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Let $p_n$ be the $n$-th prime number. Is it true that if $n$ is sufficiently large then will $$p_1×p_2×p_3×...×p_n+1$$ always be a composite number?

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    $\begingroup$ Almost certainly, the answer is not known. $\endgroup$ – quasi Jun 13 '18 at 18:23
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    $\begingroup$ It certainly doesn't have $\{p_1,\cdots, p_n\}$ as factors. This construction was used by Euclid to prove that there are infinitely many primes. $\endgroup$ – Doug M Jun 13 '18 at 18:24
  • $\begingroup$ You should precise if your question is about the first or the second of these two statements: 1) There is a natural number $N$ such that $p_1\cdot\ldots\cdot p_n+1$ is composite for every $n\ge N$. 2) For every natural number $N$ there is some $n\ge N$ such that $p_1\cdot\ldots\cdot p_n+1$ is composite. Nevertheless, as @quasi has said, the answer is probably not known in either case. $\endgroup$ – ajotatxe Jun 13 '18 at 18:24
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    $\begingroup$ This is an open question in number theory. Heuristically, if anything, "Euclid numbers" are more likely than a random nearby number to be prime as noted by @Doug oeis.org/A014545 oeis.org/A006862 $\endgroup$ – David Diaz Jun 13 '18 at 18:32
  • $\begingroup$ @DavidDiaz I didn't mean to suggest that $(p_1\times \cdots \times p_n) + 1$ is necessarily prime. The construction merely proves that for any finite list of numbers, there exists a number co-prime to all all them. And any finite list of prime numbers does not include all prime numbers. $\endgroup$ – Doug M Jun 13 '18 at 18:40
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I tried my best to explain why we always get a composite number.

Case 1: if these primes are arrange in ascending order and $p_1$ is 3 than: $$p_1×p_2×p_3×...×p_n+1$$ is always a composite number as product of $n_th$ odd numbers (here primes) will always odd and adding 1 makes it even.

Case 2: if we take $p_n$ as 2 than $$p_1×p_2×p_3×...×p_n+1$$

will never be an even number as ($p_1×p_2×p_3×...×p_n$) will be even and adding 1 makes it odd.

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