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

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?

• Can you elaborate on "the product of all primes below and the prime we are evaluating itself are the middle between two unmarked numbers"? – Gregory Grant Jul 21 '16 at 20:00
• The sieve is not used to prove there are infinitely many primes so much as to exhaustively list all primes. So I don't think it will prove it for twin primes either. – Gregory Grant Jul 21 '16 at 20:00
• @GregoryGrant Sorry for being unclear, if we have 2 * 3 * 5 * 7 ... * pi where pi is the prime we are sieving the multiples of out, that number and any multiple of that number have that number -1 and that number +1 unmarked, as the closest marked number have to be atleast 2 away because that is the smallest prime. This is because we can percieve the aforementioned product as marked by all numbers and so distances of other marked numbers should be the prime numbers in the product itself. – My Name Jul 21 '16 at 20:06
• @GregoryGrant but if we sieve out multiples of primes, and between the sieving of each prime the number cannot become finite, it has to have infinitely many primes right? – My Name Jul 21 '16 at 20:10
• @GregoryGrant Oh, now I see, thank you for showing me. – My Name Jul 21 '16 at 20:46

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