When is $p^2+1$ twice of a prime?
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When trying to solve a bigger problem, I came across the problem of characterising all primes $p,q$ such that $p^2+1=2q$. That is, are there necessary and sufficient conditions for primes $p,q$ to satisfy the equation? Even better, is there a paramatrisation of solutions (though, this is probably unlikely)? I know the solutions $(p,q)=(3,5),(5,13),(11,61),(19,181),(29,421)$ but I don't know that these are the only ones (indeed, it seems likely there are many more).
Some incomplete thoughts: I immediately made the factorisation $(p+1)(p-1)=2(q-1)$, and because $p$ is obviously odd, let $p=2p_1+1$. This leads us to the equation $2p_1(p_1+1)=q-1$. If we substituted $q=2q_1+1$ we have $p_1(p_1+1)=q_1$, so $q$ is twice of the product of two consecutive integers, plus one. It follows that $q$ is $1$ mod $4$, and $p^2$ is $1$ mod $8$. The same conclusion can be obtained simply by noting that $-1$ must be a quadratic residue mod $2q$, hence $1$ is a $4$th power residue mod $2q$. By Lagrange's Theorem we have $4midphi(2q)=q-1$, so $q equiv1$ mod $4$. Any further thoughts are appreciated!
abstract-algebra number-theory prime-numbers diophantine-equations
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add a comment |
$begingroup$
When trying to solve a bigger problem, I came across the problem of characterising all primes $p,q$ such that $p^2+1=2q$. That is, are there necessary and sufficient conditions for primes $p,q$ to satisfy the equation? Even better, is there a paramatrisation of solutions (though, this is probably unlikely)? I know the solutions $(p,q)=(3,5),(5,13),(11,61),(19,181),(29,421)$ but I don't know that these are the only ones (indeed, it seems likely there are many more).
Some incomplete thoughts: I immediately made the factorisation $(p+1)(p-1)=2(q-1)$, and because $p$ is obviously odd, let $p=2p_1+1$. This leads us to the equation $2p_1(p_1+1)=q-1$. If we substituted $q=2q_1+1$ we have $p_1(p_1+1)=q_1$, so $q$ is twice of the product of two consecutive integers, plus one. It follows that $q$ is $1$ mod $4$, and $p^2$ is $1$ mod $8$. The same conclusion can be obtained simply by noting that $-1$ must be a quadratic residue mod $2q$, hence $1$ is a $4$th power residue mod $2q$. By Lagrange's Theorem we have $4midphi(2q)=q-1$, so $q equiv1$ mod $4$. Any further thoughts are appreciated!
abstract-algebra number-theory prime-numbers diophantine-equations
$endgroup$
7
$begingroup$
oeis.org/A048161
$endgroup$
– mathlove
Dec 29 '18 at 5:51
3
$begingroup$
You should definitely look at the link in the first comment. Whether there are infinitely many such $p$ is unknown... BTW if $p>5$ then $pequiv pm 1 mod 10,$ for if $5<pequiv pm 3 mod 10$ then $ ( p^2+1)/2$ is greater than $5$ and divisible by $5.$
$endgroup$
– DanielWainfleet
Dec 29 '18 at 6:28
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@mathlove Wow, thanks. If you make that an answer I would happily accept it :)
$endgroup$
– YiFan
Dec 29 '18 at 8:50
$begingroup$
@YiFan Do you want to find all the solutions or only decide whether infinite many exist ? Besides some basic restrictions, there won't be useful methods. You will just have to check the number $frac{p^2+1}{2}$ in general. If you are interested in an evidence that infinite many solutions probably exist, I can undelete my answer.
$endgroup$
– Peter
Dec 30 '18 at 10:44
$begingroup$
@Peter I wanted to see if there was a nice necessary and sufficient criterion that could be used to describe all solutions to the equation, but I now know that's probably not going to happen. I'm certainly interested in anything related to the problem though, and your answer is welcome :)
$endgroup$
– YiFan
Dec 30 '18 at 11:21
add a comment |
$begingroup$
When trying to solve a bigger problem, I came across the problem of characterising all primes $p,q$ such that $p^2+1=2q$. That is, are there necessary and sufficient conditions for primes $p,q$ to satisfy the equation? Even better, is there a paramatrisation of solutions (though, this is probably unlikely)? I know the solutions $(p,q)=(3,5),(5,13),(11,61),(19,181),(29,421)$ but I don't know that these are the only ones (indeed, it seems likely there are many more).
Some incomplete thoughts: I immediately made the factorisation $(p+1)(p-1)=2(q-1)$, and because $p$ is obviously odd, let $p=2p_1+1$. This leads us to the equation $2p_1(p_1+1)=q-1$. If we substituted $q=2q_1+1$ we have $p_1(p_1+1)=q_1$, so $q$ is twice of the product of two consecutive integers, plus one. It follows that $q$ is $1$ mod $4$, and $p^2$ is $1$ mod $8$. The same conclusion can be obtained simply by noting that $-1$ must be a quadratic residue mod $2q$, hence $1$ is a $4$th power residue mod $2q$. By Lagrange's Theorem we have $4midphi(2q)=q-1$, so $q equiv1$ mod $4$. Any further thoughts are appreciated!
abstract-algebra number-theory prime-numbers diophantine-equations
$endgroup$
When trying to solve a bigger problem, I came across the problem of characterising all primes $p,q$ such that $p^2+1=2q$. That is, are there necessary and sufficient conditions for primes $p,q$ to satisfy the equation? Even better, is there a paramatrisation of solutions (though, this is probably unlikely)? I know the solutions $(p,q)=(3,5),(5,13),(11,61),(19,181),(29,421)$ but I don't know that these are the only ones (indeed, it seems likely there are many more).
Some incomplete thoughts: I immediately made the factorisation $(p+1)(p-1)=2(q-1)$, and because $p$ is obviously odd, let $p=2p_1+1$. This leads us to the equation $2p_1(p_1+1)=q-1$. If we substituted $q=2q_1+1$ we have $p_1(p_1+1)=q_1$, so $q$ is twice of the product of two consecutive integers, plus one. It follows that $q$ is $1$ mod $4$, and $p^2$ is $1$ mod $8$. The same conclusion can be obtained simply by noting that $-1$ must be a quadratic residue mod $2q$, hence $1$ is a $4$th power residue mod $2q$. By Lagrange's Theorem we have $4midphi(2q)=q-1$, so $q equiv1$ mod $4$. Any further thoughts are appreciated!
abstract-algebra number-theory prime-numbers diophantine-equations
abstract-algebra number-theory prime-numbers diophantine-equations
edited Jan 7 at 10:13
YiFan
asked Dec 29 '18 at 4:54
YiFanYiFan
4,7531727
4,7531727
7
$begingroup$
oeis.org/A048161
$endgroup$
– mathlove
Dec 29 '18 at 5:51
3
$begingroup$
You should definitely look at the link in the first comment. Whether there are infinitely many such $p$ is unknown... BTW if $p>5$ then $pequiv pm 1 mod 10,$ for if $5<pequiv pm 3 mod 10$ then $ ( p^2+1)/2$ is greater than $5$ and divisible by $5.$
$endgroup$
– DanielWainfleet
Dec 29 '18 at 6:28
$begingroup$
@mathlove Wow, thanks. If you make that an answer I would happily accept it :)
$endgroup$
– YiFan
Dec 29 '18 at 8:50
$begingroup$
@YiFan Do you want to find all the solutions or only decide whether infinite many exist ? Besides some basic restrictions, there won't be useful methods. You will just have to check the number $frac{p^2+1}{2}$ in general. If you are interested in an evidence that infinite many solutions probably exist, I can undelete my answer.
$endgroup$
– Peter
Dec 30 '18 at 10:44
$begingroup$
@Peter I wanted to see if there was a nice necessary and sufficient criterion that could be used to describe all solutions to the equation, but I now know that's probably not going to happen. I'm certainly interested in anything related to the problem though, and your answer is welcome :)
$endgroup$
– YiFan
Dec 30 '18 at 11:21
add a comment |
7
$begingroup$
oeis.org/A048161
$endgroup$
– mathlove
Dec 29 '18 at 5:51
3
$begingroup$
You should definitely look at the link in the first comment. Whether there are infinitely many such $p$ is unknown... BTW if $p>5$ then $pequiv pm 1 mod 10,$ for if $5<pequiv pm 3 mod 10$ then $ ( p^2+1)/2$ is greater than $5$ and divisible by $5.$
$endgroup$
– DanielWainfleet
Dec 29 '18 at 6:28
$begingroup$
@mathlove Wow, thanks. If you make that an answer I would happily accept it :)
$endgroup$
– YiFan
Dec 29 '18 at 8:50
$begingroup$
@YiFan Do you want to find all the solutions or only decide whether infinite many exist ? Besides some basic restrictions, there won't be useful methods. You will just have to check the number $frac{p^2+1}{2}$ in general. If you are interested in an evidence that infinite many solutions probably exist, I can undelete my answer.
$endgroup$
– Peter
Dec 30 '18 at 10:44
$begingroup$
@Peter I wanted to see if there was a nice necessary and sufficient criterion that could be used to describe all solutions to the equation, but I now know that's probably not going to happen. I'm certainly interested in anything related to the problem though, and your answer is welcome :)
$endgroup$
– YiFan
Dec 30 '18 at 11:21
7
7
$begingroup$
oeis.org/A048161
$endgroup$
– mathlove
Dec 29 '18 at 5:51
$begingroup$
oeis.org/A048161
$endgroup$
– mathlove
Dec 29 '18 at 5:51
3
3
$begingroup$
You should definitely look at the link in the first comment. Whether there are infinitely many such $p$ is unknown... BTW if $p>5$ then $pequiv pm 1 mod 10,$ for if $5<pequiv pm 3 mod 10$ then $ ( p^2+1)/2$ is greater than $5$ and divisible by $5.$
$endgroup$
– DanielWainfleet
Dec 29 '18 at 6:28
$begingroup$
You should definitely look at the link in the first comment. Whether there are infinitely many such $p$ is unknown... BTW if $p>5$ then $pequiv pm 1 mod 10,$ for if $5<pequiv pm 3 mod 10$ then $ ( p^2+1)/2$ is greater than $5$ and divisible by $5.$
$endgroup$
– DanielWainfleet
Dec 29 '18 at 6:28
$begingroup$
@mathlove Wow, thanks. If you make that an answer I would happily accept it :)
$endgroup$
– YiFan
Dec 29 '18 at 8:50
$begingroup$
@mathlove Wow, thanks. If you make that an answer I would happily accept it :)
$endgroup$
– YiFan
Dec 29 '18 at 8:50
$begingroup$
@YiFan Do you want to find all the solutions or only decide whether infinite many exist ? Besides some basic restrictions, there won't be useful methods. You will just have to check the number $frac{p^2+1}{2}$ in general. If you are interested in an evidence that infinite many solutions probably exist, I can undelete my answer.
$endgroup$
– Peter
Dec 30 '18 at 10:44
$begingroup$
@YiFan Do you want to find all the solutions or only decide whether infinite many exist ? Besides some basic restrictions, there won't be useful methods. You will just have to check the number $frac{p^2+1}{2}$ in general. If you are interested in an evidence that infinite many solutions probably exist, I can undelete my answer.
$endgroup$
– Peter
Dec 30 '18 at 10:44
$begingroup$
@Peter I wanted to see if there was a nice necessary and sufficient criterion that could be used to describe all solutions to the equation, but I now know that's probably not going to happen. I'm certainly interested in anything related to the problem though, and your answer is welcome :)
$endgroup$
– YiFan
Dec 30 '18 at 11:21
$begingroup$
@Peter I wanted to see if there was a nice necessary and sufficient criterion that could be used to describe all solutions to the equation, but I now know that's probably not going to happen. I'm certainly interested in anything related to the problem though, and your answer is welcome :)
$endgroup$
– YiFan
Dec 30 '18 at 11:21
add a comment |
2 Answers
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We have infinitely many primes of the form $$frac{p^2+1}{2}$$ where $p$ is itself prime, if we have infinite many positive integers $k$ such that $$2k+1$$ and $$2k^2+2k+1$$ are simultaneously prime. (in this case, just set $p=2k+1$). The Bunyakovsky conjecture implies that this is the case, so very likely infinitely many examples exist. But I am convinced that the problem is open.
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add a comment |
$begingroup$
I propose the following "kind of parametrization". Remark first, as you did, that necessarily $qequiv 1$ mod $4$ (this can be seen directly from the characterization of integers which are sums of squares, see e.g. Samuel's ANT, V-6). Then work in the Gauss ring $mathbf Z[i]$, which is a PID with units $u=pm 1,pm i$. For convenience, write $z'≈z$ for $z'=uz$. It is classically known that $2$ is totally ramified and $q$ splits completely in the Gauss ring, more precisely $2=(1+i)^2≈(1+i)(1-i)$ and $q≈(x+iy)(x-iy)$, where $xpm iy$ are prime elements of $mathbf Z[i]$. Because of unique factorization (up to units), the given equation will be equivalent to $p+i≈(1+i)(x+iy)$, hence our "pseudo-parametrization" will be as follows: 1) Take a prime $qequiv 1$ mod $4$ and decompose it in $mathbf Z[i]$; 2) Solve $p+i≈(1+i)(x+iy)$ in $mathbf Z^2$. Because of symmetry, let us solve only the equation $p+i=-(1+i)(x+iy)$. By identification, $p=y-x$ and $-1=x+y$, so $p=2y+1$. It remains however to check that $2y+1$ is a prime in $mathbf Z$, which is why our method cannot be considered as a genuine parametrization. Although we have just shown that the OP question is essentially equivalent to the Bunyakovsky conjecture evoked by @Peter, I don't know if this gives any new hint.
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add a comment |
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2 Answers
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2 Answers
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$begingroup$
We have infinitely many primes of the form $$frac{p^2+1}{2}$$ where $p$ is itself prime, if we have infinite many positive integers $k$ such that $$2k+1$$ and $$2k^2+2k+1$$ are simultaneously prime. (in this case, just set $p=2k+1$). The Bunyakovsky conjecture implies that this is the case, so very likely infinitely many examples exist. But I am convinced that the problem is open.
$endgroup$
add a comment |
$begingroup$
We have infinitely many primes of the form $$frac{p^2+1}{2}$$ where $p$ is itself prime, if we have infinite many positive integers $k$ such that $$2k+1$$ and $$2k^2+2k+1$$ are simultaneously prime. (in this case, just set $p=2k+1$). The Bunyakovsky conjecture implies that this is the case, so very likely infinitely many examples exist. But I am convinced that the problem is open.
$endgroup$
add a comment |
$begingroup$
We have infinitely many primes of the form $$frac{p^2+1}{2}$$ where $p$ is itself prime, if we have infinite many positive integers $k$ such that $$2k+1$$ and $$2k^2+2k+1$$ are simultaneously prime. (in this case, just set $p=2k+1$). The Bunyakovsky conjecture implies that this is the case, so very likely infinitely many examples exist. But I am convinced that the problem is open.
$endgroup$
We have infinitely many primes of the form $$frac{p^2+1}{2}$$ where $p$ is itself prime, if we have infinite many positive integers $k$ such that $$2k+1$$ and $$2k^2+2k+1$$ are simultaneously prime. (in this case, just set $p=2k+1$). The Bunyakovsky conjecture implies that this is the case, so very likely infinitely many examples exist. But I am convinced that the problem is open.
edited Jan 7 at 10:00
the_fox
2,90031538
2,90031538
answered Dec 29 '18 at 22:39
PeterPeter
48.9k1239136
48.9k1239136
add a comment |
add a comment |
$begingroup$
I propose the following "kind of parametrization". Remark first, as you did, that necessarily $qequiv 1$ mod $4$ (this can be seen directly from the characterization of integers which are sums of squares, see e.g. Samuel's ANT, V-6). Then work in the Gauss ring $mathbf Z[i]$, which is a PID with units $u=pm 1,pm i$. For convenience, write $z'≈z$ for $z'=uz$. It is classically known that $2$ is totally ramified and $q$ splits completely in the Gauss ring, more precisely $2=(1+i)^2≈(1+i)(1-i)$ and $q≈(x+iy)(x-iy)$, where $xpm iy$ are prime elements of $mathbf Z[i]$. Because of unique factorization (up to units), the given equation will be equivalent to $p+i≈(1+i)(x+iy)$, hence our "pseudo-parametrization" will be as follows: 1) Take a prime $qequiv 1$ mod $4$ and decompose it in $mathbf Z[i]$; 2) Solve $p+i≈(1+i)(x+iy)$ in $mathbf Z^2$. Because of symmetry, let us solve only the equation $p+i=-(1+i)(x+iy)$. By identification, $p=y-x$ and $-1=x+y$, so $p=2y+1$. It remains however to check that $2y+1$ is a prime in $mathbf Z$, which is why our method cannot be considered as a genuine parametrization. Although we have just shown that the OP question is essentially equivalent to the Bunyakovsky conjecture evoked by @Peter, I don't know if this gives any new hint.
$endgroup$
add a comment |
$begingroup$
I propose the following "kind of parametrization". Remark first, as you did, that necessarily $qequiv 1$ mod $4$ (this can be seen directly from the characterization of integers which are sums of squares, see e.g. Samuel's ANT, V-6). Then work in the Gauss ring $mathbf Z[i]$, which is a PID with units $u=pm 1,pm i$. For convenience, write $z'≈z$ for $z'=uz$. It is classically known that $2$ is totally ramified and $q$ splits completely in the Gauss ring, more precisely $2=(1+i)^2≈(1+i)(1-i)$ and $q≈(x+iy)(x-iy)$, where $xpm iy$ are prime elements of $mathbf Z[i]$. Because of unique factorization (up to units), the given equation will be equivalent to $p+i≈(1+i)(x+iy)$, hence our "pseudo-parametrization" will be as follows: 1) Take a prime $qequiv 1$ mod $4$ and decompose it in $mathbf Z[i]$; 2) Solve $p+i≈(1+i)(x+iy)$ in $mathbf Z^2$. Because of symmetry, let us solve only the equation $p+i=-(1+i)(x+iy)$. By identification, $p=y-x$ and $-1=x+y$, so $p=2y+1$. It remains however to check that $2y+1$ is a prime in $mathbf Z$, which is why our method cannot be considered as a genuine parametrization. Although we have just shown that the OP question is essentially equivalent to the Bunyakovsky conjecture evoked by @Peter, I don't know if this gives any new hint.
$endgroup$
add a comment |
$begingroup$
I propose the following "kind of parametrization". Remark first, as you did, that necessarily $qequiv 1$ mod $4$ (this can be seen directly from the characterization of integers which are sums of squares, see e.g. Samuel's ANT, V-6). Then work in the Gauss ring $mathbf Z[i]$, which is a PID with units $u=pm 1,pm i$. For convenience, write $z'≈z$ for $z'=uz$. It is classically known that $2$ is totally ramified and $q$ splits completely in the Gauss ring, more precisely $2=(1+i)^2≈(1+i)(1-i)$ and $q≈(x+iy)(x-iy)$, where $xpm iy$ are prime elements of $mathbf Z[i]$. Because of unique factorization (up to units), the given equation will be equivalent to $p+i≈(1+i)(x+iy)$, hence our "pseudo-parametrization" will be as follows: 1) Take a prime $qequiv 1$ mod $4$ and decompose it in $mathbf Z[i]$; 2) Solve $p+i≈(1+i)(x+iy)$ in $mathbf Z^2$. Because of symmetry, let us solve only the equation $p+i=-(1+i)(x+iy)$. By identification, $p=y-x$ and $-1=x+y$, so $p=2y+1$. It remains however to check that $2y+1$ is a prime in $mathbf Z$, which is why our method cannot be considered as a genuine parametrization. Although we have just shown that the OP question is essentially equivalent to the Bunyakovsky conjecture evoked by @Peter, I don't know if this gives any new hint.
$endgroup$
I propose the following "kind of parametrization". Remark first, as you did, that necessarily $qequiv 1$ mod $4$ (this can be seen directly from the characterization of integers which are sums of squares, see e.g. Samuel's ANT, V-6). Then work in the Gauss ring $mathbf Z[i]$, which is a PID with units $u=pm 1,pm i$. For convenience, write $z'≈z$ for $z'=uz$. It is classically known that $2$ is totally ramified and $q$ splits completely in the Gauss ring, more precisely $2=(1+i)^2≈(1+i)(1-i)$ and $q≈(x+iy)(x-iy)$, where $xpm iy$ are prime elements of $mathbf Z[i]$. Because of unique factorization (up to units), the given equation will be equivalent to $p+i≈(1+i)(x+iy)$, hence our "pseudo-parametrization" will be as follows: 1) Take a prime $qequiv 1$ mod $4$ and decompose it in $mathbf Z[i]$; 2) Solve $p+i≈(1+i)(x+iy)$ in $mathbf Z^2$. Because of symmetry, let us solve only the equation $p+i=-(1+i)(x+iy)$. By identification, $p=y-x$ and $-1=x+y$, so $p=2y+1$. It remains however to check that $2y+1$ is a prime in $mathbf Z$, which is why our method cannot be considered as a genuine parametrization. Although we have just shown that the OP question is essentially equivalent to the Bunyakovsky conjecture evoked by @Peter, I don't know if this gives any new hint.
edited Jan 7 at 9:14
answered Jan 7 at 8:45
nguyen quang donguyen quang do
9,0091724
9,0091724
add a comment |
add a comment |
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7
$begingroup$
oeis.org/A048161
$endgroup$
– mathlove
Dec 29 '18 at 5:51
3
$begingroup$
You should definitely look at the link in the first comment. Whether there are infinitely many such $p$ is unknown... BTW if $p>5$ then $pequiv pm 1 mod 10,$ for if $5<pequiv pm 3 mod 10$ then $ ( p^2+1)/2$ is greater than $5$ and divisible by $5.$
$endgroup$
– DanielWainfleet
Dec 29 '18 at 6:28
$begingroup$
@mathlove Wow, thanks. If you make that an answer I would happily accept it :)
$endgroup$
– YiFan
Dec 29 '18 at 8:50
$begingroup$
@YiFan Do you want to find all the solutions or only decide whether infinite many exist ? Besides some basic restrictions, there won't be useful methods. You will just have to check the number $frac{p^2+1}{2}$ in general. If you are interested in an evidence that infinite many solutions probably exist, I can undelete my answer.
$endgroup$
– Peter
Dec 30 '18 at 10:44
$begingroup$
@Peter I wanted to see if there was a nice necessary and sufficient criterion that could be used to describe all solutions to the equation, but I now know that's probably not going to happen. I'm certainly interested in anything related to the problem though, and your answer is welcome :)
$endgroup$
– YiFan
Dec 30 '18 at 11:21