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Math Help - Proving that sigma_k(n) is odd if n is a square or double a square

  1. #1
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    Proving that sigma_k(n) is odd if n is a square or double a square

    If the integer k  \geq 1, prove that \sigma_k(n) is odd if and only if n is a square or double a square.

    I tried the reverse direction first.
    Let n = p_1^{\alpha_1}p_2^{\alpha_2}\cdot\cdot\cdot p_z^{\alpha_z}
    \sigma_k(n) = (p_1^{\alpha_1k} + p_1^{\alpha_1k - k} +\cdot\cdot\cdot + 1)(p_2^{\alpha_2k} + p_2^{\alpha_2k - k} +\cdot\cdot\cdot + 1)\cdot\cdot\cdot(p_z^{\alpha_zk} + p_z^{\alpha_zk - k} +\cdot\cdot\cdot + 1)
    Now each alpha must be even if n is to be a square or double a square.
    (p_i^{\alpha_ik} + p_i^{\alpha_ik - k} +\cdot\cdot\cdot + 1) \mbox{ is odd for }0 \leq i \leq z is odd if the sums are odd. If p_i is 2, then the term becomes even, and so the whole product becomes even, which doesn't make sense, or am I doing something wrong?

    Thanks guys.
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  2. #2
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    A suggestion

    I think you've pretty much gotten the result already. As you said, for a square the exponents are all even. That means that the sum of powers

    <br />
(p^{\alpha} + p^{\alpha -1} +\cdot\cdot\cdot + p + 1)<br />

    is odd for odd p because alpha is even.

    For the factor of 2,

    <br />
(2^{\alpha} + 2^{\alpha -1} +\cdot\cdot\cdot + 2 + 1)<br />

    is odd irregardless of alpha, so the exponent could be even or odd.
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