TITLE: Every natural number $n$ can be written as $n=s-t$ with $\omega(s)=\omega(t)$
QUESTION [5 upvotes]: Can we prove the following statement ?

Let $\omega(m)$ denote the number of all distinct prime factors of a positive integer $m$.
Prove that every  natural  number $n$ can  be  written  as $n=s-t$, where $s$ and $t$ are positive integers with $\omega(s)=\omega(t)$, in other words, as the difference of two positive integers with the same number of distinct prime factors.


If $n=1$ , we can choose $\ s=3\ $ and $\ t=2$.

If $n$ is even , we can choose $\ s=2n\ $ and $\ t=n$.

REPLY [2 votes]: If $n$ is even then $n$ and $2n$ have the same number of prime factors, and so
$$n=2n−n,$$
shows that $n$ is the difference of two natural numbers having the same number of prime factors.
If $n$ is odd then let $q$ be the smallest odd prime number not dividing $n$. Then $q+1$ is even, and its odd factors all divide $n$. It follows that $qn$ and $(q+1)n$ have the same number of prime factors, and so
$$n=(q+1)n−qn,$$
shows that $n$ is the difference of two natural numbers having the same number of prime factors.