4
$\begingroup$

Prove that for any odd prime $p$ $ H_{p-1}=1+\frac{1}{2} + \cdots + \frac{1}{p-1} $ contains a multiple of $p$ in the numerator when written in reduced form, i.e. $\frac{a}{b}$ where $\mathrm{gcd}(a,b)=1$.

  • 2
    Thanks for the acceptance (within one minute of posting the answer!), but see this thread on meta http://meta.math.stackexchange.com/questions/2553/length-of-time-to-wait-before-accepting-an-answer . I think the software allows un-accepting an answer and choosing later which one (if any) to accept.2011-11-06

3 Answers 3

1

Because $p$ is odd, the indices in the sum can be grouped into $(p-1)/2$ pairs $\{ i , p-i \}$ and in each pair the sum is divisible by $p$.

Stronger statements mod $p^2$ and $p^3$ are known as Wolstenholme's congruences.

http://en.wikipedia.org/wiki/Wolstenholme%27s_theorem

4

The operation of reciprocation in $\mathbb{Z}/p \mathbb{Z}$ is a one-to-one map, thus reciprocals of all positive integers $1,2,\ldots,p-1$ would be permutations thereof. The sum of permutated numbers is the same as the sum of ordered numbers, i.e. $ \sum_{k=1}^{p-1} \frac{1}{k} \equiv \sum_{i=1}^{p-1} i \equiv p \cdot \frac{p-1}{2} \equiv 0 \mod p $

  • 0
    The answer is of course correct, the point was only that since the conclusion is false for $p=2$ while the first sentence is true for all $p$, it is interesting to "localize" where the difficulty with $2$ occurs (or at least, that question came to mind while reading the answer). Every line of the proof works for all $p$, except that $p(p-1)/2$ is no longer divisible by $p$ when $p=2$.2011-11-07
1

Write it as $H_{p-1} = \frac{\frac{(p-1)!}{1} + \dots + \frac{(p-1)!}{p-1}}{(p-1)!}.$ Then the denominator is not divisible by $p$, so it is enough to prove that the numerator is. Now $\mathbb{Z}_p$ is a field so we have actually $\frac{(p-1)!}{1} + \dots + \frac{(p-1)!}{p-1} = (p-1)!(1 + \dots + (p-1)) = 1 + \dots + (p-1) = \frac{p(p-1)}{2}$ which is $0$.

Here we have repeatedly used the fact that $\mathbb{Z}_p \setminus \{0\}$ is a group under multiplication. Taking inverses or multiplying by a group element are bijective operations.

  • 0
    @zyx: Thanks. I was writing the answer a bit hastily, and Wilson's theorem just appeared to me for some reason. Edited.2011-11-06