Find all positive integral soludions of $frac{a}{b+c} + frac{b}{a+c} + frac{c}{a+b} = 4$ [duplicate]











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  • Find answer of $frac{x}{y+z}+frac{y}{x+z}+frac{z}{y+x}=4$

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Given an equation $frac{a}{b+c} + frac{b}{a+c} + frac{c}{a+b} = 4$, how to solve this problem in positive integers?



I've tried to assume $ale ble c$ and that $b=a+k_1, c=a+k_2$. So the equation become



$$frac{a}{2a+k_1+k_2} + frac{a+k_1}{2a+k_2} + frac{a+k_2}{2a+k_1} = 4$$



or equivalently,



$$frac{1}{2+frac{k_1}{a}+frac{k_2}{a}} + frac{1+frac{k_1}{a}}{2+frac{k_2}{a}} + frac{1+frac{k_2}{a}}{2+frac{k_1}{a}} = 4$$



Now let $x= frac{k_1}{a}, y= frac{k_2}{a}$, it is sufficient to find all positive rational solutions of $frac{1}{2+x+y} + frac{1+x}{2+y} + frac{1+y}{2+x} = 4$.










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marked as duplicate by Watson, Lord Shark the Unknown, amWhy, Lord_Farin, Leucippus Nov 29 at 0:25


This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.















  • What have you attempted thus far?
    – user477343
    Feb 18 at 10:02










  • Please give your inputs on this problem.
    – Agile_Eagle
    Feb 18 at 10:23















up vote
-1
down vote

favorite
2













This question already has an answer here:




  • Find answer of $frac{x}{y+z}+frac{y}{x+z}+frac{z}{y+x}=4$

    2 answers




Given an equation $frac{a}{b+c} + frac{b}{a+c} + frac{c}{a+b} = 4$, how to solve this problem in positive integers?



I've tried to assume $ale ble c$ and that $b=a+k_1, c=a+k_2$. So the equation become



$$frac{a}{2a+k_1+k_2} + frac{a+k_1}{2a+k_2} + frac{a+k_2}{2a+k_1} = 4$$



or equivalently,



$$frac{1}{2+frac{k_1}{a}+frac{k_2}{a}} + frac{1+frac{k_1}{a}}{2+frac{k_2}{a}} + frac{1+frac{k_2}{a}}{2+frac{k_1}{a}} = 4$$



Now let $x= frac{k_1}{a}, y= frac{k_2}{a}$, it is sufficient to find all positive rational solutions of $frac{1}{2+x+y} + frac{1+x}{2+y} + frac{1+y}{2+x} = 4$.










share|cite|improve this question















marked as duplicate by Watson, Lord Shark the Unknown, amWhy, Lord_Farin, Leucippus Nov 29 at 0:25


This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.















  • What have you attempted thus far?
    – user477343
    Feb 18 at 10:02










  • Please give your inputs on this problem.
    – Agile_Eagle
    Feb 18 at 10:23













up vote
-1
down vote

favorite
2









up vote
-1
down vote

favorite
2






2






This question already has an answer here:




  • Find answer of $frac{x}{y+z}+frac{y}{x+z}+frac{z}{y+x}=4$

    2 answers




Given an equation $frac{a}{b+c} + frac{b}{a+c} + frac{c}{a+b} = 4$, how to solve this problem in positive integers?



I've tried to assume $ale ble c$ and that $b=a+k_1, c=a+k_2$. So the equation become



$$frac{a}{2a+k_1+k_2} + frac{a+k_1}{2a+k_2} + frac{a+k_2}{2a+k_1} = 4$$



or equivalently,



$$frac{1}{2+frac{k_1}{a}+frac{k_2}{a}} + frac{1+frac{k_1}{a}}{2+frac{k_2}{a}} + frac{1+frac{k_2}{a}}{2+frac{k_1}{a}} = 4$$



Now let $x= frac{k_1}{a}, y= frac{k_2}{a}$, it is sufficient to find all positive rational solutions of $frac{1}{2+x+y} + frac{1+x}{2+y} + frac{1+y}{2+x} = 4$.










share|cite|improve this question
















This question already has an answer here:




  • Find answer of $frac{x}{y+z}+frac{y}{x+z}+frac{z}{y+x}=4$

    2 answers




Given an equation $frac{a}{b+c} + frac{b}{a+c} + frac{c}{a+b} = 4$, how to solve this problem in positive integers?



I've tried to assume $ale ble c$ and that $b=a+k_1, c=a+k_2$. So the equation become



$$frac{a}{2a+k_1+k_2} + frac{a+k_1}{2a+k_2} + frac{a+k_2}{2a+k_1} = 4$$



or equivalently,



$$frac{1}{2+frac{k_1}{a}+frac{k_2}{a}} + frac{1+frac{k_1}{a}}{2+frac{k_2}{a}} + frac{1+frac{k_2}{a}}{2+frac{k_1}{a}} = 4$$



Now let $x= frac{k_1}{a}, y= frac{k_2}{a}$, it is sufficient to find all positive rational solutions of $frac{1}{2+x+y} + frac{1+x}{2+y} + frac{1+y}{2+x} = 4$.





This question already has an answer here:




  • Find answer of $frac{x}{y+z}+frac{y}{x+z}+frac{z}{y+x}=4$

    2 answers








elementary-number-theory






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edited Feb 18 at 10:16

























asked Feb 18 at 10:00









Victor Chen

572313




572313




marked as duplicate by Watson, Lord Shark the Unknown, amWhy, Lord_Farin, Leucippus Nov 29 at 0:25


This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.






marked as duplicate by Watson, Lord Shark the Unknown, amWhy, Lord_Farin, Leucippus Nov 29 at 0:25


This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.














  • What have you attempted thus far?
    – user477343
    Feb 18 at 10:02










  • Please give your inputs on this problem.
    – Agile_Eagle
    Feb 18 at 10:23


















  • What have you attempted thus far?
    – user477343
    Feb 18 at 10:02










  • Please give your inputs on this problem.
    – Agile_Eagle
    Feb 18 at 10:23
















What have you attempted thus far?
– user477343
Feb 18 at 10:02




What have you attempted thus far?
– user477343
Feb 18 at 10:02












Please give your inputs on this problem.
– Agile_Eagle
Feb 18 at 10:23




Please give your inputs on this problem.
– Agile_Eagle
Feb 18 at 10:23










1 Answer
1






active

oldest

votes

















up vote
5
down vote



accepted










I think we have to use a computer because the solutions are:
$a=color{red}{154476802108746166441951315019919837485664325669565431700026634898253202035277999}$
$b=color{orange}{36875131794129999827197811565225474825492979968971970996283137471637224634055579}$
$c=color{green}{4373612677928697257861252602371390152816537558161613618621437993378423467772036}$



This is not a joke. The three gargantuan numbers are the least positive integer that satisfies the equation.



Here you can find the demonstration about the number of digit that the solution should have in the general case of
$$frac{a}{b+c} + frac{b}{a+c} + frac{c}{a+b} = N$$
This MathOverflow link as a full discussion about $N=4$.






share|cite|improve this answer























  • If memory serves, there's a mathoverflow question answering, or trying to answer this question too.
    – Mike Miller
    Feb 18 at 10:21










  • @MikeMiller I found it. Thank you for the suggestion.
    – user507623
    Feb 18 at 10:22










  • Could you give me a link please? @Pippo
    – Victor Chen
    Feb 18 at 10:23


















1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes








up vote
5
down vote



accepted










I think we have to use a computer because the solutions are:
$a=color{red}{154476802108746166441951315019919837485664325669565431700026634898253202035277999}$
$b=color{orange}{36875131794129999827197811565225474825492979968971970996283137471637224634055579}$
$c=color{green}{4373612677928697257861252602371390152816537558161613618621437993378423467772036}$



This is not a joke. The three gargantuan numbers are the least positive integer that satisfies the equation.



Here you can find the demonstration about the number of digit that the solution should have in the general case of
$$frac{a}{b+c} + frac{b}{a+c} + frac{c}{a+b} = N$$
This MathOverflow link as a full discussion about $N=4$.






share|cite|improve this answer























  • If memory serves, there's a mathoverflow question answering, or trying to answer this question too.
    – Mike Miller
    Feb 18 at 10:21










  • @MikeMiller I found it. Thank you for the suggestion.
    – user507623
    Feb 18 at 10:22










  • Could you give me a link please? @Pippo
    – Victor Chen
    Feb 18 at 10:23















up vote
5
down vote



accepted










I think we have to use a computer because the solutions are:
$a=color{red}{154476802108746166441951315019919837485664325669565431700026634898253202035277999}$
$b=color{orange}{36875131794129999827197811565225474825492979968971970996283137471637224634055579}$
$c=color{green}{4373612677928697257861252602371390152816537558161613618621437993378423467772036}$



This is not a joke. The three gargantuan numbers are the least positive integer that satisfies the equation.



Here you can find the demonstration about the number of digit that the solution should have in the general case of
$$frac{a}{b+c} + frac{b}{a+c} + frac{c}{a+b} = N$$
This MathOverflow link as a full discussion about $N=4$.






share|cite|improve this answer























  • If memory serves, there's a mathoverflow question answering, or trying to answer this question too.
    – Mike Miller
    Feb 18 at 10:21










  • @MikeMiller I found it. Thank you for the suggestion.
    – user507623
    Feb 18 at 10:22










  • Could you give me a link please? @Pippo
    – Victor Chen
    Feb 18 at 10:23













up vote
5
down vote



accepted







up vote
5
down vote



accepted






I think we have to use a computer because the solutions are:
$a=color{red}{154476802108746166441951315019919837485664325669565431700026634898253202035277999}$
$b=color{orange}{36875131794129999827197811565225474825492979968971970996283137471637224634055579}$
$c=color{green}{4373612677928697257861252602371390152816537558161613618621437993378423467772036}$



This is not a joke. The three gargantuan numbers are the least positive integer that satisfies the equation.



Here you can find the demonstration about the number of digit that the solution should have in the general case of
$$frac{a}{b+c} + frac{b}{a+c} + frac{c}{a+b} = N$$
This MathOverflow link as a full discussion about $N=4$.






share|cite|improve this answer














I think we have to use a computer because the solutions are:
$a=color{red}{154476802108746166441951315019919837485664325669565431700026634898253202035277999}$
$b=color{orange}{36875131794129999827197811565225474825492979968971970996283137471637224634055579}$
$c=color{green}{4373612677928697257861252602371390152816537558161613618621437993378423467772036}$



This is not a joke. The three gargantuan numbers are the least positive integer that satisfies the equation.



Here you can find the demonstration about the number of digit that the solution should have in the general case of
$$frac{a}{b+c} + frac{b}{a+c} + frac{c}{a+b} = N$$
This MathOverflow link as a full discussion about $N=4$.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Feb 18 at 10:24

























answered Feb 18 at 10:15







user507623



















  • If memory serves, there's a mathoverflow question answering, or trying to answer this question too.
    – Mike Miller
    Feb 18 at 10:21










  • @MikeMiller I found it. Thank you for the suggestion.
    – user507623
    Feb 18 at 10:22










  • Could you give me a link please? @Pippo
    – Victor Chen
    Feb 18 at 10:23


















  • If memory serves, there's a mathoverflow question answering, or trying to answer this question too.
    – Mike Miller
    Feb 18 at 10:21










  • @MikeMiller I found it. Thank you for the suggestion.
    – user507623
    Feb 18 at 10:22










  • Could you give me a link please? @Pippo
    – Victor Chen
    Feb 18 at 10:23
















If memory serves, there's a mathoverflow question answering, or trying to answer this question too.
– Mike Miller
Feb 18 at 10:21




If memory serves, there's a mathoverflow question answering, or trying to answer this question too.
– Mike Miller
Feb 18 at 10:21












@MikeMiller I found it. Thank you for the suggestion.
– user507623
Feb 18 at 10:22




@MikeMiller I found it. Thank you for the suggestion.
– user507623
Feb 18 at 10:22












Could you give me a link please? @Pippo
– Victor Chen
Feb 18 at 10:23




Could you give me a link please? @Pippo
– Victor Chen
Feb 18 at 10:23



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