Is the conjecture about prime numbers true?
$begingroup$
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
asked Jun 13 '18 at 18:17
user569084user569084
513
$endgroup$
|
show 2 more comments
$begingroup$
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
asked Jun 13 '18 at 18:17
user569084user569084
513
$endgroup$
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
|
show 2 more comments
$begingroup$
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
asked Jun 13 '18 at 18:17
user569084user569084
513
$endgroup$
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
asked Jun 13 '18 at 18:17
user569084user569084
513
asked Jun 13 '18 at 18:17
user569084user569084
513
edited Jun 14 '18 at 10:35
user569084
asked Jun 13 '18 at 18:17
user569084user569084
513
asked Jun 13 '18 at 18:17
user569084user569084
513
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
|
show 2 more comments
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
|
show 2 more comments
1 Answer
1
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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.
answered Dec 19 '18 at 15:44
DynamoDynamo
104517
$endgroup$
add a comment |
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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.
answered Dec 19 '18 at 15:44
DynamoDynamo
104517
$endgroup$
add a comment |
$begingroup$
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.
answered Dec 19 '18 at 15:44
DynamoDynamo
104517
$endgroup$
add a comment |
$begingroup$
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.
answered Dec 19 '18 at 15:44
DynamoDynamo
104517
$endgroup$
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.
answered Dec 19 '18 at 15:44
DynamoDynamo
104517
answered Dec 19 '18 at 15:44
DynamoDynamo
104517
answered Dec 19 '18 at 15:44
DynamoDynamo
104517
answered Dec 19 '18 at 15:44
DynamoDynamo
104517
104517
add a comment |
add a comment |
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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