Let $f(n)$ be a quadratic polynomial .Then is the density of integers such that $|\omega(f(n+k))-\omega(f(n))|\le C$ for some constant $C$ zero? Let $f(n)$ be a quadratic polynomial with integral coefficients such that $f(n)>0$ for all natural $n$.
Let $ \omega(n)$ the number of distinct prime divisors of the positive integer $n$, and let $ k$ be a fixed positive integer.
Define $ S(N)$ to be
$S(N)=\#\{ 1 \le n \le N: |\omega(f(n+k))-\omega(f(n))| \le C\},$ for some constant $C$
Is it true that $ \displaystyle \lim_{N \to +\infty}{\frac{S(N)}{N}}=0$?
 A: Assume throughout that $k \neq 0 $ since otherwise the probability is $1$ and not $0$ as required.
A special case of Corollary 1.9 of https://arxiv.org/pdf/2001.10970.pdf says that if we have two integer polynomials $f_1, f_2$ of arbitrary degree, then if we let $c_1,c_2$ be the number of irreducible polynomials dividing $f_1$ and $f_2$ respectively and if we let $c_0$ denote the number of common irreducible polynomial factors then the vector $$ \left(\frac{\omega(f_1(n)) - c_1 \log \log n }{\sqrt{c_1\log \log n } },\frac{\omega(f_2(n)) - c_2 \log \log n }{\sqrt{c_2 \log \log n } }\right)$$ behaves as a two dimensional Gaussian with covariance matrix having $1$ in the diagonal and $c_0/\sqrt{c_1 c_2 } $ in the off-diagonal entries.
How is this relevant? If $f$ is irreducible then letting $f_1=f(n)$ and $f_2(n)=f(n+k )$ we see that both $f_i $ are irreducible and clearly coprime as polynomials. Hence $c_1=c_2=1, c_0=0$. This means that $$\frac{\omega(f (n)) -   \log \log n }{\sqrt{ \log \log n } } - \frac{\omega(f(n+k)) -   \log \log n }{\sqrt{  \log \log n } } = \frac{\omega(f (n)) - \omega(f(n+k )) }{\sqrt{ \log \log n } } $$ converges in law to a normal distribution with mean $0 $ and variance $  2 $. In other words, for every fixed $a<b $ one has $$ \lim_{x\to \infty} \frac{1}{x} \#\left\{n \in [1,x]: a< \frac{\omega(f(n)) - \omega(f(n+k)) }{\sqrt{ \log \log n } }  \leq b  \right \} = \int_a^b \frac{\exp(-t^2/4)}{4 \sqrt \pi }\mathrm d t .$$ Note that if $| \omega(f(n)) - \omega(f(n+k)) | \leq C $ then for every $\epsilon>0 $ we  have  $$   \left |  \frac{\omega(f(n)) - \omega(f(n+k)) }{\sqrt{ \log \log n } }  \right|  \leq  \epsilon  ,$$ hence, the probability that $| \omega(f(n)) - \omega(f(n+k)) | \leq C $ happens is$$ \ll \int_{-\epsilon}^\epsilon \frac{\exp(-t^2/4)}{4 \sqrt \pi }\mathrm d t  \ll \epsilon  .  $$ Choosing $\epsilon$ arbitrarily small shows that  the probability is   $0 $. Incidentally this argument proves the much stronger statement that the probability of $n $ for which $$| \omega(f(n)) - \omega(f(n+k)) | \leq \frac{\sqrt{\log \log n }}{ \log \log \log n }$$ also goes to zero.
In the case that the quadratic polynomial $f $ is reducible (and not the square of a linear polynomial) then  $c_1=2=c_2$ but $c_0 $ is not necessarily $ 0 $. For example, it can be $1$ in the case $ f(n)= n(n+k )$. If $c_1 = 0 $ then the proof is essentially the same as in the case when $f $ is irreducible. If $c_0 =1 $ then  the off-diagonal element above is $1/2$. Therefore, $$\frac{\omega(f (n)) -  2 \log \log n }{\sqrt{ 2\log \log n } } - \frac{\omega(f(n+k)) -  2 \log \log n }{\sqrt{ 2 \log \log n } } = \frac{\omega(f (n)) - \omega(f(n+k )) }{\sqrt{ 2\log \log n } } $$ converges in law to the standard normal distribution, which proves what you want again. If $ f$ is the square of a linear polynomial then the same holds for $f(n+ k ) $ and by using the fact that $\omega(t^2)=\omega(t)$ the question reduces to examining $f(n)$ linear. This is covered already by the arguments above when $ f $ was irreducible.
