proof that $\tau (n)$ is the number of positive divisors of $n$

The following is a proof that $\tau$ counts the positive divisors of its input (which must be a positive integer).

Proof.

Recall that $\tau$ behaves according to the following two rules:

1. If $p$ is a prime and $k$ is a nonnegative integer, then $\tau (p^k)=k+1$.

2. If $\gcd (a,b)=1$, then $\tau (ab)=\tau (a)\tau (b)$.

Let $p$ be a prime. Then $p^0=1$. Note that 1 is the only positive divisor of 1 and $\tau (1)=\tau (p^0)=0+1=1$.

Suppose that, for all positive integers $m$ smaller than $z\in \mathbb{Z}$ with $z>1$, the number of positive divisors of $m$ is $\tau (m)$. Since $z>1$, there exists a prime divisor $p$ of $z$. Let $k$ be a positive integer such that $p^k$ exactly divides $z$. Let $a$ be a positive integer such that $z=p^{k}a$. Then $\gcd (a,p)=1$. Thus, $\gcd (a,p^k)=1$. Since , by the induction hypothesis, there are $\tau (a)$ positive divisors of $a$.

Let $d$ be a positive divisor of $z$. Let $y$ be a nonnegative integer such that $p^y$ exactly divides $d$. Thus, $0\le y\le x$, and there are $k+1$ choices for $y$. Let $c$ be a positive integer such that $d=p^{y}c$. Then $\gcd (c,p)=1$. Since $c$ divides $d$ and $d$ divides $z$, we conclude that $c$ divides $z$. Since $c$ divides $p^{k}a$ and $\gcd (c,p)=1$, it must be the case that $c$ divides $a$. Thus, there are $\tau (a)$ choices for $c$. Since there are $k+1$ choices for $y$ and there are $\tau (a)$ choices for $c$, there are $(k+1)\tau (a)$ choices for $d$. Hence, there are $(k+1)\tau (a)$ positive divisors of $z$. Since $\tau (z)=\tau (p^{k}a)=\tau (p^k)\tau (a)=(k+1)\tau (a)$, it follows that, for every positive integer $n$, the number of positive divisors of $n$ is $\tau (n)$.∎