Wolstenholme

**Cairo** · May 9th 2009, 10:09 PM

Let

1 + 1/2 + 1/3 +...+1/823 = r/s*823.

Without calculating the LHS, prove that r= s (mod 823^3).

**Cairo** · May 11th 2009, 09:12 AM

Put another way,

How do I show that p^3 divides r-s in this instance?

**Media_Man** · May 11th 2009, 09:30 AM

Well, if $r=823s\sum_{n=1}^{823}\frac{1}{n}$ , then $r-s=823s\sum_{n=1}^{822}\frac{1}{n}$ . Since $823$ is prime, by Wolstenholme's Theorem, the numerator of $\sum_{n=1}^{822}\frac{1}{n}$ is divisible by $823^2$ , and by definition, $s$ is the lowest positive integer ensuring the RHS is a whole number, essentially doing the job of canceling all the denominators: $r-s=823s\sum_{n=1}^{822}\frac{1}{n}=823s\frac{N}{D}= 823s\frac{823^2k}{D}=823^3\frac{sk}{D}=823^3s'k$ , for some $s'$ and $k$ .

Therefore, the entire RHS is divisible by $823^3$ , hence $r \equiv s (\bmod 823^3)$

*Are you asking us to prove Wolstenholme's Theorem or to simply connect the dots enough to make use of this theorem?

Thread: Wolstenholme

LinkBack

Thread Tools

Display

Wolstenholme

Wolstenholme's Theorem

Search Tags