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Math Help - Sums of squares of positive integers prime to n

  1. #1
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    Sums of squares of positive integers prime to n

    Let S(n) denote the sum of the squares of the positive integers \leq n and prime to n.
    Prove that
    \sum_{j = 1}^n{j^2}  = \sum_{d \mid n}{d^2S\left(\frac{n}{d}\right)} = \sum_{d \mid n}{\frac{n^2}{d^2}S(d)}

    I have trouble with separating the integers  \leq n into classes, so that all integers k such that (k,n) = d are in the same class.
    Thanks for the help.
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  2. #2
    Senior Member Shanks's Avatar
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    Let N=\{1^2,2^2,....n^2\}, for each d|n, define x,n)=d\}" alt="N_d=\{x^2x,n)=d\}" />, then \{N_d:d|n\} is a partion on N.
    And for each N_d, Since (\frac{x}{d},\frac{n}{d})=1, By the definition of S(.), the sum of elements in N_d is d^2\times S(\frac{n}{d}),
    Thus,the first equality of <br />
\sum_{j = 1}^n{j^2} = \sum_{d \mid n}{d^2S\left(\frac{n}{d}\right)} = \sum_{d \mid n}{\frac{n^2}{d^2}S(d)}
    is proved.
    When d exhaust all the divisors of n, so do \frac{n}{d}, Thus the second equality obviously hold.
    Last edited by Shanks; November 24th 2009 at 03:10 AM.
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