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
>correct.
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,
And:
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.
So,
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.
Qed.
thanks,
Leroy Quet
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