Why can the sieve of eratosthenes not be used to confirm the twin primes conjecture?

Question

I have been having fun thinking about sieves and more particularly the twin prime conjecture. As I am fairly new to this type of mathematics, I am wondering, if we use the sieve of erastothenes, aka marking of multiples of primes, and we can see that between each step of the sieve the product of all primes below and the prime we are evaluating itself are the middle between two unmarked numbers. So seeing that the number of unmarked numbers stays infinite, won't that make for infinitely many twin primes?

Answer 1

No, because although the number of unmarked numbers stay infinite, the sieve of Eratosthenes algorithm is very limited in predicting what numbers will stay unmarked before it gets to those numbers.

Let's say $n$ is a very large even integer, bigger than a googolplex. Clearly $n$ is not prime, because it is a nontrivial multiple of $2$. Maybe both $n - 1$ and $n + 1$ are primes, which would mean a twin prime pair. Or maybe either $n - 1$ or $n + 1$ is divisible by some large prime $p < \sqrt{n}$ (and no smaller primes), while the other number is itself prime. But if that's the case, the algorithm won't know until it gets around to evaluating the multiples of $p$.

The $n$ I'm suggesting is too large for your typical computer to do much with. Therefore, for an exercise, I suggest you try $n = 3 \times 2^{128}$ instead. You should find that neither $n - 1$ nor $n + 1$ is prime, and that one of them has a least prime factor you would not immediately think of, while the other one does.

Answer 2

Note that $2\cdot 3\cdot 5\cdot 7\pm1$ is not a pair of twin primes: $2\cdot 3\cdot 5\cdot 7-1=11\cdot 19$.