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Math Help - A couple proofs.

  1. #1
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    A couple proofs.

    Prove that:

    (1) \sum_{d|n}\tau^3(d) = \left(\sum_{d|n}\tau(d)\right)^2


    (2) \prod_{d|n}d = n^{\frac{\tau(n)}{2}}

    These come from my Arithmetic functions section in my book. \tau is a divisor function and is multiplicative.

    Thanks for any help!
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  2. #2
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    Quote Originally Posted by Proof_of_life View Post
    (1) \sum_{d|n}\tau^3(d) = \left(\sum_{d|n}\tau(d)\right)^2
    Let f(n) = \sum_{d|n} \tau^3(d) and g(n) = \left( \sum_{d|n}\tau (d) \right)^3.

    The functions f(n),g(n) are weakly multiplicative meaning f(ab)=f(a)f(b) \text{ and }g(ab) = g(a)g(b) whenever \gcd(a,b) = 1.

    This means it remains to prove f(p^k) = g(p^k) for all primes p and positive integers k.

    f(p^k) = \tau^3 (1) + \tau^3 (p) + ... + \tau^3 (p^k) = 1^3 + 2^3 + ... + (k+1)^3
    g(p^k) = ( 1 + \tau (p) + ... + \tau (p^k) )^2 = (1+2+...+(k+1))^2.

    This if of course true using the identity, (1+2+...+n)^2 = 1^3+2^2+...+n^3.

    (2) \prod_{d|n}d = n^{\frac{\tau(n)}{2}}
    If n is not a square then among its factors d_1,d_2,...,d_m for any d_i we can find a mate d_j so that d_id_j=n. Thus, we get n^{\tau(n)/2}. If n is a square we need to modify the proof a little bit.

    This is Mine 95th Post!!!
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  3. #3
    Super Member PaulRS's Avatar
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    We have: \sum_{d|n}{\ln(d)}=\sum_{d|n}{\ln\left(\frac{n}{d}  \right)}

    \sum_{d|n}{\ln(d)}=\sum_{d|n}{\ln(n)}-\sum_{d|n}{\ln(d)}

    Thus: 2\cdot{\sum_{d|n}{\ln(d)}}=\tau(n)\cdot{\ln(n)}

    And then (2) follows easily
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