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Math Help - Wolstenholme

  1. #1
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    Wolstenholme

    Let

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

    Without calculating the LHS, prove that r= s (mod 823^3).
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  2. #2
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    Put another way,

    How do I show that p^3 divides r-s in this instance?
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  3. #3
    Senior Member
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    Apr 2009
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    Atlanta, GA
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    Wolstenholme's Theorem

    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?
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