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?

  • 4
    $\begingroup$ Almost certainly, the answer is not known. $\endgroup$ – quasi Jun 13 '18 at 18:23
  • 3
    $\begingroup$ 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
  • 2
    $\begingroup$ 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

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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.