1. ## Divisibility by prime

Let $\frac{m}{n} = 1 + \frac{1}{2} + \frac{1}{3} + \mbox{...} + \frac{1}{p-1}$ . $p$ is a prime number > 2 . Prove that $m$ is divisible by $p$. If $p > 3$, Prove that $m$ is divisible by $p^2$.

I've done the first bit, I can't seem to get ideas for the 2nd part of the problem. Please help!

2. Originally Posted by sashikanth
Let $\frac{m}{n} = 1 + \frac{1}{2} + \frac{1}{3} + \mbox{...} + \frac{1}{p-1}$ . $p$ is a prime number > 2 . Prove that $m$ is divisible by $p$. If $p > 3$, Prove that $m$ is divisible by $p^2$.

I've done the first bit, I can't seem to get ideas for the 2nd part of the problem. Please help!

Check the following:

http://people.brandeis.edu/~gessel/h...pers/wolst.pdf

http://math.uci.edu/~tchoi/notes/wolstenholme.pdf

Wolstenholme's theorem - Wikipedia, the free encyclopedia

This is a very interesting, non-trivial, result.

Tonio

3. Thanks a lot! I was trying to solve the 2nd bit in a manner I solved the first bit, which is a bit different from what was given on the website and I made no headway. Thank you, this is a beautiful solution.

4. Originally Posted by sashikanth
Thanks a lot! I was trying to solve the 2nd bit in a manner I solved the first bit, which is a bit different from what was given on the website and I made no headway. Thank you, this is a beautiful solution.

Would you guys mind sharing the way to do the first bit of the problem?
The way I did was - gouped the ith and (p-i)th term and when I summed it p came out from all the grouping.

Is there a better way to do it?

5. Originally Posted by aman_cc
Would you guys mind sharing the way to do the first bit of the problem?
The way I did was - gouped the ith and (p-i)th term and when I summed it p came out from all the grouping.

Is there a better way to do it?
Let $m$ be our numerator.

Then, $m\equiv \sum_{i=1}^{p-1} (p-1)! i^{-1} \bmod p$, where $i^{-1}$ is inverse modulo $p$.

But $\sum_{i=1}^{p-1} (p-1)! i^{-1} \equiv (p-1)! \sum_{i=1}^{p-1}i\bmod{p}$ since $i^{-1}$ is a permuation of $\{1,2,...,p-1\}$.

Now note $\sum_{i=1}^{p-1} i = \frac{p(p-1)}{2}$.
Therefore, $m$ is divisible by $p$.

6. The above post outlines my method for the 1st bit.