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edited Jan 7 2011 at 19:25 |
Can the number of solutions $xy(x-y-1)=n$ for $x,y,n \in Z$ be unbounded as n varies?
x,y are integral points on an Elliptic Curve and are easy to find using enumeration of divisors of n (assuming n can be factored).
If yes, will large number of solutions give moderate rank EC?
If one drops $-1$ i.e. $xy(x-y)=n$ the number of solutions can be unbounded via multiples of rational point(s) and then multiplying by a cube. (EDIT): Explanation): Another unbounded case for varying $a , n$ is $xy(x-y-a)=n$. If $(x,y)$ is on the curve then $(d x,d y)$ is on $xy(x-y-a d)=n d^3$. Find many rational points and multiply by a suitable $d$. Not using the group law seems quite tricky for me. The constant $-1$ was included on purpose in the initial post.
I would be interested in this computational experiment: find $n$ that gives a lot of solutions, say $100$ (I can't do it), check which points are linearly independent and this is a lower bound on the rank.
What I find intriguing is that all integral points in this model come from factorization/divisors only.
Current record is n=179071200 with 22 solutions with positive x,y. Due to Matthew Conroy.
Current record is n=391287046550400 with 26 solutions with positive x,y. Due to Aaron Meyerowitz
Current record is n=8659883232000 with 28 solutions with positive x,y. Found by Tapio Rajala.
Current record is n=2597882099904000 with 36 solutions with positive x,y. Found by Tapio Rajala.
EDIT: $ab(a+b+9)=195643523275200$ has 48 positive integer points. – Aaron Meyerowitz (note this is a different curve and 7 <= rank <= 13)
EDIT:
A variation: $(x^2-x-17)^2 - y^2 = n$ appears to be eligible for the same question. The quartic model is a difference of two squares and checking if the first square is of the form $x^2-x-17$ is easy.
Is it possible some relation in the primes or primes or divisors of certain form to produce records: Someone is trying in $\mathbb{Z}[t]$ Can the number of solutions xy(x−y−1)=n for x,y,n∈Z[t] be unbounded as n varies? ? Read an article I didn't quite understand about maximizing the Selmer rank by chosing the primes carefully.
EDIT: The curve was chosen at random just to give a clear computational challenge.
EDIT: On second thought, can a symbolic approach work? Set $n=d_1 d_2 ... d_k$ where d_i are variables. Pick, well, ?some 100? ($d_i$, $y_i$) for ($x$,$y$) (or a product of $d_i$ for $x$). The result is a nonlinear system (last time I tried this I failed to make it work in practice).
EDIT: Related search seems "thue mahler" equation'
Related: unboundedness of number of integral points on elliptic curves?
Crossposted on MATH.SE: http://math.stackexchange.com/questions/14932/can-the-number-of-solutions-xyx-y-1-n-for-x-y-n-in-z-be-unbounded-as-n
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edited Jan 3 2011 at 21:33 |
Can the number of solutions $xy(x-y-1)=n$ for $x,y,n \in Z$ be unbounded as n varies?
x,y are integral points on an Elliptic Curve and are easy to find using enumeration of divisors of n (assuming n can be factored).
If yes, will large number of solutions give moderate rank EC?
If one drops $-1$ i.e. $xy(x-y)=n$ the number of solutions can be unbounded via multiples of rational point(s) and then multiplying by a cube. (EDIT): Another unbounded case for varying $a , n$ is $xy(x-y-a)=n$. If $(x,y)$ is on the curve then $(d x,d y)$ is on $xy(x-y-a d)=n d^3$. Find many rational points and multiply by a suitable $d$. Not using the group law seems quite tricky for me. The constant $-1$ was included on purpose in the initial post.
I would be interested in this computational experiment: find $n$ that gives a lot of solutions, say $100$ (I can't do it), check which points are linearly independent and this is a lower bound on the rank.
What I find intriguing is that all integral points in this model come from factorization/divisors only.
Current record is n=179071200 with 22 solutions with positive x,y. Due to Matthew Conroy.
Current record is n=391287046550400 with 26 solutions with positive x,y. Due to Aaron Meyerowitz
Current record is n=8659883232000 with 28 solutions with positive x,y. Found by Tapio Rajala.
Current record is n=2597882099904000 with 36 solutions with positive x,y. Found by Tapio Rajala.
EDIT: $ab(a+b+9)=195643523275200$ has 48 positive integer points. – Aaron Meyerowitz (_note note this is a different curve and 7 <= rank <= 13_)
EDIT: A variation: $(x^2-x-17)^2 - y^2 = n$ appears to be eligible for the same question. The quartic model is a difference of two squares and checking if the first square is of the form $x^2-x-17$ is easy.
Is it possible some relation in the primes or primes of certain form to produce records? Read an article I didn't quite understand about maximizing the Selmer rank by chosing the primes carefully.
EDIT: The curve was chosen at random just to give a clear computational challenge.
EDIT: On second thought, can a symbolic approach work? Set $n=d_1 d_2 ... d_k$ where d_i are variables. Pick, well, ?some 100? ($d_i$, $y_i$) for ($x$,$y$) (or a product of $d_i$ for $x$). The result is a nonlinear system (last time I tried this I failed to make it work in practice).
EDIT: Related search seems "thue mahler" equation'
Related: unboundedness of number of integral points on elliptic curves?
Crossposted on MATH.SE: http://math.stackexchange.com/questions/14932/can-the-number-of-solutions-xyx-y-1-n-for-x-y-n-in-z-be-unbounded-as-n
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edited Jan 1 2011 at 16:45 |
Can the number of solutions $xy(x-y-1)=n$ for $x,y,n \in Z$ be unbounded as n varies?
x,y are integral points on an Elliptic Curve and are easy to find using enumeration of divisors of n (assuming n can be factored).
If yes, will large number of solutions give moderate rank EC?
If one drops $-1$ i.e. $xy(x-y)=n$ the number of solutions can be unbounded via multiples of rational point(s) and then multiplying by a cube. (EDIT): Another unbounded case for varying $a , n$ is $xy(x-y-a)=n$. If $(x,y)$ is on the curve then $(d x,d y)$ is on $xy(x-y-a d)=n d^3$. Find many rational points and multiply by a suitable $d$. Not using the group law seems quite tricky for me. The constant $-1$ was included on purpose in the initial post.
I would be interested in this computational experiment: find $n$ that gives a lot of solutions, say $100$ (I can't do it), check which points are linearly independent and this is a lower bound on the rank.
What I find intriguing is that all integral points in this model come from factorization/divisors only.
Current record is n=179071200 with 22 solutions with positive x,y. Due to Matthew Conroy.
Current record is n=391287046550400 with 26 solutions with positive x,y. Due to Aaron Meyerowitz
Current record is n=8659883232000 with 28 solutions with positive x,y. Found by Tapio Rajala.
Current record is n=2597882099904000 with 36 solutions with positive x,y. Found by Tapio Rajala.
EDIT: $ab(a+b+9)=195643523275200$ has 48 positive integer points. – Aaron Meyerowitz (_note this is a different curve and 7 <= rank <= 13 _)
EDIT: A variation: $(x^2-x-17)^2 - y^2 = n$ appears to be eligible for the same question. The quartic model is a difference of two squares and checking if the first square is of the form $x^2-x-17$ is easy.
Is it possible some relation in the primes or primes of certain form to produce records? Read an article I didn't quite understand about maximizing the Selmer rank by chosing the primes carefully.
EDIT: The curve was chosen at random just to give a clear computational challenge.
EDIT: On second thought, can a symbolic approach work? Set $n=d_1 d_2 ... d_k$ where d_i are variables. Pick, well, ?some 100? ($d_i$, $y_i$) for ($x$,$y$) (or a product of $d_i$ for $x$). The result is a nonlinear system (last time I tried this I failed to make it work in practice).
Related: unboundedness of number of integral points on elliptic curves?
Crossposted on MATH.SE: http://math.stackexchange.com/questions/14932/can-the-number-of-solutions-xyx-y-1-n-for-x-y-n-in-z-be-unbounded-as-n
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edited Dec 31 2010 at 7:04 |
Can the number of solutions $xy(x-y-1)=n$ for $x,y,n \in Z$ be unbounded as n varies?
x,y are integral points on an Elliptic Curve and are easy to find using enumeration of divisors of n (assuming n can be factored).
If yes, will large number of solutions give moderate rank EC?
If one drops $-1$ i.e. $xy(x-y)=n$ the number of solutions can be unbounded via multiples of rational point(s) and then multiplying by a cube. (EDIT): Another unbounded case for varying $a , n$ is $xy(x-y-a)=n$. If $(x,y)$ is on the curve then $(d x,d y)$ is on $xy(x-y-a d)=n d^3$. Find many rational points and multiply by a suitable $d$. Not using the group law seems quite tricky for me. The constant $-1$ was included on purpose in the initial post.
I would be interested in this computational experiment: find $n$ that gives a lot of solutions, say $100$ (I can't do it), check which points are linearly independent and this is a lower bound on the rank.
What I find intriguing is that all integral points in this model come from factorization/divisors only.
Current record is n=179071200 with 22 solutions with positive x,y. Due to Matthew Conroy.
Current record is n=391287046550400 with 26 solutions with positive x,y. Due to Aaron Meyerowitz
Current record is n=8659883232000 with 28 solutions with positive x,y. Found by Tapio Rajala.
EDIT: $ab(a+b+9)=195643523275200$ has 48 positive integer points. – Aaron Meyerowitz (_note this is a different curve and 7 <= rank <= 13 _)
EDIT: A variation: $(x^2-x-17)^2 - y^2 = n$ appears to be eligible for the same question. The quartic model is a difference of two squares and checking if the first square is of the form $x^2-x-17$ is easy.
Is it possible some relation in the primes or primes of certain form to produce records? Read an article I didn't quite understand about maximizing the Selmer rank by chosing the primes carefully.
EDIT: The curve was chosen at random just to give a clear computational challenge.
EDIT: On second thought, can a symbolic approach work? Set $n=d_1 d_2 ... d_k$ where d_i are variables. Pick, well, ?some 100? ($d_i$, $y_i$) for ($x$,$y$) (or a product of $d_i$ for $x$). The result is a nonlinear system (last time I tried this I failed to make it work in practice).
Related: unboundedness of number of integral points on elliptic curves?
Crossposted on MATH.SE: http://math.stackexchange.com/questions/14932/can-the-number-of-solutions-xyx-y-1-n-for-x-y-n-in-z-be-unbounded-as-n
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edited Dec 30 2010 at 8:43 |
Can the number of solutions $xy(x-y-1)=n$ for $x,y,n \in Z$ be unbounded as n varies?
x,y are integral points on an Elliptic Curve and are easy to find using enumeration of divisors of n (assuming n can be factored).
If yes, will large number of solutions give moderate rank EC?
If one drops $-1$ i.e. $xy(x-y)=n$ the number of solutions can be unbounded via multiples of rational point(s) and then multiplying by a cube.
I would be interested in this computational experiment: find $n$ that gives a lot of solutions, say $100$ (I can't do it), check which points are linearly independent and this is a lower bound on the rank.
What I find intriguing is that all integral points in this model come from factorization/divisors only.
Current record is n=179071200 with 22 solutions with positive x,y. Due to Matthew Conroy.
Current record is n=391287046550400 with 26 solutions with positive x,y. Due to Aaron Meyerowitz
Current record is n=8659883232000 with 28 solutions with positive x,y. Found by Tapio Rajala.
EDIT: $ab(a+b+9)=195643523275200$ has 48 positive integer points. – Aaron Meyerowitz (note _note this is a different curve and 7 <= rank <= 13 _)
EDIT: A variation: $(x^2-x-17)^2 - y^2 = n$ appears to be eligible for the same question. The quartic model is a difference of two squares and checking if the first square is of the form $x^2-x-17$ is easy.
Is it possible some relation in the primes or primes of certain form to produce records? Read an article I didn't quite understand about maximizing the Selmer rank by chosing the primes carefully.
EDIT: The curve was chosen at random just to give a clear computational challenge.
EDIT: On second thought, can a symbolic approach work? Set $n=d_1 d_2 ... d_k$ where d_i are variables. Pick, well, ?some 100? ($d_i$, $y_i$) for ($x$,$y$) (or a product of $d_i$ for $x$). The result is a nonlinear system (last time I tried this I failed to make it work in practice).
Related: unboundedness of number of integral points on elliptic curves?
Crossposted on MATH.SE: http://math.stackexchange.com/questions/14932/can-the-number-of-solutions-xyx-y-1-n-for-x-y-n-in-z-be-unbounded-as-n
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edited Dec 30 2010 at 7:50 |
Can the number of solutions $xy(x-y-1)=n$ for $x,y,n \in Z$ be unbounded as n varies?
x,y are integral points on an Elliptic Curve and are easy to find using enumeration of divisors of n (assuming n can be factored).
If yes, will large number of solutions give moderate rank EC?
If one drops $-1$ i.e. $xy(x-y)=n$ the number of solutions can be unbounded via multiples of rational point(s) and then multiplying by a cube.
I would be interested in this computational experiment: find $n$ that gives a lot of solutions, say $100$ (I can't do it), check which points are linearly independent and this is a lower bound on the rank.
What I find intriguing is that all integral points in this model come from factorization/divisors only.
Current record is n=179071200 with 22 solutions with positive x,y. Due to Matthew Conroy.
Current record is n=391287046550400 with 26 solutions with positive x,y. Due to Aaron Meyerowitz
Current record is n=8659883232000 with 28 solutions with positive x,y. Found by Tapio Rajala.
EDIT: $ab(a+b+9)=195643523275200$ has 48 positive integer points. – Aaron Meyerowitz (note this is a different curve)
EDIT: A variation: $(x^2-x-17)^2 - y^2 = n$ appears to be eligible for the same question. The quartic model is a difference of two squares and checking if the first square is of the form $x^2-x-17$ is easy.
Is it possible some relation in the primes or primes of certain form to produce records? Read an article I didn't quite understand about maximizing the Selmer rank by chosing the primes carefully.
EDIT: The curve was chosen at random just to give a clear computational challenge.
EDIT: On second thought, can a symbolic approach work? Set $n=d_1 d_2 ... d_k$ where d_i are variables. Pick, well, ?some 100? ($d_i$, $y_i$) for ($x$,$y$) (or a product of $d_i$ for $x$). The result is a nonlinear system (last time I tried this I failed to make it work in practice).
Related: unboundedness of number of integral points on elliptic curves?
Crossposted on MATH.SE: http://math.stackexchange.com/questions/14932/can-the-number-of-solutions-xyx-y-1-n-for-x-y-n-in-z-be-unbounded-as-n
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edited Dec 30 2010 at 7:34 |
Can the number of solutions $xy(x-y-1)=n$ for $x,y,n \in Z$ be unbounded as n varies?
x,y are integral points on an Elliptic Curve and are easy to find using enumeration of divisors of n (assuming n can be factored).
If yes, will large number of solutions give moderate rank EC?
If one drops $-1$ i.e. $xy(x-y)=n$ the number of solutions can be unbounded via multiples of rational point(s) and then multiplying by a cube.
I would be interested in this computational experiment: find $n$ that gives a lot of solutions, say $100$ (I can't do it), check which points are linearly independent and this is a lower bound on the rank.
What I find intriguing is that all integral points in this model come from factorization/divisors only.
Current record is n=179071200 with 22 solutions with positive x,y. Due to Matthew Conroy.
Current record is n=391287046550400 with 26 solutions with positive x,y. Due to Aaron Meyerowitz
Current record is n=8659883232000 with 28 solutions with positive x,y. Found by Tapio Rajala.
EDIT: $ab(a+b+9)=195643523275200$ has 48 positive integer points. – Aaron Meyerowitz (note this is a different curve)
EDIT: A variation: $(x^2-x-17)^2 - y^2 = n$ appears to be eligible for the same question. The quartic model is a difference of two squares and checking if the first square is of the form $x^2-x-17$ is easy.
Is it possible some relation in the primes or primes of certain form to produce records? Read an article I didn't quite understand about maximizing the Selmer rank by chosing the primes carefully.
EDIT: The curve was chosen at random just to give a clear computational challenge.
EDIT: On second thought, can a symbolic approach work? Set $n=d_1 d_2 ... d_k$ where d_i are variables. Pick, well, ?some 100? ($d_i$, $y_i$) for ($x$,$y$) (or a product of $d_i$ for $x$). The result is a nonlinear system (last time I tried this I failed to make it work in practice).
Crossposted on MATH.SE: http://math.stackexchange.com/questions/14932/can-the-number-of-solutions-xyx-y-1-n-for-x-y-n-in-z-be-unbounded-as-n
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edited Dec 28 2010 at 15:03 |
Can the number of solutions $xy(x-y-1)=n$ for $x,y,n \in Z$ be unbounded as n varies?
x,y are integral points on an Elliptic Curve and are easy to find using enumeration of divisors of n (assuming n can be factored).
If yes, will large number of solutions give moderate rank EC?
If one drops $-1$ i.e. $xy(x-y)=n$ the number of solutions can be unbounded via multiples of rational point(s) and then multiplying by a cube.
I would be interested in this computational experiment: find $n$ that gives a lot of solutions, say $100$ (I can't do it), check which points are linearly independent and this is a lower bound on the rank.
What I find intriguing is that all integral points in this model come from factorization/divisors only.
Current record is n=179071200 with 22 solutions with positive x,y. Due to Matthew Conroy.
Current record is n=391287046550400 with 26 solutions with positive x,y. Due to Aaron Meyerowitz
EDIT: $ab(a+b+9)=195643523275200$ has 48 positive integer points. – Aaron Meyerowitz (note this is a different curve)
EDIT: A variation: $(x^2-x-17)^2 - y^2 = n$ appears to be eligible for the same question. The quartic model is a difference of two squares and checking if the first square is of the form $x^2-x-17$ is easy.
Is it possible some relation in the primes or primes of certain form to produce records? Read an article I didn't quite understand about maximizing the Selmer rank by chosing the primes carefully.
EDIT: The curve was chosen at random just to give a clear computational challenge.
EDIT: On second thought, can a symbolic approach work? Set $n=d_1 d_2 ... d_k$ where d_i are variables. Pick, well, ?some 100? (d_i, d_j) $d_i$, $y_i$) for ($x$,$y$) (or a product of d_i). $d_i$ for $x$). The result is a nonlinear system (last time I tried this I failed to make it work in practice).
Crossposted on MATH.SE: http://math.stackexchange.com/questions/14932/can-the-number-of-solutions-xyx-y-1-n-for-x-y-n-in-z-be-unbounded-as-n
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edited Dec 28 2010 at 9:54 |
Can the number of solutions $xy(x-y-1)=n$ for $x,y,n \in Z$ be unbounded as n varies?
x,y are integral points on an Elliptic Curve and are easy to find using enumeration of divisors of n (assuming n can be factored).
If yes, will large number of solutions give moderate rank EC?
If one drops $-1$ i.e. $xy(x-y)=n$ the number of solutions can be unbounded via multiples of rational point(s) and then multiplying by a cube.
I would be interested in this computational experiment: find $n$ that gives a lot of solutions, say $100$ (I can't do it), check which points are linearly independent and this is a lower bound on the rank.
What I find intriguing is that all integral points in this model come from factorization/divisors only.
Current record is n=179071200 with 22 solutions with positive x,y. Due to Matthew Conroy.
Current record is n=391287046550400 with 26 solutions with positive x,y. Due to Aaron Meyerowitz
EDIT: $ab(a+b+9)=195643523275200$ has 48 positive integer points. – Aaron Meyerowitz (note this is a different curve)
EDIT: A variation: $(x^2-x-17)^2 - y^2 = n$ appears to be eligible for the same question. The quartic model is a difference of two squares and checking if the first square is of the form $x^2-x-17$ is easy.
Is it possible some relation in the primes or primes of certain form to produce records? Read an article I didn't quite understand about maximizing the Selmer rank by chosing the primes carefully.
EDIT: The curve was chosen at random just to give a clear computational challenge.
EDIT: On second thought, can a symbolic approach work? Set $n=d_1 d_2 ... d_k$ where d_i are variables. Pick, well, ?some 100? (d_i, d_j) for ($x$,$y$) (or a product of d_i). The result is a nonlinear system (last time I tried this I failed to make it work in practice).
Crossposted on MATH.SE: http://math.stackexchange.com/questions/14932/can-the-number-of-solutions-xyx-y-1-n-for-x-y-n-in-z-be-unbounded-as-n
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edited Dec 28 2010 at 9:18 |
Can the number of solutions $xy(x-y-1)=n$ for $x,y,n \in Z$ be unbounded as n varies?
x,y are integral points on an Elliptic Curve and are easy to find using enumeration of divisors of n (assuming n can be factored).
If yes, will large number of solutions give moderate rank EC?
If one drops $-1$ i.e. $xy(x-y)=n$ the number of solutions can be unbounded via multiples of rational point(s) and then multiplying by a cube.
I would be interested in this computational experiment: find $n$ that gives a lot of solutions, say $100$ (I can't do it), check which points are linearly independent and this is a lower bound on the rank.
What I find intriguing is that all integral points in this model come from factorization/divisors only.
Current record is n=179071200 with 22 solutions with positive x,y. Due to Matthew Conroy.
Current record is n=391287046550400 with 26 solutions with positive x,y. Due to Aaron Meyerowitz
EDIT: $ab(a+b+9)=195643523275200$ has 48 positive integer points. – Aaron Meyerowitz (note this is a different curve)
EDIT: A variation: $(x^2-x-17)^2 - y^2 = n$ appears to be eligible for the same question. The quartic model is a difference of two squares and checking if the first square is of the form $x^2-x-17$ is easy.
Is it possible some relation in the primes or primes of certain form to produce records? Read an article I didn't quite understand about maximizing the Selmer rank by chosing the primes carefully.
EDIT: The curve was chosen at random just to give a clear computational challenge.
Crossposted on MATH.SE: http://math.stackexchange.com/questions/14932/can-the-number-of-solutions-xyx-y-1-n-for-x-y-n-in-z-be-unbounded-as-n
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edited Dec 28 2010 at 6:37 |
Can the number of solutions $xy(x-y-1)=n$ for $x,y,n \in Z$ be unbounded as n varies?
x,y are integral points on an Elliptic Curve and are easy to find using enumeration of divisors of n (assuming n can be factored).
If yes, will large number of solutions give moderate rank EC?
If one drops $-1$ i.e. $xy(x-y)=n$ the answer is "yes" number of solutions can be unbounded via multiples of rational point(s) and then multiplying by a cube.
I would be interested in this computational experiment: find $n$ that gives a lot of solutions, say $100$ (I can't do it), check which points are linearly independent and this is a lower bound on the rank.
What I find intriguing is that all integral points in this model come from factorization/divisors only.
Current record is n=179071200 with 22 solutions with positive x,y. Due to Matthew Conroy.
Current record is n=391287046550400 with 26 solutions with positive x,y. Due to Aaron Meyerowitz
Is it possible some relation in the primes or primes of certain form to produce records? Read an article I didn't quite understand about maximizing the Selmer rank by chosing the primes carefully.
EDIT: The curve was chosen at random just to give a clear computational challenge.
Crossposted on MATH.SE: http://math.stackexchange.com/questions/14932/can-the-number-of-solutions-xyx-y-1-n-for-x-y-n-in-z-be-unbounded-as-n
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edited Dec 27 2010 at 9:15 |
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asked Dec 27 2010 at 8:33 |
Can the number of solutions $xy(x-y-1)=n$ for $x,y,n \in Z$ be unbounded as n varies?
Can the number of solutions $xy(x-y-1)=n$ for $x,y,n \in Z$ be unbounded as n varies?
x,y are integral points on an Elliptic Curve and are easy to find using enumeration of divisors of n (assuming n can be factored).
If yes, will large number of solutions give moderate rank EC?
If one drops $-1$ i.e. $xy(x-y)=n$ the answer is "yes" via multiples of rational point(s) and then multiplying by a cube.
I would be interested in this computational experiment: find $n$ that gives a lot of solutions, say $100$ (I can't do it), check which points are linearly independent and this is a lower bound on the rank.
What I find intriguing is that all integral points in this model come from factorization/divisors only.
Current record is n=179071200 with 22 solutions with positive x,y. Due to Matthew Conroy.
Is it possible some relation in the primes or primes of certain form to produce records? Read an article I didn't quite understand about maximizing the Selmer rank by chosing the primes carefully.
Crossposted on MATH.SE: http://math.stackexchange.com/questions/14932/can-the-number-of-solutions-xyx-y-1-n-for-x-y-n-in-z-be-unbounded-as-n
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