Numerators of harmonic numbers
johndcook.com
johndcook.com
ONE
Multiply through by P := 1.2.3...(p-1); note that if a fraction's numerator is a multiple of p after doing this, it must have been before as well. Now what we have is obviously an integer; it's P/1 + P/2 + P/3 + ... + P/(p-1).
Clearly none of these terms is a multiple of p. And no two of these terms can be equal mod p, because if P/a-P/b is a multiple of p then so is ab(P/a-P/b) = (b-a)P, but that's impossible: neither b-a nor P is a multiple of p, and p is prime.
So their values mod p must be 1,2,3,...,p-1 in some order, and when you add those up you get a multiple of p. We're done.
(For an audience familiar with the relevant ideas, this can be considerably abbreviated: "Mod p, this is the sum of the inverses of 1,2,...,p-1, and these are just 1,2,...,p-1 in some order, whose sum is 0, QED.")
TWO
Group the fractions in pairs: 1/k and 1/(p-k). The sum of each pair is p/k(p-k) and of course when we add these up the numerator is still a multiple of p (because there are no p's in the denominator). Done.
(I think the first of these is really the simpler and more informative of the two, even though the second is shorter.)
That means that if a number fails this test then it's definitely not prime, but if it passes then it may or may not be prime.
However, this link (as included in another comment) gives more information:
https://en.wikipedia.org/wiki/Wolstenholme%27s_theorem#Gener...
Today I found something called Wolstenholme’s
theorem, which says:
A number n (> 3) is prime if the numerator
of H(n-1) is a multiple of n², where
H(n) = 1/1 + 1/2 + ... 1/n
The theorem says:
For a prime p > 3, the numerator of H_{p-1}
is divisible by p^2.
I've tried to comment on your blog, but it requires a login using any of several techniques, none of which I use, and none of which I'm willing to create just for this. So I didn't.
https://en.wikipedia.org/wiki/Wolstenholme%27s_theorem#Gener...