Show that if $n>0$, then there exists $x$ such that $\tau (nx) = n$.

Question

Show that if $n>0$, then there exists $x$ such that $\tau (nx) = n$.

Note: $\tau (a)$ is the number of divisors of $a$.

I tried working with the prime factorization of the numbers in the equation but that did not get me anywhere.

Answer 1

Since $n>0$ we have that $n=p_1^{a_1}\cdot p_2^{a_2}\cdot\ldots\cdot p_k^{a_k}$, for some non-negative integer $k$, distinct primes $p_1,\ldots,p_k$, and some positive integers $a_1,\ldots,a_k$.

Let $x$ be defined as follows:

$$x=p_1^{p_1^{a_1}-a_1-1}\cdot p_2^{p_2^{a_2}-a_2-1}\cdot\ldots\cdot p_k^{p_k^{a_k}-a_k-1}.$$

Note that $x$ will be a positive integer since for each positive integer $m$ and each prime $p$ we have that $p^m-1\ge m$, with equality holding iff $p=2$ and $m=1$. And we also have the following:

$$\begin{align*} \tau(n\cdot x) &= \tau\left(p_1^{a_1}\cdot p_2^{a_2}\cdot\ldots\cdot p_k^{a_k}\cdot p_1^{p_1^{a_1}-a_1-1}\cdot p_2^{p_2^{a_2}-a_2-1}\cdot\ldots\cdot p_k^{p_k^{a_k}-a_k-1}\right) \\ &= \tau\left(p_1^{p_1^{a_1}-1}\cdot p_2^{p_2^{a_2}-1}\cdot\ldots\cdot p_k^{p_k^{a_k}-1}\right) \\ &= p_1^{a_1}\cdot p_2^{a_2}\cdot\ldots\cdot p_k^{a_k} \\ &= n \end{align*}$$

Answer 2

To find a multiple of $n$ with exactly $n$ divisors, simply replace each prime power $p^a$ in the factorization of $n$ with $p^{p^a-1}$.