Mixing
Are there infinitely many primes of this form?
Clash Royale CLAN TAG#URR8PPP
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The semiprime $87 = 3*29$ has a curious property: it's the fact that both
$87^2 + 29^2 + 3^2 = 8419$
and
$87^2 - 29^2 - 3^2 = 6719$
are prime numbers.
This intrigued me and led me to wonder if there are other semiprimes with the same property, and I found that
$21 = 3*7$ is another example, since both
$21^2 + 7^2 + 3^2= 499$
and
$21^2 - 7^2 - 3^2 = 383$ are prime numbers
So the following question arises: Are there infinitely many prime numbers $p$ and $q$, with $p neq q$, such that both
$(pq)^2 + p^2 + q^2$
$(pq)^2 - p^2 - q^2$
are also primes?
Does this follows from some known theorem or conjecture?
nt.number-theory analytic-number-theory prime-numbers
edited Aug 13 at 23:17
isaacg
1155
asked Aug 13 at 20:19
E.F
411
add a comment |Â
up vote
8
down vote
favorite
The semiprime $87 = 3*29$ has a curious property: it's the fact that both
$87^2 + 29^2 + 3^2 = 8419$
and
$87^2 - 29^2 - 3^2 = 6719$
are prime numbers.
This intrigued me and led me to wonder if there are other semiprimes with the same property, and I found that
$21 = 3*7$ is another example, since both
$21^2 + 7^2 + 3^2= 499$
and
$21^2 - 7^2 - 3^2 = 383$ are prime numbers
So the following question arises: Are there infinitely many prime numbers $p$ and $q$, with $p neq q$, such that both
$(pq)^2 + p^2 + q^2$
$(pq)^2 - p^2 - q^2$
are also primes?
Does this follows from some known theorem or conjecture?
nt.number-theory analytic-number-theory prime-numbers
edited Aug 13 at 23:17
isaacg
1155
asked Aug 13 at 20:19
E.F
411
Does it help anyhow to consider a second degree equation $ X^2-SX+P=0 $ with $ P=(a+b)(a-b) $ , $ a=(pq)^2=S/2 $, $b=p^2+q^2 $ as well as the equation $Y^2-bY+a=0 $?
– Sylvain JULIEN
Aug 13 at 21:41
I don't know if it helps, but by some basic algebra massaging you can show your two polynomials to be equal to $(pq-1)^2 + (p+q)^2 - 1$ and $(pq+1)^2 - (p+q)^2 -1$ respectively. The latter implies that the composite number $(p-1)(q-1)(p+1)(q+1)$ must be one more than a prime number, while the former (interestingly?) implies that the Gaussian composite $(p-i)(q-i)(p+i)(q+i)$ must be one more than a prime number (they are the same statement but with $p, q mapsto ip, ip$).
– Michael Cromer
Aug 14 at 2:18
As usual with this kind of question about primes, I dare say that the answer is "probably, yes".
– Eric Duminil
Aug 14 at 12:10
See Bateman-Horn conjecture
– Kevin Casto
Aug 14 at 19:22
add a comment |Â
up vote
8
down vote
favorite
up vote
8
down vote
favorite
The semiprime $87 = 3*29$ has a curious property: it's the fact that both
$87^2 + 29^2 + 3^2 = 8419$
and
$87^2 - 29^2 - 3^2 = 6719$
are prime numbers.
This intrigued me and led me to wonder if there are other semiprimes with the same property, and I found that
$21 = 3*7$ is another example, since both
$21^2 + 7^2 + 3^2= 499$
and
$21^2 - 7^2 - 3^2 = 383$ are prime numbers
So the following question arises: Are there infinitely many prime numbers $p$ and $q$, with $p neq q$, such that both
$(pq)^2 + p^2 + q^2$
$(pq)^2 - p^2 - q^2$
are also primes?
Does this follows from some known theorem or conjecture?
nt.number-theory analytic-number-theory prime-numbers
edited Aug 13 at 23:17
isaacg
1155
asked Aug 13 at 20:19
E.F
411
The semiprime $87 = 3*29$ has a curious property: it's the fact that both
$87^2 + 29^2 + 3^2 = 8419$
and
$87^2 - 29^2 - 3^2 = 6719$
are prime numbers.
This intrigued me and led me to wonder if there are other semiprimes with the same property, and I found that
$21 = 3*7$ is another example, since both
$21^2 + 7^2 + 3^2= 499$
and
$21^2 - 7^2 - 3^2 = 383$ are prime numbers
So the following question arises: Are there infinitely many prime numbers $p$ and $q$, with $p neq q$, such that both
$(pq)^2 + p^2 + q^2$
$(pq)^2 - p^2 - q^2$
are also primes?
Does this follows from some known theorem or conjecture?
nt.number-theory analytic-number-theory prime-numbers
edited Aug 13 at 23:17
isaacg
1155
asked Aug 13 at 20:19
E.F
411
edited Aug 13 at 23:17
isaacg
1155
edited Aug 13 at 23:17
isaacg
1155
edited Aug 13 at 23:17
isaacg
1155
1155
asked Aug 13 at 20:19
E.F
411
asked Aug 13 at 20:19
E.F
411
asked Aug 13 at 20:19
E.F
411
411
Does it help anyhow to consider a second degree equation $ X^2-SX+P=0 $ with $ P=(a+b)(a-b) $ , $ a=(pq)^2=S/2 $, $b=p^2+q^2 $ as well as the equation $Y^2-bY+a=0 $?
– Sylvain JULIEN
Aug 13 at 21:41
I don't know if it helps, but by some basic algebra massaging you can show your two polynomials to be equal to $(pq-1)^2 + (p+q)^2 - 1$ and $(pq+1)^2 - (p+q)^2 -1$ respectively. The latter implies that the composite number $(p-1)(q-1)(p+1)(q+1)$ must be one more than a prime number, while the former (interestingly?) implies that the Gaussian composite $(p-i)(q-i)(p+i)(q+i)$ must be one more than a prime number (they are the same statement but with $p, q mapsto ip, ip$).
– Michael Cromer
Aug 14 at 2:18
As usual with this kind of question about primes, I dare say that the answer is "probably, yes".
– Eric Duminil
Aug 14 at 12:10
See Bateman-Horn conjecture
– Kevin Casto
Aug 14 at 19:22
add a comment |Â
Does it help anyhow to consider a second degree equation $ X^2-SX+P=0 $ with $ P=(a+b)(a-b) $ , $ a=(pq)^2=S/2 $, $b=p^2+q^2 $ as well as the equation $Y^2-bY+a=0 $?
– Sylvain JULIEN
Aug 13 at 21:41
I don't know if it helps, but by some basic algebra massaging you can show your two polynomials to be equal to $(pq-1)^2 + (p+q)^2 - 1$ and $(pq+1)^2 - (p+q)^2 -1$ respectively. The latter implies that the composite number $(p-1)(q-1)(p+1)(q+1)$ must be one more than a prime number, while the former (interestingly?) implies that the Gaussian composite $(p-i)(q-i)(p+i)(q+i)$ must be one more than a prime number (they are the same statement but with $p, q mapsto ip, ip$).
– Michael Cromer
Aug 14 at 2:18
As usual with this kind of question about primes, I dare say that the answer is "probably, yes".
– Eric Duminil
Aug 14 at 12:10
See Bateman-Horn conjecture
– Kevin Casto
Aug 14 at 19:22
Does it help anyhow to consider a second degree equation $ X^2-SX+P=0 $ with $ P=(a+b)(a-b) $ , $ a=(pq)^2=S/2 $, $b=p^2+q^2 $ as well as the equation $Y^2-bY+a=0 $?
– Sylvain JULIEN
Aug 13 at 21:41
Does it help anyhow to consider a second degree equation $ X^2-SX+P=0 $ with $ P=(a+b)(a-b) $ , $ a=(pq)^2=S/2 $, $b=p^2+q^2 $ as well as the equation $Y^2-bY+a=0 $?
– Sylvain JULIEN
Aug 13 at 21:41
I don't know if it helps, but by some basic algebra massaging you can show your two polynomials to be equal to $(pq-1)^2 + (p+q)^2 - 1$ and $(pq+1)^2 - (p+q)^2 -1$ respectively. The latter implies that the composite number $(p-1)(q-1)(p+1)(q+1)$ must be one more than a prime number, while the former (interestingly?) implies that the Gaussian composite $(p-i)(q-i)(p+i)(q+i)$ must be one more than a prime number (they are the same statement but with $p, q mapsto ip, ip$).
– Michael Cromer
Aug 14 at 2:18
I don't know if it helps, but by some basic algebra massaging you can show your two polynomials to be equal to $(pq-1)^2 + (p+q)^2 - 1$ and $(pq+1)^2 - (p+q)^2 -1$ respectively. The latter implies that the composite number $(p-1)(q-1)(p+1)(q+1)$ must be one more than a prime number, while the former (interestingly?) implies that the Gaussian composite $(p-i)(q-i)(p+i)(q+i)$ must be one more than a prime number (they are the same statement but with $p, q mapsto ip, ip$).
– Michael Cromer
Aug 14 at 2:18
As usual with this kind of question about primes, I dare say that the answer is "probably, yes".
– Eric Duminil
Aug 14 at 12:10
As usual with this kind of question about primes, I dare say that the answer is "probably, yes".
– Eric Duminil
Aug 14 at 12:10
See Bateman-Horn conjecture
– Kevin Casto
Aug 14 at 19:22
See Bateman-Horn conjecture
– Kevin Casto
Aug 14 at 19:22
add a comment |Â
1 Answer
1
active
oldest
votes
up vote
17
down vote
If both $p$ and $q$ are $pm 1mod 6$, then $(pq)^2 + p^2 + q^2$ is divisible by 3, so that there is only a possibility if one of the primes is 2 or 3. For $q = 2$ (and $p ne 3$), $(pq)^2 + p^2 + q^2 = 5p^2 + 4$ also is divisible by 3. The question then becomes, are there infinitely many primes $p$ so that both $10p^2 + 9$ and $8p^2 - 9$ are prime?
Since it is not known if there is any polynomial of degree greater than 1 that produces infinitely many primes, let alone a pair of polynomials that in an infinite number of primes produce prime numbers, I guess it will be extremely hard to prove that there are infinitely many such pairs $(p, q = 3)$. It may be possible to prove that they don't, though I expect that would be hard as well.
answered Aug 13 at 21:18
doetoe
44047
1
There are 26 such primes $p$ up to 10,000: 7, 29, 83, 181, 197, 337, 601, 631, 1303, 1847, 2029, 3023, 3109, 3359, 4591, 4649, 4831, 6397, 6791, 7489, 7559, 7573, 7951, 8609, 8933, and 9857 (provided you trust Maple and my programming skills).
– Gerry Myerson
Aug 13 at 23:11
5
I should have checked OEIS first. They're tabulated at oeis.org/A079796 and called "nonomatic primes".
– Gerry Myerson
Aug 13 at 23:13
The first $10,000$ such primes are given at the OEIS and a comment that the number is likely infinite. It should be possible to calculate how many such primes one would expect up to $N$ and use that data to see how well that seems to work.
– Aaron Meyerowitz
Aug 14 at 6:18
@GerryMyerson great find! Do you have any idea why they are called that way?
– doetoe
Aug 14 at 7:17
No idea. Was wondering myself.
– Gerry Myerson
Aug 14 at 7:41
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
17
down vote
If both $p$ and $q$ are $pm 1mod 6$, then $(pq)^2 + p^2 + q^2$ is divisible by 3, so that there is only a possibility if one of the primes is 2 or 3. For $q = 2$ (and $p ne 3$), $(pq)^2 + p^2 + q^2 = 5p^2 + 4$ also is divisible by 3. The question then becomes, are there infinitely many primes $p$ so that both $10p^2 + 9$ and $8p^2 - 9$ are prime?
Since it is not known if there is any polynomial of degree greater than 1 that produces infinitely many primes, let alone a pair of polynomials that in an infinite number of primes produce prime numbers, I guess it will be extremely hard to prove that there are infinitely many such pairs $(p, q = 3)$. It may be possible to prove that they don't, though I expect that would be hard as well.
answered Aug 13 at 21:18
doetoe
44047
1
There are 26 such primes $p$ up to 10,000: 7, 29, 83, 181, 197, 337, 601, 631, 1303, 1847, 2029, 3023, 3109, 3359, 4591, 4649, 4831, 6397, 6791, 7489, 7559, 7573, 7951, 8609, 8933, and 9857 (provided you trust Maple and my programming skills).
– Gerry Myerson
Aug 13 at 23:11
5
I should have checked OEIS first. They're tabulated at oeis.org/A079796 and called "nonomatic primes".
– Gerry Myerson
Aug 13 at 23:13
The first $10,000$ such primes are given at the OEIS and a comment that the number is likely infinite. It should be possible to calculate how many such primes one would expect up to $N$ and use that data to see how well that seems to work.
– Aaron Meyerowitz
Aug 14 at 6:18
@GerryMyerson great find! Do you have any idea why they are called that way?
– doetoe
Aug 14 at 7:17
No idea. Was wondering myself.
– Gerry Myerson
Aug 14 at 7:41
add a comment |Â
up vote
17
down vote
If both $p$ and $q$ are $pm 1mod 6$, then $(pq)^2 + p^2 + q^2$ is divisible by 3, so that there is only a possibility if one of the primes is 2 or 3. For $q = 2$ (and $p ne 3$), $(pq)^2 + p^2 + q^2 = 5p^2 + 4$ also is divisible by 3. The question then becomes, are there infinitely many primes $p$ so that both $10p^2 + 9$ and $8p^2 - 9$ are prime?
Since it is not known if there is any polynomial of degree greater than 1 that produces infinitely many primes, let alone a pair of polynomials that in an infinite number of primes produce prime numbers, I guess it will be extremely hard to prove that there are infinitely many such pairs $(p, q = 3)$. It may be possible to prove that they don't, though I expect that would be hard as well.
answered Aug 13 at 21:18
doetoe
44047
1
There are 26 such primes $p$ up to 10,000: 7, 29, 83, 181, 197, 337, 601, 631, 1303, 1847, 2029, 3023, 3109, 3359, 4591, 4649, 4831, 6397, 6791, 7489, 7559, 7573, 7951, 8609, 8933, and 9857 (provided you trust Maple and my programming skills).
– Gerry Myerson
Aug 13 at 23:11
5
I should have checked OEIS first. They're tabulated at oeis.org/A079796 and called "nonomatic primes".
– Gerry Myerson
Aug 13 at 23:13
The first $10,000$ such primes are given at the OEIS and a comment that the number is likely infinite. It should be possible to calculate how many such primes one would expect up to $N$ and use that data to see how well that seems to work.
– Aaron Meyerowitz
Aug 14 at 6:18
@GerryMyerson great find! Do you have any idea why they are called that way?
– doetoe
Aug 14 at 7:17
No idea. Was wondering myself.
– Gerry Myerson
Aug 14 at 7:41
add a comment |Â
up vote
17
down vote
up vote
17
down vote
If both $p$ and $q$ are $pm 1mod 6$, then $(pq)^2 + p^2 + q^2$ is divisible by 3, so that there is only a possibility if one of the primes is 2 or 3. For $q = 2$ (and $p ne 3$), $(pq)^2 + p^2 + q^2 = 5p^2 + 4$ also is divisible by 3. The question then becomes, are there infinitely many primes $p$ so that both $10p^2 + 9$ and $8p^2 - 9$ are prime?
Since it is not known if there is any polynomial of degree greater than 1 that produces infinitely many primes, let alone a pair of polynomials that in an infinite number of primes produce prime numbers, I guess it will be extremely hard to prove that there are infinitely many such pairs $(p, q = 3)$. It may be possible to prove that they don't, though I expect that would be hard as well.
answered Aug 13 at 21:18
doetoe
44047
If both $p$ and $q$ are $pm 1mod 6$, then $(pq)^2 + p^2 + q^2$ is divisible by 3, so that there is only a possibility if one of the primes is 2 or 3. For $q = 2$ (and $p ne 3$), $(pq)^2 + p^2 + q^2 = 5p^2 + 4$ also is divisible by 3. The question then becomes, are there infinitely many primes $p$ so that both $10p^2 + 9$ and $8p^2 - 9$ are prime?
Since it is not known if there is any polynomial of degree greater than 1 that produces infinitely many primes, let alone a pair of polynomials that in an infinite number of primes produce prime numbers, I guess it will be extremely hard to prove that there are infinitely many such pairs $(p, q = 3)$. It may be possible to prove that they don't, though I expect that would be hard as well.
answered Aug 13 at 21:18
doetoe
44047
answered Aug 13 at 21:18
doetoe
44047
answered Aug 13 at 21:18
doetoe
44047
answered Aug 13 at 21:18
doetoe
44047
44047
1
There are 26 such primes $p$ up to 10,000: 7, 29, 83, 181, 197, 337, 601, 631, 1303, 1847, 2029, 3023, 3109, 3359, 4591, 4649, 4831, 6397, 6791, 7489, 7559, 7573, 7951, 8609, 8933, and 9857 (provided you trust Maple and my programming skills).
– Gerry Myerson
Aug 13 at 23:11
5
I should have checked OEIS first. They're tabulated at oeis.org/A079796 and called "nonomatic primes".
– Gerry Myerson
Aug 13 at 23:13
The first $10,000$ such primes are given at the OEIS and a comment that the number is likely infinite. It should be possible to calculate how many such primes one would expect up to $N$ and use that data to see how well that seems to work.
– Aaron Meyerowitz
Aug 14 at 6:18
@GerryMyerson great find! Do you have any idea why they are called that way?
– doetoe
Aug 14 at 7:17
No idea. Was wondering myself.
– Gerry Myerson
Aug 14 at 7:41
add a comment |Â
1
There are 26 such primes $p$ up to 10,000: 7, 29, 83, 181, 197, 337, 601, 631, 1303, 1847, 2029, 3023, 3109, 3359, 4591, 4649, 4831, 6397, 6791, 7489, 7559, 7573, 7951, 8609, 8933, and 9857 (provided you trust Maple and my programming skills).
– Gerry Myerson
Aug 13 at 23:11
5
I should have checked OEIS first. They're tabulated at oeis.org/A079796 and called "nonomatic primes".
– Gerry Myerson
Aug 13 at 23:13
The first $10,000$ such primes are given at the OEIS and a comment that the number is likely infinite. It should be possible to calculate how many such primes one would expect up to $N$ and use that data to see how well that seems to work.
– Aaron Meyerowitz
Aug 14 at 6:18
@GerryMyerson great find! Do you have any idea why they are called that way?
– doetoe
Aug 14 at 7:17
No idea. Was wondering myself.
– Gerry Myerson
Aug 14 at 7:41
1
1
There are 26 such primes $p$ up to 10,000: 7, 29, 83, 181, 197, 337, 601, 631, 1303, 1847, 2029, 3023, 3109, 3359, 4591, 4649, 4831, 6397, 6791, 7489, 7559, 7573, 7951, 8609, 8933, and 9857 (provided you trust Maple and my programming skills).
– Gerry Myerson
Aug 13 at 23:11
There are 26 such primes $p$ up to 10,000: 7, 29, 83, 181, 197, 337, 601, 631, 1303, 1847, 2029, 3023, 3109, 3359, 4591, 4649, 4831, 6397, 6791, 7489, 7559, 7573, 7951, 8609, 8933, and 9857 (provided you trust Maple and my programming skills).
– Gerry Myerson
Aug 13 at 23:11
5
5
I should have checked OEIS first. They're tabulated at oeis.org/A079796 and called "nonomatic primes".
– Gerry Myerson
Aug 13 at 23:13
I should have checked OEIS first. They're tabulated at oeis.org/A079796 and called "nonomatic primes".
– Gerry Myerson
Aug 13 at 23:13
The first $10,000$ such primes are given at the OEIS and a comment that the number is likely infinite. It should be possible to calculate how many such primes one would expect up to $N$ and use that data to see how well that seems to work.
– Aaron Meyerowitz
Aug 14 at 6:18
The first $10,000$ such primes are given at the OEIS and a comment that the number is likely infinite. It should be possible to calculate how many such primes one would expect up to $N$ and use that data to see how well that seems to work.
– Aaron Meyerowitz
Aug 14 at 6:18
@GerryMyerson great find! Do you have any idea why they are called that way?
– doetoe
Aug 14 at 7:17
@GerryMyerson great find! Do you have any idea why they are called that way?
– doetoe
Aug 14 at 7:17
No idea. Was wondering myself.
– Gerry Myerson
Aug 14 at 7:41
No idea. Was wondering myself.
– Gerry Myerson
Aug 14 at 7:41
add a comment |Â
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Does it help anyhow to consider a second degree equation $ X^2-SX+P=0 $ with $ P=(a+b)(a-b) $ , $ a=(pq)^2=S/2 $, $b=p^2+q^2 $ as well as the equation $Y^2-bY+a=0 $?
– Sylvain JULIEN
Aug 13 at 21:41
I don't know if it helps, but by some basic algebra massaging you can show your two polynomials to be equal to $(pq-1)^2 + (p+q)^2 - 1$ and $(pq+1)^2 - (p+q)^2 -1$ respectively. The latter implies that the composite number $(p-1)(q-1)(p+1)(q+1)$ must be one more than a prime number, while the former (interestingly?) implies that the Gaussian composite $(p-i)(q-i)(p+i)(q+i)$ must be one more than a prime number (they are the same statement but with $p, q mapsto ip, ip$).
– Michael Cromer
Aug 14 at 2:18
As usual with this kind of question about primes, I dare say that the answer is "probably, yes".
– Eric Duminil
Aug 14 at 12:10
See Bateman-Horn conjecture
– Kevin Casto
Aug 14 at 19:22