Harmonic-Number-Related Divisibility

Leroy Quet qq-quet at mindspring.com
Sun Mar 7 22:48:22 CET 2004

I wrote:

>Let H(m) be the m_th harmonic number:
>H(m) = 1 + 1/2 + 1/3 + ...+ 1/m.
>If we represent H(m) as a reduced fraction (and even if we do not reduce, 
>for that matter...), I once proved that:
>If p = odd prime, then
>p divides (2*numerator(H(p-3)) -3*denominator(H(p-3))).
>But I forgot now the specifics of the proof (or "proof"), which I hope is 

Perhaps I can improvise an easy alternative proof of this.

First, we have:

n(k)/d(k) = H(k), where n and d are integers and GCD(n(k),d(k))=1,

p^2|n(p-1) by Wolstenholme's theorem.
(All we need here is p^1|n(p-1).)

And H(p-3) = H(p-1) -1/(p-2) -1/(p-1) =

((p-1)(p-2)n(p-1) -d(p-1)(2p -3))/(d(p-1)(p-1)(p-2)).

So, n(p-3)*m = (p-1)(p-2)n(p-1)-d(p-1)(2p-3),
and d(p-3)*m = d(p-1)(p-1)(p-2),   (*)
for some m.

2*n(p-3) -3*d(p-3) =

(1/m)*(2(p-1)(p-2) n(p-1)-2d(p-1) (2p-3) -3 d(p-1)(p-1)(p-2)) 

= (1/m) (k*p +6 d(p-1) -6 d(p-1)),
for some integer k,
which is a multiple of p if m is coprime with p.

But since GCD(d(p-1),p) = GCD(p-1,p) = GCD(p-2,p) = 1
(GCD(d(p-1),p)=1 because p|n(p-1)),
then, from (*), GCD(d(p-3)*m,p) = 1,
and so GCD(m,p) =1.


Leroy Quet

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