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

**bictory** · November 23rd 2009, 08:55 PM

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.

**qmech** · November 24th 2009, 08:17 AM

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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