Every natural number $n$ can be written as $n=s-t$ with $\omega(s)=\omega(t)$ 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 $\ t=n$.
 A: There is a way to prove this directly though, and it is pretty simple:
Case 1: Both 2 and 3 divide $n$, or neither 2 nor 3 divide $n$: Then let $s=3n$ and $t=2n$. 
Case 2: 3 does not divide $n$ but $2$ does: Then let $s=2n$ and $t=n$.
Case 3: 3 divides $n$ but 2 does not: Then let $p$ be the smallest odd prime that does not divide $n$. Then let $s=pn$ and let $t=(p-1)n$. [Note that $p-1$ is the product of smaller odd primes, all of which divide $n$ [by def'n of $p$], and 2, which does not. So $\omega(pn) = \omega(n) +1$ [because $p$ doesn't divide $n$], while $\omega((p-1)n) = \omega(n) +1$ as well, because the one prime factor of $p-1$ that does not divide $n$ is 2.]
A: 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.
A: The Bunyakovsky-conjecture implies that we always find a solution.
To show this, assume that there are infinite many positive integers $p$, such that $$2p+1$$ $$2p+3$$ $$2p^2+4p+1$$ are simultaneously prime. Then, we can choose $p>n$ satisfying this property. Then, $$(2p+1)(2p+3)n-2(2p^2+4p+1)n=n$$ is a solution, whenever $n$ is odd (the even case has already been solved). 
There are plenty of other possibilities to choose the expressions, so the given statement is in fact much weaker than Bunyakovsky's conjecture.
A: *

*If $\ n\ $ is even, see user3733558's comments or Servaes's answer for a much simply argument than the one I originally gave.


*If $\ n\ $ is odd, let $  q\ $ be the smallest odd prime not dividing $\ n\ $.  Then all the odd prime factors of $\ q-1\ $ must be factors of $\ n\ $, and both $\ qn\ $ and $\ (q-1)n\ $ must therefore have exactly one more prime factor than $\ n\ $.
A: If $d$ is any even number, Schinzel's hypothesis implies the existence of a prime $p$, such that $p+d$ is prime as well. In this case , the pair $(p,p+d)$ does the job.
Let $d$ be an odd positive integer.
$$[dx(x+y),2d\frac{x^2+xy+1}{2}]$$ is a suitable pair with difference $d$ , if all the numbers $$x,x+y,\frac{x^2+xy+1}{2}$$ are odd distinct prime numbers , all larger than the largest prime factor of $d$.
With $$p=x,q=\frac{x^2+1}{2},y=2k$$ where $p$ is an odd prime , we get the numbers $$p,p+2k,q+pk$$ Note that $q$ is an integer, if $p$ is an odd prime. Schinzel's hypothesis predicts infinite many $k$ such that all the $3$ numbers are prime because $p$ and $q$ must be coprime. Since $p$ is arbitary , we can establish large enough primes and hence get the desired solution.
So, if Schinzel's hypothesis holds, we can find a pair with every given difference.
