Why does Solve lock up when trying to solve the quadratic equation with large integers?
$begingroup$
Why does Solve lock up when trying to solve the equation
Solve[(x^2+y^2)+(x+y)==2^511 && x>0 && y>0,{x,y},Integers]
It works up 2^185, but at higher powers of 2, it seems to stop processing. The program says it's running, but there is no solution after running overnight. Running Mathematica 11.3 on Windows 32-bit OS.
equation-solving
edited 6 hours ago
mikado
6,6471929
asked 8 hours ago
user63373user63373
233
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user63373 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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add a comment |
$begingroup$
Why does Solve lock up when trying to solve the equation
Solve[(x^2+y^2)+(x+y)==2^511 && x>0 && y>0,{x,y},Integers]
It works up 2^185, but at higher powers of 2, it seems to stop processing. The program says it's running, but there is no solution after running overnight. Running Mathematica 11.3 on Windows 32-bit OS.
equation-solving
edited 6 hours ago
mikado
6,6471929
asked 8 hours ago
user63373user63373
233
New contributor
user63373 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
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1
$begingroup$
You can format inline code and code blocks by selecting the code and clicking the{}button above the edit window. The edit window help button?is useful for learning how to format your questions and answers. You may also find this meta Q&A helpful
$endgroup$
– Michael E2
8 hours ago
1
$begingroup$
I quickly get solutions for2^257(V11.3.0, macos).
$endgroup$
– Michael E2
8 hours ago
$begingroup$
Sorry, there was a typo - it should be 2^511
$endgroup$
– user63373
7 hours ago
1
$begingroup$
Probably it's combinatorial blowup.n = 185gives 32 solutions butn = 257already gives 1024. Might be more work and memory to store the symbolics than your CPU can handle.
$endgroup$
– b3m2a1
7 hours ago
2
$begingroup$
@b3m2a1 I think the reasons probably have to do with number theory.n = 323produces 8192 solutions in 2.3s andn = 325produces only 128 solutions in 500s. The memory growth is quite low. I think forn = 511, you just have to wait long enough, and I can't predict how long that is.
$endgroup$
– Michael E2
7 hours ago
add a comment |
$begingroup$
Why does Solve lock up when trying to solve the equation
Solve[(x^2+y^2)+(x+y)==2^511 && x>0 && y>0,{x,y},Integers]
It works up 2^185, but at higher powers of 2, it seems to stop processing. The program says it's running, but there is no solution after running overnight. Running Mathematica 11.3 on Windows 32-bit OS.
equation-solving
edited 6 hours ago
mikado
6,6471929
asked 8 hours ago
user63373user63373
233
New contributor
user63373 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
Why does Solve lock up when trying to solve the equation
Solve[(x^2+y^2)+(x+y)==2^511 && x>0 && y>0,{x,y},Integers]
It works up 2^185, but at higher powers of 2, it seems to stop processing. The program says it's running, but there is no solution after running overnight. Running Mathematica 11.3 on Windows 32-bit OS.
equation-solving
equation-solving
edited 6 hours ago
mikado
6,6471929
asked 8 hours ago
user63373user63373
233
New contributor
user63373 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited 6 hours ago
mikado
6,6471929
asked 8 hours ago
user63373user63373
233
New contributor
user63373 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited 6 hours ago
mikado
6,6471929
edited 6 hours ago
mikado
6,6471929
edited 6 hours ago
mikado
6,6471929
6,6471929
asked 8 hours ago
user63373user63373
233
New contributor
user63373 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 8 hours ago
user63373user63373
233
asked 8 hours ago
user63373user63373
233
233
New contributor
user63373 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
user63373 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
user63373 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
1
$begingroup$
You can format inline code and code blocks by selecting the code and clicking the{}button above the edit window. The edit window help button?is useful for learning how to format your questions and answers. You may also find this meta Q&A helpful
$endgroup$
– Michael E2
8 hours ago
1
$begingroup$
I quickly get solutions for2^257(V11.3.0, macos).
$endgroup$
– Michael E2
8 hours ago
$begingroup$
Sorry, there was a typo - it should be 2^511
$endgroup$
– user63373
7 hours ago
1
$begingroup$
Probably it's combinatorial blowup.n = 185gives 32 solutions butn = 257already gives 1024. Might be more work and memory to store the symbolics than your CPU can handle.
$endgroup$
– b3m2a1
7 hours ago
2
$begingroup$
@b3m2a1 I think the reasons probably have to do with number theory.n = 323produces 8192 solutions in 2.3s andn = 325produces only 128 solutions in 500s. The memory growth is quite low. I think forn = 511, you just have to wait long enough, and I can't predict how long that is.
$endgroup$
– Michael E2
7 hours ago
add a comment |
1
$begingroup$
You can format inline code and code blocks by selecting the code and clicking the{}button above the edit window. The edit window help button?is useful for learning how to format your questions and answers. You may also find this meta Q&A helpful
$endgroup$
– Michael E2
8 hours ago
1
$begingroup$
I quickly get solutions for2^257(V11.3.0, macos).
$endgroup$
– Michael E2
8 hours ago
$begingroup$
Sorry, there was a typo - it should be 2^511
$endgroup$
– user63373
7 hours ago
1
$begingroup$
Probably it's combinatorial blowup.n = 185gives 32 solutions butn = 257already gives 1024. Might be more work and memory to store the symbolics than your CPU can handle.
$endgroup$
– b3m2a1
7 hours ago
2
$begingroup$
@b3m2a1 I think the reasons probably have to do with number theory.n = 323produces 8192 solutions in 2.3s andn = 325produces only 128 solutions in 500s. The memory growth is quite low. I think forn = 511, you just have to wait long enough, and I can't predict how long that is.
$endgroup$
– Michael E2
7 hours ago
1
1
$begingroup$
You can format inline code and code blocks by selecting the code and clicking the
{} button above the edit window. The edit window help button ? is useful for learning how to format your questions and answers. You may also find this meta Q&A helpful$endgroup$
– Michael E2
8 hours ago
$begingroup$
You can format inline code and code blocks by selecting the code and clicking the
{} button above the edit window. The edit window help button ? is useful for learning how to format your questions and answers. You may also find this meta Q&A helpful$endgroup$
– Michael E2
8 hours ago
1
1
$begingroup$
I quickly get solutions for
2^257 (V11.3.0, macos).$endgroup$
– Michael E2
8 hours ago
$begingroup$
I quickly get solutions for
2^257 (V11.3.0, macos).$endgroup$
– Michael E2
8 hours ago
$begingroup$
Sorry, there was a typo - it should be 2^511
$endgroup$
– user63373
7 hours ago
$begingroup$
Sorry, there was a typo - it should be 2^511
$endgroup$
– user63373
7 hours ago
1
1
$begingroup$
Probably it's combinatorial blowup.
n = 185 gives 32 solutions but n = 257 already gives 1024. Might be more work and memory to store the symbolics than your CPU can handle.$endgroup$
– b3m2a1
7 hours ago
$begingroup$
Probably it's combinatorial blowup.
n = 185 gives 32 solutions but n = 257 already gives 1024. Might be more work and memory to store the symbolics than your CPU can handle.$endgroup$
– b3m2a1
7 hours ago
2
2
$begingroup$
@b3m2a1 I think the reasons probably have to do with number theory.
n = 323 produces 8192 solutions in 2.3s and n = 325 produces only 128 solutions in 500s. The memory growth is quite low. I think for n = 511, you just have to wait long enough, and I can't predict how long that is.$endgroup$
– Michael E2
7 hours ago
$begingroup$
@b3m2a1 I think the reasons probably have to do with number theory.
n = 323 produces 8192 solutions in 2.3s and n = 325 produces only 128 solutions in 500s. The memory growth is quite low. I think for n = 511, you just have to wait long enough, and I can't predict how long that is.$endgroup$
– Michael E2
7 hours ago
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Here's a guess:
The Diophantine problem
$$ x^2+y^2+x+y=a$$
is equivalent, via $u=2x+1,v=2y+1$ to finding the odd solutions to
$$u^2+v^2=2+4a ,.$$
Whether Solve makes this transformation or not,
solving the Pythagorean equation can be done from the prime factorization of $2+4a$.
How long Solve takes thus might depend on how long it takes to factor $2+4a$.
This is not hard to verify:
Block[{FactorInteger = (Print["FactorInteger"[##]]; Abort) &},
PrintTemporary@Dynamic@Clock@Infinity;
Print[2 + 4 2^325];
Solve[(x^2 + y^2) + (x + y) == 2^325 && x > 0 && y > 0, {x, y}, Integers] // AbsoluteTiming
]
2734063405978764905465627783897026706691461788616515545532213258012441248999219
90402939147127881730
FactorInteger[
2734063405978764905465627783897026706691461788616515545532213258012441248999219
90402939147127881730]
$Aborted
Well, it turns out it takes about 500 sec. to factor 2 + 4 * 2^325, which is about how long it takes the Solve above to run.
answered 6 hours ago
Michael E2Michael E2
148k12198478
$endgroup$
$begingroup$
Thank you very much. This seems to be exactly what is happening. Much appreciated.
$endgroup$
– user63373
6 hours ago
$begingroup$
@user63373 You're welcome. :)
$endgroup$
– Michael E2
5 hours ago
add a comment |
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1 Answer
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1 Answer
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oldest
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votes
$begingroup$
Here's a guess:
The Diophantine problem
$$ x^2+y^2+x+y=a$$
is equivalent, via $u=2x+1,v=2y+1$ to finding the odd solutions to
$$u^2+v^2=2+4a ,.$$
Whether Solve makes this transformation or not,
solving the Pythagorean equation can be done from the prime factorization of $2+4a$.
How long Solve takes thus might depend on how long it takes to factor $2+4a$.
This is not hard to verify:
Block[{FactorInteger = (Print["FactorInteger"[##]]; Abort) &},
PrintTemporary@Dynamic@Clock@Infinity;
Print[2 + 4 2^325];
Solve[(x^2 + y^2) + (x + y) == 2^325 && x > 0 && y > 0, {x, y}, Integers] // AbsoluteTiming
]
2734063405978764905465627783897026706691461788616515545532213258012441248999219
90402939147127881730
FactorInteger[
2734063405978764905465627783897026706691461788616515545532213258012441248999219
90402939147127881730]
$Aborted
Well, it turns out it takes about 500 sec. to factor 2 + 4 * 2^325, which is about how long it takes the Solve above to run.
answered 6 hours ago
Michael E2Michael E2
148k12198478
$endgroup$
$begingroup$
Thank you very much. This seems to be exactly what is happening. Much appreciated.
$endgroup$
– user63373
6 hours ago
$begingroup$
@user63373 You're welcome. :)
$endgroup$
– Michael E2
5 hours ago
add a comment |
$begingroup$
Here's a guess:
The Diophantine problem
$$ x^2+y^2+x+y=a$$
is equivalent, via $u=2x+1,v=2y+1$ to finding the odd solutions to
$$u^2+v^2=2+4a ,.$$
Whether Solve makes this transformation or not,
solving the Pythagorean equation can be done from the prime factorization of $2+4a$.
How long Solve takes thus might depend on how long it takes to factor $2+4a$.
This is not hard to verify:
Block[{FactorInteger = (Print["FactorInteger"[##]]; Abort) &},
PrintTemporary@Dynamic@Clock@Infinity;
Print[2 + 4 2^325];
Solve[(x^2 + y^2) + (x + y) == 2^325 && x > 0 && y > 0, {x, y}, Integers] // AbsoluteTiming
]
2734063405978764905465627783897026706691461788616515545532213258012441248999219
90402939147127881730
FactorInteger[
2734063405978764905465627783897026706691461788616515545532213258012441248999219
90402939147127881730]
$Aborted
Well, it turns out it takes about 500 sec. to factor 2 + 4 * 2^325, which is about how long it takes the Solve above to run.
answered 6 hours ago
Michael E2Michael E2
148k12198478
$endgroup$
$begingroup$
Thank you very much. This seems to be exactly what is happening. Much appreciated.
$endgroup$
– user63373
6 hours ago
$begingroup$
@user63373 You're welcome. :)
$endgroup$
– Michael E2
5 hours ago
add a comment |
$begingroup$
Here's a guess:
The Diophantine problem
$$ x^2+y^2+x+y=a$$
is equivalent, via $u=2x+1,v=2y+1$ to finding the odd solutions to
$$u^2+v^2=2+4a ,.$$
Whether Solve makes this transformation or not,
solving the Pythagorean equation can be done from the prime factorization of $2+4a$.
How long Solve takes thus might depend on how long it takes to factor $2+4a$.
This is not hard to verify:
Block[{FactorInteger = (Print["FactorInteger"[##]]; Abort) &},
PrintTemporary@Dynamic@Clock@Infinity;
Print[2 + 4 2^325];
Solve[(x^2 + y^2) + (x + y) == 2^325 && x > 0 && y > 0, {x, y}, Integers] // AbsoluteTiming
]
2734063405978764905465627783897026706691461788616515545532213258012441248999219
90402939147127881730
FactorInteger[
2734063405978764905465627783897026706691461788616515545532213258012441248999219
90402939147127881730]
$Aborted
Well, it turns out it takes about 500 sec. to factor 2 + 4 * 2^325, which is about how long it takes the Solve above to run.
answered 6 hours ago
Michael E2Michael E2
148k12198478
$endgroup$
Here's a guess:
The Diophantine problem
$$ x^2+y^2+x+y=a$$
is equivalent, via $u=2x+1,v=2y+1$ to finding the odd solutions to
$$u^2+v^2=2+4a ,.$$
Whether Solve makes this transformation or not,
solving the Pythagorean equation can be done from the prime factorization of $2+4a$.
How long Solve takes thus might depend on how long it takes to factor $2+4a$.
This is not hard to verify:
Block[{FactorInteger = (Print["FactorInteger"[##]]; Abort) &},
PrintTemporary@Dynamic@Clock@Infinity;
Print[2 + 4 2^325];
Solve[(x^2 + y^2) + (x + y) == 2^325 && x > 0 && y > 0, {x, y}, Integers] // AbsoluteTiming
]
2734063405978764905465627783897026706691461788616515545532213258012441248999219
90402939147127881730
FactorInteger[
2734063405978764905465627783897026706691461788616515545532213258012441248999219
90402939147127881730]
$Aborted
Well, it turns out it takes about 500 sec. to factor 2 + 4 * 2^325, which is about how long it takes the Solve above to run.
answered 6 hours ago
Michael E2Michael E2
148k12198478
answered 6 hours ago
Michael E2Michael E2
148k12198478
answered 6 hours ago
Michael E2Michael E2
148k12198478
answered 6 hours ago
Michael E2Michael E2
148k12198478
148k12198478
$begingroup$
Thank you very much. This seems to be exactly what is happening. Much appreciated.
$endgroup$
– user63373
6 hours ago
$begingroup$
@user63373 You're welcome. :)
$endgroup$
– Michael E2
5 hours ago
add a comment |
$begingroup$
Thank you very much. This seems to be exactly what is happening. Much appreciated.
$endgroup$
– user63373
6 hours ago
$begingroup$
@user63373 You're welcome. :)
$endgroup$
– Michael E2
5 hours ago
$begingroup$
Thank you very much. This seems to be exactly what is happening. Much appreciated.
$endgroup$
– user63373
6 hours ago
$begingroup$
Thank you very much. This seems to be exactly what is happening. Much appreciated.
$endgroup$
– user63373
6 hours ago
$begingroup$
@user63373 You're welcome. :)
$endgroup$
– Michael E2
5 hours ago
$begingroup$
@user63373 You're welcome. :)
$endgroup$
– Michael E2
5 hours ago
add a comment |
user63373 is a new contributor. Be nice, and check out our Code of Conduct.
user63373 is a new contributor. Be nice, and check out our Code of Conduct.
user63373 is a new contributor. Be nice, and check out our Code of Conduct.
user63373 is a new contributor. Be nice, and check out our Code of Conduct.
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$begingroup$
You can format inline code and code blocks by selecting the code and clicking the
{}button above the edit window. The edit window help button?is useful for learning how to format your questions and answers. You may also find this meta Q&A helpful$endgroup$
– Michael E2
8 hours ago
1
$begingroup$
I quickly get solutions for
2^257(V11.3.0, macos).$endgroup$
– Michael E2
8 hours ago
$begingroup$
Sorry, there was a typo - it should be 2^511
$endgroup$
– user63373
7 hours ago
1
$begingroup$
Probably it's combinatorial blowup.
n = 185gives 32 solutions butn = 257already gives 1024. Might be more work and memory to store the symbolics than your CPU can handle.$endgroup$
– b3m2a1
7 hours ago
2
$begingroup$
@b3m2a1 I think the reasons probably have to do with number theory.
n = 323produces 8192 solutions in 2.3s andn = 325produces only 128 solutions in 500s. The memory growth is quite low. I think forn = 511, you just have to wait long enough, and I can't predict how long that is.$endgroup$
– Michael E2
7 hours ago