wu :: forums - Print Page


    
      
        wu :: forums
        (http://www.ocf.berkeley.edu/~wwu/cgi-bin/yabb/YaBB.cgi)
      

        riddles >> putnam exam (pure math) >> Interesting Property of Sum of Reciprocals
        
(Message started by: Barukh on Jul 13^th, 2004, 9:27am)

Title: Interesting Property of Sum of Reciprocals
Post by Barukh on Jul 13^th, 2004, 9:27am

Prove the following statements.

1. If p is a prime number greater than 3, then the numerator of the reduced fraction 1 + 1/2 + 1/3 + ... + 1/(p-1) is divisible by p².

2. If p is a prime number greater than 3, then the numerator of the reduced fraction 1 + 1/4 + 1/9 + ... + 1/(p-1)² is divisible by p.

Title: Re: Interesting Property of Sum of Reciprocals
Post by Aryabhatta on Jul 15^th, 2004, 11:19am

Here is a nice proof of 1) which I read somewhere. I dont just don't remember where. (perhaps the newsgroup sci.math)

Define f(x) = (x-1)(x-2)....(x-(p-1)) = [sum]_{i = 0 to p-1} a_ixⁱ

a₀ = (p-1)!
We are interested in showing that p[sup2] divides a₁.

It can be shown that p divides a₂ by showing that a₂ = 0 looking at the group (Z/p)* (incidentally this is how we can prove part 2) about the square reciprocals)

Now f(0) = f(p).
Consider H(x) = (f(x) - f(0))/x = [sum] _{i = 1 to p-1} a_ix^i-1

Now H(p) = 0. Also H(p) = a₂p + a₁ (mod p[sup2])

Therefore a₂p + a₁ = 0 mod p[sup2].
since p divides a₂, p[sup2] divides a₁. Which is what we wanted.

I think this was called Wolstenholmes' theorem or something like that.

Title: Re: Interesting Property of Sum of Reciprocals
Post by Barukh on Jul 16^th, 2004, 2:05am

on 07/15/04 at 11:19:23, Aryabhatta wrote:

Here is a nice proof of 1) which I read somewhere. I dont just don't remember where. (perhaps the newsgroup sci.math)

Nice indeed.

on 07/15/04 at 11:19:23, Aryabhatta wrote:

I think this was called Wolstenholmes' theorem or something like that.

;D You may want to see my reaction in another thread, same section several minutes ago. ;D