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Consider Euclid's proof of the infinitude of the prime: suppose we have a list of $n$ primes $p_1, \dots, p_n$. Then $p_1 \dots p_n + 1$ is coprime to them all, hence gives a new prime factor to add to our list.

Now, a well known mistake is to say that $p_1 ... p_n + 1$ must itself be prime, this not true in general, $30031 = (2)(3)(5)(7)(11)(13) + 1$ is a counterexample, divisible by $59$.

Can this mistake be "fixed"? In the sense that in fact $p_1\dots p_n + 1$ is prime infinitely often, so that this does give a proof of their infinitude? If it makes it easier, I am okay with starting with any prime sequence $(p_n)$, including all of them, or the primes produced as the factors by Euclid's proof starting with 2.

I apologize if this has a well-known answer, I searched MSE and MO but could not find anything. It is for idle curiosity (it came up as a thought when discussing prime numbers with a non-math friend).

Bill Dubuque
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Alex Pawelko
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    The question for the "all of them" version is "(is it known that) there are infinitely many primorial primes?", which is addressed here. A good tip for getting the right terminology to search is to find a relevant sequence on OEIS by just generating a few terms of what you're interested in - you're asking "is https://oeis.org/A018239 infinite?" – Izaak van Dongen Jun 18 '25 at 07:56
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    Ah thank you! I am essentially never near number theory or combinatorics, so I forget about things like the OEIS... – Alex Pawelko Jun 18 '25 at 07:58
  • Questions much simpler than this are still open. For example, we don't know whether there are infinitely many primes of the form $n^2+1$. Then again, how can one decide which of the two open questions is simpler? – Ivan Neretin Jun 18 '25 at 08:13
  • "Can this mistake be "fixed"?" Don't fix what's not broken. – fleablood Jun 18 '25 at 14:48
  • "In the sense that in fact p1…pn+1 is prime infinitely often" How on earth would you know that without doing a bootstrapping argument that ... is actually faultier If $q_i = p_1p_2...+1$ are a list of such primes then construct $q_1q_2...a_3 + 1$ need not be a prime of that form because it's prime factors could all be regular primes not of that form. And even if it were valid (it isn't) it's still prone to the same mistake. – fleablood Jun 18 '25 at 14:56

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