Why does Solve lock up when trying to solve the quadratic equation with large integers?

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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







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$endgroup$







  • 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
    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 = 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




    $begingroup$
    @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















4












$begingroup$


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







share|improve this question











$endgroup$







  • 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
    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 = 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




    $begingroup$
    @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













4












4








4


1



$begingroup$


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







share|improve this question











$endgroup$




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






share|improve this question















share|improve this question













share|improve this question




share|improve this question








edited Mar 11 at 6:47









J. M. is away♦

98.9k10311467




98.9k10311467










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 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 = 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




    $begingroup$
    @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












  • 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
    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 = 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




    $begingroup$
    @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







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
Mar 10 at 18:27




$begingroup$
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




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










1 Answer
1






active

oldest

votes


















8












$begingroup$

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.






share|improve this answer









$endgroup$












  • $begingroup$
    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











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1 Answer
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active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









8












$begingroup$

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.






share|improve this answer









$endgroup$












  • $begingroup$
    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















8












$begingroup$

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.






share|improve this answer









$endgroup$












  • $begingroup$
    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













8












8








8





$begingroup$

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.






share|improve this answer









$endgroup$



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.







share|improve this answer












share|improve this answer



share|improve this answer










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
















  • $begingroup$
    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

















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