# Is the conjecture about prime numbers true?

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?

• Almost certainly, the answer is not known. – quasi Jun 13 '18 at 18:23
• 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. – Doug M Jun 13 '18 at 18:24
• 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. – ajotatxe Jun 13 '18 at 18:24
• 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 – David Diaz Jun 13 '18 at 18:32
• @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. – Doug M Jun 13 '18 at 18:40

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.