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
    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 = 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




    $begingroup$
    @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
















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
    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 = 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




    $begingroup$
    @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














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 6 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 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 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 = 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




    $begingroup$
    @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














  • 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 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 = 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




    $begingroup$
    @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








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










1 Answer
1






active

oldest

votes


















7












$begingroup$

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
    6 hours ago










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











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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









7












$begingroup$

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
    6 hours ago










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
















7












$begingroup$

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
    6 hours ago










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














7












7








7





$begingroup$

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 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


















  • $begingroup$
    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










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










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濃尾地震

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濃尾地震 『岐阜市街大地震之図』 歌川国利画 本震 発生日 1891年（明治24年）10月28日 発生時刻 6時38分50秒（JST） 震央 日本 岐阜県本巣郡西根尾村（現・本巣市） 北緯35度35分 東経136度20分（ 地図 ） 規模    マグニチュード（M）8.0 最大震度    震度7： 注1 福井県今立郡鯖江町、愛知県葉栗郡大田島村、東春日井郡勝川村 地震の種類 直下型地震 被害 死傷者数 死者7,273人、負傷者17,175人 注1：当時の震度階級では「激烈」 プロジェクト:地球科学 プロジェクト:災害 テンプレートを表示 濃尾地震 （のうびじしん）は、1891年（明治24年）10月28日に濃尾地方で発生した、日本史上最大の内陸地殻内地震。「 美濃・尾張地震 （みの・おわりじしん）」とも呼ばれている。辛卯の年に発生したことから 辛卯震災 と呼んでいる報告書もある。 目次 1 概要 1.1 震源断層 2 被害 3 各地の震度 4 前兆現象 5 報道 6 学術的な意義 7 地震防災 8 脚注 9 関連項目 10 外部リンク 概要 濃尾地震発生当時の根尾谷断層 濃尾地震を引き起こした根尾谷断層 写真中央を斜めに走る段差が根尾谷断層 濃尾地震は、1891年10月28日6時38分50秒に発生した。震源は、岐阜県本巣郡西根尾村（現・本巣市）、北緯35度35分、東経136度20分付近。河角廣 (1951) は岐阜市付近（北緯35.6度、東経136.6度）に震央を仮定し規模 M K = 7.0 を与え [1] 、マグニチュードは M = 8.4 に換算されているが、明治・大正期の地震については0.5程度大きく見積もられているとされる [2] 。また、震央距離と震度との関係など当時のデータから後にM8.0 [3] とも推定される。アメリカ地質調査所 (USGS) でも8.0としている [4] 。「根尾谷断層帯」が活動した典型的な内陸地殻内地震（いわゆる直下型地震）であり、これは記録が残っている日本の内陸...
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