# Numerator of p-1th Harmonic Number is Divisible by Prime p/Proof 1

## Theorem

Let $p$ be an odd prime.

Consider the harmonic number $H_{p - 1}$ expressed in canonical form.

The numerator of $H_{p - 1}$ is divisible by $p$.

## Proof

Add the terms of $H_{p - 1}$ using the definition of rational addition to obtain $\dfrac m n$.

Do not cancel common prime factors from $m$ and $n$.

It is seen that $n = \paren {p - 1}!$

Hence $p$ is not a divisor of $n$.

The numerator $m$ is seen to be:

$m = \dfrac {\paren {p - 1}!} 1 + \dfrac {\paren {p - 1}!} 2 + \cdots + \dfrac {\paren {p - 1}!} {p - 1}$

Thus it is sufficient to show that $m$ is a multiple of $p$.

Each term in this sum is an integer of the form $\dfrac {\paren {p - 1}!} k$.

For each $k \in \set {1, 2, \ldots, p - 1}$, define $k'= - \dfrac {\paren {p - 1}!} k \bmod p$.

$k k' \equiv -\paren {p - 1}! \equiv 1 \pmod p$

Therefore

$k' \equiv k^{-1} \pmod p$
$\struct {\Z'_p, \times}$ is an abelian group.

Since Inverse in Group is Unique, the set:

$\set {1', 2', \ldots, \paren {p - 1}'}$

is merely the set:

$\set {1, 2, \ldots, p - 1}$

in a different order.

Thus

 $\displaystyle m$ $=$ $\displaystyle \dfrac {\paren {p - 1}!} 1 + \dfrac {\paren {p - 1}!} 2 + \cdots + \dfrac {\paren {p - 1}!} {p - 1}$ $\displaystyle$ $\equiv$ $\displaystyle 1 + 2 + \cdots + p - 1$ $\displaystyle \pmod p$ $\displaystyle$ $\equiv$ $\displaystyle \frac {p \paren {p - 1} } 2$ $\displaystyle \pmod p$ Closed Form for Triangular Numbers $\displaystyle$ $\equiv$ $\displaystyle 0$ $\displaystyle \pmod p$

$\blacksquare$