# 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.

• Those are the only ways to have $\omega(n(n+2))=2$, right? So if you could prove $\omega(n(n+2))=2$ infinitely often, you would have proved that at least one of the two notorious conjectures is true. So, I don't think you're going to make any progress on this problem, not without some outstanding new insight. Jan 14 at 3:52
• I don't know if this is known (surely no), but you can check out this question and the literature surrounding Chen's theorem: mathoverflow.net/questions/217554/… Jan 14 at 3:53
• @GerryMyerson any time $n=p^k-2$ is prime, with $p$ prime, you get an example. So $n=7,79$ are other examples. Jan 14 at 3:53
• Oh thanks everyone! So this question is equivalent with twin prime conj. $\vee$ Mersenne prime conj. $\vee$ ~Pillai's conj.? And their three are all unsolved...? Jan 14 at 4:04
• @Nightflight You're welcome. However, note Pillai's conjecture is only for differences of perfect powers, i.e., where $e_1 \gt 1$ and $e_2 \gt 1$ in my earlier comment. Thus, there's still the case where exactly one of $e_1$ and $e_2$ is $1$ left to consider. Jan 14 at 4:07

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 $$n$$.