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The euclidean theorem says that if we have a limited prime numbers and we added 1 it cant be divided by any prime numbers enter image description here

I notice that it work in some cases with lower number but when I added a couple of them things turn off.

M=2 * 3 * 5 * 7 * 11 * 13 * 17 * 19 =9.699.690 => Then we plus +1 => M= 9.699.691

When I use this website to corroborate if it is a prime number it says that it is not. enter image description here

I know there are infnite primes numbers. But why does this theorem have another conditions. Or it does sometimes doesn't work?

Thank you

RodParedes
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    If you take a finite list of primes, multiply them then add one, the result is not divisible by any prime on the list. Any prime factor is new, that is, not on your list. The new primes are $347$ and $27953$ – Will Jagy Nov 27 '22 at 20:44
  • The constructed number need not be prime. But it must have at least one prime factor and this cannot be in the list, so the list cannot be complete. – Peter Nov 27 '22 at 21:37

2 Answers2

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“If we had limited prime numbers” then the product of all primes, plus 1, would have no prime factor and therefore would be prime.

If you multiply all primes from 2 to P and add 1, that number has indeed no prime factor up to P. But the prime numbers are not limited, so it can have factors greater than P.

gnasher729
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That step of the proof is not saying that $M$ is prime. It's only saying that $M$ isn't divisible by any of the prime numbers in your finite list. Since all numbers are divisible by some prime numbers that allows you to add a prime to the list, whether or not $M$ is that prime.

In the case of the list of primes you used there, the argument goes:

Is $\{2,3,5,7,11,13,17,19 \}$ the list of all primes?

No. $2 \times 3 \times 5 \times 7 \times 11 \times 13 \times 17 \times 19 + 1 = 9699691$ is a number that is not divisible by any of our list of primes. $9699691$ is divisible by the prime $347$. $347$ is a prime number that isn't in our list.

Chessanator
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