Why does Solve lock up when trying to solve the quadratic equation with large integers?
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$begingroup$
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 number-theory diophantine-equations
edited Mar 11 at 6:47
J. M. is away♦
98.9k10311467
asked Mar 10 at 18:23
user63373user63373
233
$endgroup$
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 number-theory diophantine-equations
edited Mar 11 at 6:47
J. M. is away♦
98.9k10311467
asked Mar 10 at 18:23
user63373user63373
233
$endgroup$
1
$begingroup$
You can format inline code and code blocks by selecting the code and clicking thebutton 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
Mar 10 at 18:27
1
$begingroup$
I quickly get solutions for2^257(V11.3.0, macos).
$endgroup$
– Michael E2
Mar 10 at 18:29
$begingroup$
Sorry, there was a typo - it should be 2^511
$endgroup$
– user63373
Mar 10 at 19:21
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
Mar 10 at 19:38
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
Mar 10 at 19:57
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 number-theory diophantine-equations
edited Mar 11 at 6:47
J. M. is away♦
98.9k10311467
asked Mar 10 at 18:23
user63373user63373
233
$endgroup$
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 number-theory diophantine-equations
equation-solving number-theory diophantine-equations
edited Mar 11 at 6:47
J. M. is away♦
98.9k10311467
asked Mar 10 at 18:23
user63373user63373
233
edited Mar 11 at 6:47
J. M. is away♦
98.9k10311467
asked Mar 10 at 18:23
user63373user63373
233
edited Mar 11 at 6:47
J. M. is away♦
98.9k10311467
edited Mar 11 at 6:47
J. M. is away♦
98.9k10311467
edited Mar 11 at 6:47
J. M. is away♦
98.9k10311467
98.9k10311467
asked Mar 10 at 18:23
user63373user63373
233
asked Mar 10 at 18:23
user63373user63373
233
asked Mar 10 at 18:23
user63373user63373
233
233
1
$begingroup$
You can format inline code and code blocks by selecting the code and clicking thebutton 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
Mar 10 at 18:27
1
$begingroup$
I quickly get solutions for2^257(V11.3.0, macos).
$endgroup$
– Michael E2
Mar 10 at 18:29
$begingroup$
Sorry, there was a typo - it should be 2^511
$endgroup$
– user63373
Mar 10 at 19:21
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
Mar 10 at 19:38
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
Mar 10 at 19:57
add a comment |
1
$begingroup$
You can format inline code and code blocks by selecting the code and clicking thebutton 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
Mar 10 at 18:27
1
$begingroup$
I quickly get solutions for2^257(V11.3.0, macos).
$endgroup$
– Michael E2
Mar 10 at 18:29
$begingroup$
Sorry, there was a typo - it should be 2^511
$endgroup$
– user63373
Mar 10 at 19:21
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
Mar 10 at 19:38
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
Mar 10 at 19:57
1
1
$begingroup$
You can format inline code and code blocks by selecting the code and clicking the
? is useful for learning how to format your questions and answers. You may also find this meta Q&A helpful$endgroup$
– Michael E2
Mar 10 at 18:27
$begingroup$
You can format inline code and code blocks by selecting the code and clicking the
? is useful for learning how to format your questions and answers. You may also find this meta Q&A helpful$endgroup$
– Michael E2
Mar 10 at 18:27
1
1
$begingroup$
I quickly get solutions for
2^257 (V11.3.0, macos).$endgroup$
– Michael E2
Mar 10 at 18:29
$begingroup$
I quickly get solutions for
2^257 (V11.3.0, macos).$endgroup$
– Michael E2
Mar 10 at 18:29
$begingroup$
Sorry, there was a typo - it should be 2^511
$endgroup$
– user63373
Mar 10 at 19:21
$begingroup$
Sorry, there was a typo - it should be 2^511
$endgroup$
– user63373
Mar 10 at 19:21
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
Mar 10 at 19:38
$begingroup$
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
Mar 10 at 19:38
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
Mar 10 at 19:57
$begingroup$
@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
Mar 10 at 19:57
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 Mar 10 at 20:38
Michael E2Michael E2
150k12203482
$endgroup$
$begingroup$
Thank you very much. This seems to be exactly what is happening. Much appreciated.
$endgroup$
– user63373
Mar 10 at 21:19
$begingroup$
@user63373 You're welcome. :)
$endgroup$
– Michael E2
Mar 10 at 22:10
add a comment |
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1 Answer
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1 Answer
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$begingroup$
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 Mar 10 at 20:38
Michael E2Michael E2
150k12203482
$endgroup$
$begingroup$
Thank you very much. This seems to be exactly what is happening. Much appreciated.
$endgroup$
– user63373
Mar 10 at 21:19
$begingroup$
@user63373 You're welcome. :)
$endgroup$
– Michael E2
Mar 10 at 22:10
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 Mar 10 at 20:38
Michael E2Michael E2
150k12203482
$endgroup$
$begingroup$
Thank you very much. This seems to be exactly what is happening. Much appreciated.
$endgroup$
– user63373
Mar 10 at 21:19
$begingroup$
@user63373 You're welcome. :)
$endgroup$
– Michael E2
Mar 10 at 22:10
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 Mar 10 at 20:38
Michael E2Michael E2
150k12203482
$endgroup$
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 Mar 10 at 20:38
Michael E2Michael E2
150k12203482
answered Mar 10 at 20:38
Michael E2Michael E2
150k12203482
answered Mar 10 at 20:38
Michael E2Michael E2
150k12203482
answered Mar 10 at 20:38
Michael E2Michael E2
150k12203482
150k12203482
$begingroup$
Thank you very much. This seems to be exactly what is happening. Much appreciated.
$endgroup$
– user63373
Mar 10 at 21:19
$begingroup$
@user63373 You're welcome. :)
$endgroup$
– Michael E2
Mar 10 at 22:10
add a comment |
$begingroup$
Thank you very much. This seems to be exactly what is happening. Much appreciated.
$endgroup$
– user63373
Mar 10 at 21:19
$begingroup$
@user63373 You're welcome. :)
$endgroup$
– Michael E2
Mar 10 at 22:10
$begingroup$
Thank you very much. This seems to be exactly what is happening. Much appreciated.
$endgroup$
– user63373
Mar 10 at 21:19
$begingroup$
Thank you very much. This seems to be exactly what is happening. Much appreciated.
$endgroup$
– user63373
Mar 10 at 21:19
$begingroup$
@user63373 You're welcome. :)
$endgroup$
– Michael E2
Mar 10 at 22:10
$begingroup$
@user63373 You're welcome. :)
$endgroup$
– Michael E2
Mar 10 at 22:10
add a comment |
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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
Mar 10 at 18:27
1
$begingroup$
I quickly get solutions for
2^257(V11.3.0, macos).$endgroup$
– Michael E2
Mar 10 at 18:29
$begingroup$
Sorry, there was a typo - it should be 2^511
$endgroup$
– user63373
Mar 10 at 19:21
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
Mar 10 at 19:38
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
Mar 10 at 19:57