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Thread: A couple proofs.

  1. #1
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    A couple proofs.

    Prove that:

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


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

    These come from my Arithmetic functions section in my book. $\displaystyle \tau$ 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) $\displaystyle \sum_{d|n}\tau^3(d) = \left(\sum_{d|n}\tau(d)\right)^2$
    Let $\displaystyle f(n) = \sum_{d|n} \tau^3(d)$ and $\displaystyle g(n) = \left( \sum_{d|n}\tau (d) \right)^3$.

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

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

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

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

    (2)$\displaystyle \prod_{d|n}d = n^{\frac{\tau(n)}{2}}$
    If $\displaystyle n$ is not a square then among its factors $\displaystyle d_1,d_2,...,d_m$ for any $\displaystyle d_i$ we can find a mate $\displaystyle d_j$ so that $\displaystyle d_id_j=n$. Thus, we get $\displaystyle n^{\tau(n)/2}$. If $\displaystyle n$ 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: $\displaystyle \sum_{d|n}{\ln(d)}=\sum_{d|n}{\ln\left(\frac{n}{d} \right)}$

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

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

    And then (2) follows easily
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