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

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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 1

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 $k'= - \dfrac {\paren {p - 1}!} k \bmod p$.

By Wilson's Theorem

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

Therefore

$k' \equiv k^{-1} \pmod p$

From the corollary to Reduced Residue System under Multiplication forms Abelian Group:

$\struct {\Z'_p, \times}$ 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

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

$\blacksquare$


Proof 2

Note that for any integer $x$:

\(\ds x^p - x\) \(\equiv\) \(\ds 0\) \(\ds \pmod p\) Corollary $1$ to Fermat's Little Theorem
\(\ds \) \(\equiv\) \(\ds x^{\overline p}\) \(\ds \pmod p\) Divisibility of Product of Consecutive Integers
\(\ds \) \(\equiv\) \(\ds \sum_k {p \brack k} x^k\) \(\ds \pmod p\) Sum over k of Unsigned Stirling Numbers of First Kind by x^k

By comparing coefficients:

$\ds {p \brack p} \equiv 1 \pmod p$
$\ds {p \brack 1} \equiv -1 \pmod p$
$\ds {p \brack k} \equiv 0 \pmod p$ for $k \ne 1, p$

or in a more compact form:

$\ds {p \brack k} \equiv \delta_{k p} - \delta _{k 1} \pmod p$

where:

$\ds {p \brack k}$ denotes an unsigned Stirling number of the first kind
$\delta$ is the Kronecker delta.

From Harmonic Number as Unsigned Stirling Number of First Kind over Factorial:

$\ds H_{p - 1} = \frac {p \brack 2} {\paren {p - 1}!}$

From the above we have:

$\ds p \divides {p \brack 2}$

By Prime iff Coprime to all Smaller Positive Integers we also have:

$p \nmid \paren {p - 1}!$

Hence the numerator of $H_{p - 1}$, when expressed in canonical form, is divisible by $p$.

$\blacksquare$


Historical Note

Donald E. Knuth reports in his The Art of Computer Programming: Volume 1: Fundamental Algorithms, 3rd ed. ($1997$) that this result was established by Edward Waring in $1782$, but he gives no further details.


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