Are there infinity twinprimes?

According to my information the evidence, that there are infinity primes is reasoned through the Euclid-Numbers which always are prime: \(P_{1} \times P_{2} \times ... \times P_{n} = N⇒[N+ 1] ∈ Primes\). Since you always have the opportunity, to multiply all the previous primes and add 1, there must be infinity primes.

Now I've heard 2 interesting facts:

  1. Euclid-Numbers aren't always prime and
  2. It hasn't been proven yet, that there are infinity twinprimes.

It makes good sense, that not all Euclid-Numbers necessarily have to be prime. It's proven, that the Euclid-Numbers aren't divisible by any number between P1P_{1} and PnP_{n} (because there will always be a remainder of 1), but what about Pn+1P_{n+1}? There still are other primes, that could be factors of N+1N+1. The first Euclid-Number, that isn't a prime is 30031 = 59 × 509 (it's the 6th Euclid-Number). Here the trick doesn't work anymore, because N+1N+1 took primes between PnP_{n} and NN. So why is this still a prove? Of course it seems very unlikely, that there is an end for the primes, but is it still a prove?

The reason, why this Euclid-Numbers-Trick works is, because the formula makes sure, that the most often used (least) primes can't be a factor of N+1N+1 and this makes the probability to get a composite number much worse. But actually we can do the same trick with 1-1 instead of +1+1. if we do N1N - 1, there will be a remainder of 1-1 when we divide it by any integer between P1P_{1} and PnP_{n}. So P1×P2×...×Pn=N[(N+1),(N1)]PrimesP_{1} \times P_{2} \times ... \times P_{n} = N ⇒ [(N + 1), (N - 1)]∈Primes But why isn't this a prove, that there are infinity twinprimes, though it uses the principle of the proof, that there are infinity primes (or am I a professor now 🙂)?

Note by Sedrick Heiniger
5 months, 1 week ago

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As shown in the first part, if there is a prime factor to an Euclid number (never knew they were called like that) it is surely bigger than all the primes taken into account (in calculating the Euclid number). If they are no prime factors to the Euclid number then it is a prime itself which completes the proof that there will be always a prime greater than all the primes we discover. But for the second part N + 1 and N - 1 may not be primes themselves, just having prime factors greater than the primes used to calculate N. There is no guarantee that as we approach bigger numbers, N + 1 and N - 1 will be prime. ( I don’t think u became a professor :) )

Jason Gomez - 5 months, 1 week ago

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Thank you! I've finally found the error in reasoning 😁 (I'm so stupid xD).

Sedrick Heiniger - 5 months, 1 week ago

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Lol no, we all get excited sometimes, I used to feel I could square a circle before, although knowing that it was mathematically proven that it was impossible xD

Jason Gomez - 5 months, 1 week ago

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