# Wolstenholme’s theorem

We want to show first that the harmonic number

 $H_{n}\;=:\;1+\frac{1}{2}+\frac{1}{3}+\ldots+\frac{1}{n}\,\;=\;\int_{0}^{1}% \frac{1\!-\!x^{n}}{1\!-\!x}\,dx$

is never an integer ($n>1$).

Denote by $p$ the greatest prime number not exceeding $n$.  By Bertrand’s postulate there is a prime $q$ with  $p.  Therefore we have  $n<2p$.  If $H_{n}$ were an integer, then the sum

 $n!H_{n}\;=\;\sum_{i=1}^{n}\frac{n!}{i}$

had to be divisible by $p$.  However its addend $\displaystyle\frac{n!}{p}$ is not divisible by $p$ but all other addends are, whence the sum cannot be divisible by $p$.  The contradictory situation means that $H_{n}$ is not integer when  $n>1$.

Theorem (Wolstenholme).  If $p$ is a prime number greater than 3, then the numerator of the harmonic number

 $H_{p-1}\;=\;1+\frac{1}{2}+\frac{1}{3}+\ldots+\frac{1}{p-1}$

is always divisible by $p^{2}$.

Proof.  Consider the polynomial

 $f(x)\;=:\;(x\!-\!1)(x\!-\!2)\cdots(x\!-\!p\!+\!1).$

One has

 $\displaystyle f(0)\;=\;(p\!-\!1)!\;=\;f(p)$ (1)

and

 $\displaystyle f(x)\;=\;x^{p-1}+a_{1}x^{p-2}+a_{2}x^{p-3}+\ldots+a_{p-2}x+(p\!-% 1\!)!$ (2)

where $a_{1},\,a_{2},\,\ldots,\,a_{p-2}$ are integers.  Because $1,\,2,\,\ldots,\,p\!-\!1$ form a set of all modulo $p$ incongruent roots of the Fermat’s congruence (http://planetmath.org/FermatsTheorem)  $x^{p-1}\equiv 1\pmod{p}$,  one may write the identical congruence

 $\displaystyle x^{p-1}\!-\!1\;\equiv\;x^{p-1}+a_{1}x^{p-2}+a_{2}x^{p-3}+\ldots+% a_{p-2}x+(p\!-1\!)!\pmod{p}.$ (3)

It may be written by Wilson’s theorem  $(p\!-1\!)!\equiv-1\pmod{p}$  as

 $\displaystyle a_{1}x^{p-2}+a_{2}x^{p-3}+\ldots+a_{p-2}x\;\equiv\;0\pmod{p},$ (4)

being thus true for any integer $x$.  From (4) one can successively infer that $p$ divides all coefficients $a_{i}$, i.e. that (4) actually is a formal congruence.

For the derivative of the polynomial $f(x)$ one has

 $f^{\prime}(x)\;=\;(x\!-\!2)\cdots(x\!-\!p\!+\!1)+\ldots+(x\!-\!1)\cdots(x\!-\!% p\!+\!2)$

and thus

 $\displaystyle f^{\prime}(0)\;=\;-2\cdot 3\cdots(p\!-\!1)-\ldots-1\cdot 2\cdots% (p\!-\!2).$ (5)

The Taylor series (Taylor polynomial) of $f(x)$ coincides with $f(x)$:

 $f(x)\;=\;f(0)+\frac{f^{\prime}(0)}{1!}x+\frac{f^{\prime\prime}(0)}{2!}x^{2}+% \ldots+\frac{f^{(p-1)}(0)}{(p\!-\!1)!}x^{p-1}$

By (1), this equation implies

 $\displaystyle 0\;=\;f^{\prime}(0)+\frac{f^{\prime\prime}(0)}{2}p+\ldots+\frac{% f^{(p-1)}(0)}{(p\!-\!1)!}p^{p-2}$ (6)

Since  $p\mid a_{p-3}=\frac{f^{\prime\prime}(0)}{2}$,  one has  $p\mid f^{\prime\prime}(0)$.  It then follows by (6) that  $p^{2}\mid f^{\prime}(0)$.  And since (5) divided by $-(p\!-\!1)!$ gives

 $1+\frac{1}{2}+\frac{1}{3}+\ldots+\frac{1}{p-1}\;=\;-\frac{f^{\prime}(0)}{(p\!-% \!1)!},$

the assertion has been proved.

## References

• 1 L. Kuipers: “Der Wolstenholmesche Satz”.  – Elemente der Mathematik 35 (1980).
Title Wolstenholme’s theorem WolstenholmesTheorem 2013-03-22 19:14:06 2013-03-22 19:14:06 pahio (2872) pahio (2872) 10 pahio (2872) Theorem msc 11C08 msc 11A07 HarmonicNumber