{N}}

ewcommand{\R}{\mathbb{R}}

ewcommand{\C}{\mathbb{C}}

ewcommand{\F}{\mathbb{F}}

ewcommand{e}{\,\mathrm{Re}\,}

ewcommand{\ord}{\mathrm{ord}}

ewcommand{
}{\mathrm{N}}

ewcommand{\legendre}[2]{\genfrac{(}{)}{}{}{#1}{#2}}

\begin{proof}
As the result is false for $p=2$, we must suppose $p>2$, so $p$ is odd.

$1+ \frac{1}{2} + \frac{1}{3} + \ldots + \frac{1}{p-1} = \frac{N}{D}$, where
$$ N = (p-1)! + \frac{(p-1)!}{2}+ \cdots+\frac{(p-1)!}{p-1}, \qquad D= (p-1)!.$$
From Wilson's theorem, $(p-1)! \equiv -1 \pmod p$, so in the field $\Z/p\Z$,
$$ \overline{N} = (-\overline{1})(\overline{1}^{-1} + \overline{2}^{-1}+\cdots+\overline{p-1}^{-1}).$$
Since the application $\varphi : (\Z/p\Z)^* 	o (\Z/p\Z)^*, x \mapsto x^{-1}$ is bijective (it's an involution), 
$$\overline{1}^{-1} + \overline{2}^{-1}+\cdots+\overline{p-1}^{-1} = \overline{1} + \overline{2}+\cdots+\overline{p-1} = \overline{p} 	imes \overline{\left(  \frac{p-1}{2}ight)} = \overline{0}.$$
So $p \mid N $, and $p \wedge (p-1)! = 1$, that is $p\wedge D = 1$. Thus $p$ divides the numerator of the reduced fraction of $N/D$.
\end{proof}

Exercise 3.15

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Exercise 3.15

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