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












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







share|improve this question









New contributor




user63373 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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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
    4 hours ago






  • 1




    $begingroup$
    I quickly get solutions for 2^257 (V11.3.0, macos).
    $endgroup$
    – Michael E2
    4 hours ago










  • $begingroup$
    Sorry, there was a typo - it should be 2^511
    $endgroup$
    – user63373
    4 hours ago






  • 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
    3 hours ago








  • 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
    3 hours ago
















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







share|improve this question









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.







$endgroup$








  • 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
    4 hours ago






  • 1




    $begingroup$
    I quickly get solutions for 2^257 (V11.3.0, macos).
    $endgroup$
    – Michael E2
    4 hours ago










  • $begingroup$
    Sorry, there was a typo - it should be 2^511
    $endgroup$
    – user63373
    4 hours ago






  • 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
    3 hours ago








  • 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
    3 hours ago














4












4








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







share|improve this question









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






share|improve this question









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.











share|improve this question









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.









share|improve this question




share|improve this question








edited 2 hours ago









mikado

6,6471929




6,6471929






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 5 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
    4 hours ago






  • 1




    $begingroup$
    I quickly get solutions for 2^257 (V11.3.0, macos).
    $endgroup$
    – Michael E2
    4 hours ago










  • $begingroup$
    Sorry, there was a typo - it should be 2^511
    $endgroup$
    – user63373
    4 hours ago






  • 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
    3 hours ago








  • 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
    3 hours ago














  • 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
    4 hours ago






  • 1




    $begingroup$
    I quickly get solutions for 2^257 (V11.3.0, macos).
    $endgroup$
    – Michael E2
    4 hours ago










  • $begingroup$
    Sorry, there was a typo - it should be 2^511
    $endgroup$
    – user63373
    4 hours ago






  • 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
    3 hours ago








  • 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
    3 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
4 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
4 hours ago




1




1




$begingroup$
I quickly get solutions for 2^257 (V11.3.0, macos).
$endgroup$
– Michael E2
4 hours ago




$begingroup$
I quickly get solutions for 2^257 (V11.3.0, macos).
$endgroup$
– Michael E2
4 hours ago












$begingroup$
Sorry, there was a typo - it should be 2^511
$endgroup$
– user63373
4 hours ago




$begingroup$
Sorry, there was a typo - it should be 2^511
$endgroup$
– user63373
4 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
3 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
3 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
3 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
3 hours ago










1 Answer
1






active

oldest

votes


















6












$begingroup$

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
    2 hours ago










  • $begingroup$
    @user63373 You're welcome. :)
    $endgroup$
    – Michael E2
    1 hour ago











Your Answer





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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









6












$begingroup$

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
    2 hours ago










  • $begingroup$
    @user63373 You're welcome. :)
    $endgroup$
    – Michael E2
    1 hour ago
















6












$begingroup$

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
    2 hours ago










  • $begingroup$
    @user63373 You're welcome. :)
    $endgroup$
    – Michael E2
    1 hour ago














6












6








6





$begingroup$

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 2 hours ago









Michael E2Michael E2

148k12198478




148k12198478












  • $begingroup$
    Thank you very much. This seems to be exactly what is happening. Much appreciated.
    $endgroup$
    – user63373
    2 hours ago










  • $begingroup$
    @user63373 You're welcome. :)
    $endgroup$
    – Michael E2
    1 hour ago


















  • $begingroup$
    Thank you very much. This seems to be exactly what is happening. Much appreciated.
    $endgroup$
    – user63373
    2 hours ago










  • $begingroup$
    @user63373 You're welcome. :)
    $endgroup$
    – Michael E2
    1 hour ago
















$begingroup$
Thank you very much. This seems to be exactly what is happening. Much appreciated.
$endgroup$
– user63373
2 hours ago




$begingroup$
Thank you very much. This seems to be exactly what is happening. Much appreciated.
$endgroup$
– user63373
2 hours ago












$begingroup$
@user63373 You're welcome. :)
$endgroup$
– Michael E2
1 hour ago




$begingroup$
@user63373 You're welcome. :)
$endgroup$
– Michael E2
1 hour ago










user63373 is a new contributor. Be nice, and check out our Code of Conduct.










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格利澤436b

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格利泽436b 太陽系外行星 太陽系外行星列表 艺术家笔下的格利泽436b 母恆星 母恆星 Gliese 436 星座 獅子座 赤經 ( α ) 11 h  42 m  11.0941 s [1] 赤緯 ( δ ) +26° 42′ 23.652″ [1] 距離 33.4 ly (10.2 pc) 光譜類型 M2.5 V [1] 軌道參數 半长轴 ( a ) 0.0291±0.0004 [2] AU 軌道離心率 ( e ) 0.150±0.012 [2] 公轉週期 ( P ) 2.643904±0.000005 [3] d (0.00723849 y) 軌道傾角 ( i ) 85.8 +0.21 −0.25 [3] ° 角距 ( θ ) 2.794 mas 近星點時間 ( T 0 ) 2,451,551.716 ±0.01 JD 半振幅 ( K ) 18.68±0.8 m/s 物理性质 质量 ( m ) 22.2±1.0 [2] M ⊕ 半徑 ( r ) 4.327±0.183 [2] [4] R ⊕ 密度 ( ρ ) 1.51 g cm -3 表面重力 ( g ) 1.18 g 溫度 ( T ) 712±36 [2] K 發現 發現時間 2004 發現者 巴特勒、沃格特、 馬西 et al. 發現方法 徑向速度、凌日法 發現地點 美國加利福尼亞州 發表論文 出版 其他名稱 Ross 905 b, GJ 436 b, [5] LTT 13213 b, GCTP 2704.10 b, LHS 310, AC+27:28217 b, Vyssotsky 616 b, HIP 57087 b, GEN# +9.80120068 b, LP 319-75 b, G 121-7 b, LSPM J1142+2642 b, 1RXS J114211.9+264328 b, ASCC 683818 b, G 147-68 b, UCAC2 41198281 b, BP...
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