Show that for each $k \in \mathbb N$ with $k \geq 4$ there exists a prime number $p$ such that $\Omega (N_p) = k.$ 
For $N \in \mathbb N,$ let $\Omega (N)$ denote the number of prime factors of $N$ counting multiplicities. For a prime number $p,$ let $N_p = p (p + 2) (p + 4).$ Show that for any $k \in \mathbb N$ with $k \geq 4$ there exists a prime number $p$ such that $\Omega (N_p) = k.$

What I can see is that $N_p$ has at least four prime factors (counting multiplicities) for $p \gt 3.$ But I don't see how it helps in solving the problem. Any help in this regard would be warmly appreciated.
Thanks for your time.
 A: OK, I think I have a (possible) solution. It stands on somewhat shaky ground, I'll admit.
We know from Legendre that any arithmetic sequence $ak+d$ where $(a,d)=1$ contains an infinite number of primes. Consider the sequences $15ak$, $15ak-2$ and $15ak-4$, with $2 \nmid ak$. The first sequence contains no primes; the other two contain infinitely many primes.
Now, choose an odd $a$ such that $\Omega(15a) = z \ge 2$; we can do this for an arbitrarily large $z$ by adding more factors to $a$. We can restrict $k$ to the odd primes so that $\Omega(15ak) = z+1$.
(Note that we can also use $a=k=1$ to get $(11,13,15)$ for which $\Omega(N_p)=4$. Similarly, $a=3, k=1 \implies \Omega(N_p) = 5$.)
For any odd $a$, the set $R = \{15ak-4 \mid k \in \mathbb{P}, k \ne 2\}$ contains infinitely many primes. Then via Chen's theorem, there are infinitely many* $r \in R$ where $\Omega(r)=1$ or $\Omega(r)=2$.
Then there must exist* some $k \in \mathbb{P}$ such that $p = 15k-4 \in \mathbb{P}$ and $\Omega(p+2) \in \{1,2\}$. $\Omega(p+2) = 2$ will almost certainly be easier to find. The remaining factors come from $15ak$, with $\Omega(N_p) = 3 + \Omega(15ak)$.

*I feel these two implications may be shaky. I'm firmer about Chen's theorem, as I'm all but certain it works for arithmetic sequences. The last implication I'm less certain of. If we allow composite $k$ it would certainly be true. But then $\Omega(p+4)$ becomes unstable and unpredictable.
If both of these are untrue, this does still solve the case of $\Omega(p(p+2))$ or $\Omega(p(p+4))$, but not all three factors.
