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Topic: Interesting Property of Sum of Reciprocals (Read 1075 times) |
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Barukh
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Interesting Property of Sum of Reciprocals
« on: Jul 13th, 2004, 9:27am » |
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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.
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Aryabhatta
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Re: Interesting Property of Sum of Reciprocals
« Reply #1 on: Jul 15th, 2004, 11:19am » |
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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 aixi
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 aixi-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.
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Barukh
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Re: Interesting Property of Sum of Reciprocals
« Reply #2 on: Jul 16th, 2004, 2:05am » |
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on Jul 15th, 2004, 11:19am, 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) |
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Nice indeed.
on Jul 15th, 2004, 11:19am, Aryabhatta wrote:I think this was called Wolstenholmes' theorem or something like that.
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 You may want to see my reaction in another thread, same section several minutes ago.
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