Divisibility by prime

**sashikanth** · April 15th 2010, 11:50 PM

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!

**tonio** · April 16th 2010, 03:16 AM

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

**sashikanth** · April 16th 2010, 04:43 AM

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.

**aman_cc** · April 16th 2010, 05:26 AM

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?

**chiph588@** · April 16th 2010, 06:53 AM

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 $p$ .

**sashikanth** · April 16th 2010, 09:57 AM

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

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