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$?



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?
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.


