Are there infinitely many $n$ such that $\omega(n^2+2n)=2$? $\omega:\Bbb N\to\Bbb N$ is prime omega function such that $\omega(n)=\sum_{p\mid n}1$. Is there are infinitely many $n$ such that $\omega(n^2+2n)=2$?
If twin prime conjecture is true, then there exists infinitely many $p\in\Bbb P$ such that $p+2\in\Bbb P$, so $\omega(p^2+2p)=2$.
And also Mersenne prime conjecture is true, then there exists infinitely many $p\in\Bbb P$ such that $2^p-1\in\Bbb P$. So when I let $n=2(2^p-1)$ for Mersenne prime $2^p-1$, then $n+2=2^{p+1}$. So $n^2+2n=2^{p+2}(2^p-1)$, that means $\omega(n^2+2n)=2$.
Since twin prime conjecture and Mersenne prime conjectures are both likely true, so I suspect there are infinitely many $n$ such that $\omega(n^2+2n)=2$, but I cannot go further.
I'll appriciate with your help.
+) There are another way to make $\omega(n^2+2n)=2$ as $n=p^r, n+2=q^s$ as $n=7, 25\cdots$. I'm try to prove this part, but totally stuck.

If gcd$(m,n)=1$, $\omega(mn)=\omega(m)+\omega(n). $
If $n$ is odd:
then gcd$(n,n+2)=1$, so $\omega(n^2+2n)=\omega(n)+\omega(n+2). $
$\omega(n)=0 \iff n=1$. $\omega(3) <> 2 \implies n>1, \omega(n)>0$ .
So if $n$ odd, then $n$ and $n+2$ must both be prime or powers of primes,  the former case$\ \omega(n^2+2n)=2 $ if $ n$ is lesser of two twin primes.
If n is even:
$2=$gcd$(n,n+2)$. $n=2a$. $n=2(b-1) \implies a+1=b$
$\omega(n^2+2n)=\omega(4a(a+1))$
So only one odd prime can divide $a(a+1)$. This can only happen if $a$ or $a+1$ is a power of $2$.
If $a+1=2^p$ then $a=2^p-1$ and $n^2+2n=2^{p+2}(2^p-1)$
, the Mersenne Prime scenario you mention. Interestingly, this is also $8$ times a Perfect Number. This should also work if $2^p-1$ is a power of a prime, i.e. $2^q-p^z=1 \iff 2^q \equiv 1 \pmod{p^z}$. So $q \equiv 0 \pmod{\phi(p^z)} $ by Euler's Totient rule. $\phi(p^z)=p^z-p^{z-1}=p^{z-1}(p-1)$ So $q=kp^{z-1}(p-1)$.
A third possible scenario:
Suppose $a=2^q$. Then $n^2+2n=2^{q+2}(2^q+1)$
$0=4 \cdot 2^{2q}+4\cdot2^q-n(n+2)$
$2^q=\frac{-4+4\sqrt{1+n(n+2)}}{8}=n/2$
so $n=2^{q+1}$
We need $2^q+1$  be a power of a prime. This gives us 24, 80, 288, 1088,...
We need solutions of the form $p^z-2^q=1$
$-1\equiv2^q \pmod{p^z}, z>=1 $
Does such a $q$ only  exist if $2$ is a primitive root for prime $p$?
Looks like there are 4 scenarios, twin primes and twin power of primes for odd $n$, Mersenne Primes/Perfect numbers, and powers of primes 1 greater than a power of 2 for even $n$.
