Divisor Count Function is Odd Iff Argument is Square

Theorem
Let $\tau: \Z \to \Z$ be the divisor counting function.

Then $\map \tau n$ is odd $n$ is square.

Proof
Let $n$ be an integer such that $n \ge 2$.

Let the prime decomposition of $n$ be:
 * $n = p_1^{k_1} p_2^{k_2} \ldots p_r^{k_r}$

Then from Divisor Counting Function from Prime Decomposition we have that:
 * $\ds \map \tau n = \prod_{i \mathop = 1}^r \paren {k_i + 1}$

Sufficient Condition
Let $\map \tau n$ be odd.

Then all factors of $\ds \prod_{i \mathop = 1}^r \paren {k_i + 1}$ are odd (and of course $\ge 3$).

Therefore all factors of $\ds \prod_{i \mathop = 1}^r \paren {k_i}$ are even.

Thus $n = p_1^{2 s_1} p_2^{2 s_2} \ldots p_r^{2 s_r}$ for $r_i = k_i / 2$ for all $i$.

Hence $n = \paren {p_1^{s_1} p_2^{s_2} \ldots p_r^{s_r} }^2$ and therefore is square.

Necessary Condition
Now suppose $n$ is square.

The above argument reverses, and we see that all factors of $\ds \prod_{i \mathop = 1}^r \paren {k_i + 1}$ are odd.

Hence $\map \tau n$ is itself odd.