For any integer $n>1$ exist integers $a$ and $b$ so that $\tau(a)+\tau(b)=n$ How to prove that for any integer $n>1$ exist integers $a$ and $b$ so that $$\tau(a)+\tau(b)=n$$ 
Remark: $\tau(n)$ is the number of positive divisors of $n$.
 A: Constructive Proof:
$$\tau(2^{n-2}) + \tau(1) = n$$
Because the divisors of $2^{n-2}$ are:
$$\{2^0, 2^1, \dots, 2^{n-2}\}$$
A: Here is a simple proof:
If $n$ is even, i.e. $n=2k$, then
$$
n=k+k=\tau(a)+\tau(b),
$$
with $a=b=p^{k-1}$, and $p$ a prime.
If $n$ is odd, i.e. $n=2k+1$, then
$$
n=k+(k+1)=\tau(a)+\tau(b),
$$
with $b=p^k=pa$, and $p$ a prime.
A: Given a positive integer $n>1$ there exists a positive number $c$ such that $\tau(c)=n.$ From the fundamental theorem of arithmetic, we can write $c=\prod_{1 \leq i\leq s} p_{i}^{\alpha_{i}},$ where $s=\omega(c)$ is the number of different prime factors of $c.$ Now, let $1\leq k<s,$ set $$A=\left\{p_{i}^{\beta_{i}},\quad 1\leq \beta_{i}\leq \alpha_{i},\quad  1\leq i \leq k\right\} $$
and $$B=\left\{p_{i}^{\beta_{i}},\quad 1\leq \beta_{i}\leq \alpha_{i} ,\quad k+1\leq i \leq s\right\}.$$ Since $\tau(m)=|\left\{d, \quad d|m ,\right\}|$ we can take  $a$ and $b$ such that $\tau(a)=|A|$ and $\tau(b)=|B|.$ Obviously, we have $\tau(a)+\tau(b)=\tau(c)=n.$
