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 $frac{1}{2+x+y} + frac{1+x}{2+y} + frac{1+y}{2+x} = 4$.
elementary-number-theory
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.
add a comment |
up vote
-1
down vote
favorite
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 $frac{1}{2+x+y} + frac{1+x}{2+y} + frac{1+y}{2+x} = 4$.
elementary-number-theory
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
add a comment |
up vote
-1
down vote
favorite
up vote
-1
down vote
favorite
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 $frac{1}{2+x+y} + frac{1+x}{2+y} + frac{1+y}{2+x} = 4$.
elementary-number-theory
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
elementary-number-theory
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
add a comment |
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
add a comment |
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 $N=4$.
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
add a comment |
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 $N=4$.
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
add a comment |
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 $N=4$.
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
add a comment |
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 $N=4$.
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 $N=4$.
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
add a comment |
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
add a comment |
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