Exercise 2.7.10 (Translation of Wolstenholme's congruence)

Question

Write $1/1 + 1/2  + \cdots + 1/(p-1) = a/b$ with $(a,b) = 1$. Show that $p^2 \mid a$ if $p\geq 5$.

Richard Ganaye · Accepted Answer

\begin{proof} 
If we reduce the fractions  of the left member to the same denominator, we obtain
$$\frac{a}{b} = \frac{\frac{(p-1)!}{1} + \frac{(p-1)!}{2}+ \cdots + \frac{(p-1)!}{p-1}}{(p-1)!}.$$
The $\sigma_i$ are defined by
$$(x-1)(x-2)\cdots (x- p+ 1) = \sum_{i=0}^{p-1} (-1)^i \sigma_i x^{p-1-i},$$
thus
$$\sigma_j = \sum_{1\leq i_1<i_2< \cdots < i_j \leq p-1} i_1 i_2 \cdots i_j.$$
If $j = p-2$, a $p-2$-tuple $(i_1,i_2,\ldots,i_{p-2})$ such that $1\leq i_1<i_2< \cdots < i_{p-2} \leq p-1$ contains all integers of $[\![1, p-1]\!]$, except one. Therefore
\begin{align*}
\sigma_{p-2} &= \sum_{1\leq i_1<i_2< \cdots < i_{p-2} \leq p-1} i_1 i_2 \cdots i_{p-2}\
&=\frac{(p-1)!}{1} + \frac{(p-1)!}{2}+ \cdots + \frac{(p-1)!}{p-1}.
\end{align*}
This gives
$$\frac{a}{b} = \frac{\sigma_{p-2}}{(p-1)!}.$$
By Wolstenholme's congruence, if $p \geq 5$, $p^2 \mid \sigma_{p-2}$, therefore
$$p^2 \mid b \sigma_{p-2} = a(p-1)!.$$
Since $p^2 \wedge (p-1)! = 1$, $$p^2 \mid a.$$
\end{proof}

Exercise 2.7.10 (Translation of Wolstenholme's congruence)

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Exercise 2.7.10 (Translation of Wolstenholme's congruence)

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