A simple quadratic reformulation of the twin prime conjecture?

Question

Due to https://oeis.org/A024702 we have that $p^2 - 1 ≡ 0$ (mod 24).

For twin primes, we then must also have that in this case $(p+2)^2 - 1 ≡ 0$ (mod 24), which is the same as saying that $p^2 + 4p + 3 ≡ 0$ (mod 24).

Adding these and dividing both sides by 2, one would then have a simple quadratic reformulation of the twin primes conjecture like so:

There are infinitely many primes p such that $p^2 + 2p + 1 ≡ 0$ (mod 12).

Is that correct? Would it possibly be of any practical benefit, like extending it to prime gaps of arbitrary size that have been proven already, and then "proving backwards" towards a gap of size 2?

Answer 1

No, it's not equivalent to the twin prime conjecture. Any $p$ such that $p \equiv 5$ or $11 \mod 12$ satisfies $p^2+2p+1 \equiv 0 \mod 12$. But most of them are not twin primes. Counterexamples are easy to find.

Answer 2

The issue here is that all numbers of the form $n=6k\pm 1$, not just primes, satisfy the relationship $n^2-1 \equiv 0 \bmod 24$. Such pairs of numbers may contain $0,1,\text{ or }2$ primes, e.g. $(119,121),(23,25),(47,49),(11,13)$. So even if there are infinitely many numbers such that $n^2+2n+1\equiv 0 \bmod 12$, this is not equivalent to the assertion that there are infinitely many prime numbers with that property.

Answer 3

No use , a slightly better reformulation is: There exists infinitely many semiprimes of form $p^2+2p$ , where $p$ is prime. It's still basically useless though...