Divisor Sum is Odd iff Argument is Square or Twice Square

Theorem
Let $$\sigma: \Z \to \Z$$ be the sigma function.

Then $$\sigma \left({n}\right)$$ is odd iff $$n$$ is either square or twice a square.

Proof
Let $$n$$ be an integer such that $$n \ge 2$$, with prime decomposition $$n = p_1^{k_1} p_2^{k_2} \ldots p_r^{k_r}$$.

Then from Sigma of an Integer we have that $$\sigma \left({n}\right) = \prod_{1 \le i \le r} \frac {p_i^{k_i + 1} - 1} {p_i - 1}$$.

That is, $$\sigma \left({n}\right) = \prod_{1 \le i \le r} \left({1 + p_i + p_i^2 + \ldots + p_i^{k_i}}\right)$$.

Let $$\sigma \left({n}\right)$$ be odd.

Then all factors of $$\prod_{i=1}^r \left({1 + p_i + p_i^2 + \ldots + p_i^{k_i}}\right)$$ are odd (and of course $$\ge 3$$).

For $$1 + p_i + p_i^2 + \ldots + p_i^{k_i}$$ to be odd, one of two conditions must hold:
 * $$p_i$$ is even (so that all terms of $$1 + p_i + p_i^2 + \ldots + p_i^{k_i}$$ are even except the $$1$$);
 * $$k_i$$ is even (so that $$1 + p_i + p_i^2 + \ldots + p_i^{k_i}$$ has an odd number of odd terms).

In the first case, that means $$p_i^{k_i}$$ is a power of $$2$$.

In the second case, that means $$p_i^{k_i}$$ is a square.

The result follows.

The argument reverses.