Perfect squares and divisors

Pairing. For each divisor $d$ of $n$, $n/d$ is also a divisor. Fixed points: $d=n/d$, i.e. $d^2=n$.

Count. If $n$ is not a square, all divisors pair up, $\tau(n)$ even. If $n=k^2$, all divisors except $k$ pair up, $\tau(n)$ odd.

Conclusion. $\tau(n)$ is odd iff $n$ is a perfect square.