Is the conjecture about prime numbers true?












4












$begingroup$


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?






prime-numbers prime-factorization







share|cite|improve this question











$endgroup$








  • 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 $p_1cdotldotscdot p_n+1$ is composite for every $nge N$. 2) For every natural number $N$ there is some $nge N$ such that $p_1cdotldotscdot p_n+1$ 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 $(p_1times cdots times p_n) + 1$ 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
















4












$begingroup$


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?






prime-numbers prime-factorization







share|cite|improve this question











$endgroup$








  • 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 $p_1cdotldotscdot p_n+1$ is composite for every $nge N$. 2) For every natural number $N$ there is some $nge N$ such that $p_1cdotldotscdot p_n+1$ 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 $(p_1times cdots times p_n) + 1$ 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














4












4








4


1



$begingroup$


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?






prime-numbers prime-factorization







share|cite|improve this question











$endgroup$




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?






prime-numbers prime-factorization




prime-numbers prime-factorization






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jun 14 '18 at 10:35







user569084

















asked Jun 13 '18 at 18:17









user569084user569084

513




513








  • 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 $p_1cdotldotscdot p_n+1$ is composite for every $nge N$. 2) For every natural number $N$ there is some $nge N$ such that $p_1cdotldotscdot p_n+1$ 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 $(p_1times cdots times p_n) + 1$ 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














  • 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 $p_1cdotldotscdot p_n+1$ is composite for every $nge N$. 2) For every natural number $N$ there is some $nge N$ such that $p_1cdotldotscdot p_n+1$ 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 $(p_1times cdots times p_n) + 1$ 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








4




4




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




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




3




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




$begingroup$
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 $p_1cdotldotscdot p_n+1$ is composite for every $nge N$. 2) For every natural number $N$ there is some $nge N$ such that $p_1cdotldotscdot p_n+1$ is composite. Nevertheless, as @quasi has said, the answer is probably not known in either case.
$endgroup$
– ajotatxe
Jun 13 '18 at 18:24






$begingroup$
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 $p_1cdotldotscdot p_n+1$ is composite for every $nge N$. 2) For every natural number $N$ there is some $nge N$ such that $p_1cdotldotscdot p_n+1$ 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




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






$begingroup$
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 $(p_1times cdots times p_n) + 1$ 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




$begingroup$
@DavidDiaz I didn't mean to suggest that $(p_1times cdots times p_n) + 1$ 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










1 Answer
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$begingroup$

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.






share|cite|improve this answer









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    1 Answer
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    0












    $begingroup$

    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.






    share|cite|improve this answer









    $endgroup$


















      0












      $begingroup$

      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.






      share|cite|improve this answer









      $endgroup$
















        0












        0








        0





        $begingroup$

        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.






        share|cite|improve this answer









        $endgroup$



        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.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 19 '18 at 15:44









        DynamoDynamo

        104517




        104517






























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