Same amount of divisors

Question

"Find all pairs of natural numbers ($a$,$b$) such that for a natural number $n$, $n \times a$ and $n \times b$ have the same amount of divisors." I found that $a$=$b$ aswell as prime numbers work. What would a general solution look like? Is it connected to the number of ways to combine the prime factors to a distinct product?

Answer 1

The number of divisors of a number $m$ is determined by the exponents in the prime expansion of $m$. If the prime expansion of $m$ is

$m=p_1^{a_1}p_2^{a_2}\dots p_k^{a_k}$

then $m$ has $(a_1+1)(a_2+1)\dots (a_k+1)$ divisors. So $na$ and $nb$ have to have the same exponents in their prime expansions. That should give you a clue about the conditions that $a$ and $b$ must satisfy.