Determine all triples $(m,n,p)$ of positive rational numbers.












0












$begingroup$


Determine all triples $(m,n,p)$ of positive rational numbers such that the numbers $m+frac{1}{np}, n+frac{1}{pm}, p+frac{1}{mn}$ are integers.



I have no idea how to go about. Please help.






elementary-number-theory rational-numbers







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$endgroup$












  • $begingroup$
    $m+frac{1}{n}p$ or $m+frac{1}{np}$?
    $endgroup$
    – Dietrich Burde
    Jan 12 at 16:42










  • $begingroup$
    The latter one.
    $endgroup$
    – Yellow
    Jan 12 at 16:45










  • $begingroup$
    @Yellow, could we speak in chat? chat.stackexchange.com/rooms/88182/room-for-haran-and-yellow
    $endgroup$
    – Haran
    Jan 12 at 17:19










  • $begingroup$
    @Yellow are you online?
    $endgroup$
    – Haran
    Jan 13 at 5:08
















0












$begingroup$


Determine all triples $(m,n,p)$ of positive rational numbers such that the numbers $m+frac{1}{np}, n+frac{1}{pm}, p+frac{1}{mn}$ are integers.



I have no idea how to go about. Please help.






elementary-number-theory rational-numbers







share|cite|improve this question











$endgroup$












  • $begingroup$
    $m+frac{1}{n}p$ or $m+frac{1}{np}$?
    $endgroup$
    – Dietrich Burde
    Jan 12 at 16:42










  • $begingroup$
    The latter one.
    $endgroup$
    – Yellow
    Jan 12 at 16:45










  • $begingroup$
    @Yellow, could we speak in chat? chat.stackexchange.com/rooms/88182/room-for-haran-and-yellow
    $endgroup$
    – Haran
    Jan 12 at 17:19










  • $begingroup$
    @Yellow are you online?
    $endgroup$
    – Haran
    Jan 13 at 5:08














0












0








0


1



$begingroup$


Determine all triples $(m,n,p)$ of positive rational numbers such that the numbers $m+frac{1}{np}, n+frac{1}{pm}, p+frac{1}{mn}$ are integers.



I have no idea how to go about. Please help.






elementary-number-theory rational-numbers







share|cite|improve this question











$endgroup$




Determine all triples $(m,n,p)$ of positive rational numbers such that the numbers $m+frac{1}{np}, n+frac{1}{pm}, p+frac{1}{mn}$ are integers.



I have no idea how to go about. Please help.






elementary-number-theory rational-numbers




elementary-number-theory rational-numbers






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 13 at 5:04









Haran

1,144424




1,144424










asked Jan 12 at 16:39









YellowYellow

16011




16011












  • $begingroup$
    $m+frac{1}{n}p$ or $m+frac{1}{np}$?
    $endgroup$
    – Dietrich Burde
    Jan 12 at 16:42










  • $begingroup$
    The latter one.
    $endgroup$
    – Yellow
    Jan 12 at 16:45










  • $begingroup$
    @Yellow, could we speak in chat? chat.stackexchange.com/rooms/88182/room-for-haran-and-yellow
    $endgroup$
    – Haran
    Jan 12 at 17:19










  • $begingroup$
    @Yellow are you online?
    $endgroup$
    – Haran
    Jan 13 at 5:08


















  • $begingroup$
    $m+frac{1}{n}p$ or $m+frac{1}{np}$?
    $endgroup$
    – Dietrich Burde
    Jan 12 at 16:42










  • $begingroup$
    The latter one.
    $endgroup$
    – Yellow
    Jan 12 at 16:45










  • $begingroup$
    @Yellow, could we speak in chat? chat.stackexchange.com/rooms/88182/room-for-haran-and-yellow
    $endgroup$
    – Haran
    Jan 12 at 17:19










  • $begingroup$
    @Yellow are you online?
    $endgroup$
    – Haran
    Jan 13 at 5:08
















$begingroup$
$m+frac{1}{n}p$ or $m+frac{1}{np}$?
$endgroup$
– Dietrich Burde
Jan 12 at 16:42




$begingroup$
$m+frac{1}{n}p$ or $m+frac{1}{np}$?
$endgroup$
– Dietrich Burde
Jan 12 at 16:42












$begingroup$
The latter one.
$endgroup$
– Yellow
Jan 12 at 16:45




$begingroup$
The latter one.
$endgroup$
– Yellow
Jan 12 at 16:45












$begingroup$
@Yellow, could we speak in chat? chat.stackexchange.com/rooms/88182/room-for-haran-and-yellow
$endgroup$
– Haran
Jan 12 at 17:19




$begingroup$
@Yellow, could we speak in chat? chat.stackexchange.com/rooms/88182/room-for-haran-and-yellow
$endgroup$
– Haran
Jan 12 at 17:19












$begingroup$
@Yellow are you online?
$endgroup$
– Haran
Jan 13 at 5:08




$begingroup$
@Yellow are you online?
$endgroup$
– Haran
Jan 13 at 5:08










1 Answer
1






active

oldest

votes


















2












$begingroup$

First, by multiplying, we can see that $frac{(mnp+1)^3}{(mnp)^2}$ is an integer. Consider $mnp = frac{X}{Y}$, where $gcd(X,Y)=1$. Then, you can see that $X=Y=1$. This gives you $mnp=1$. Replace $m=frac{w}{x}$, $n=frac{y}{z}$, $p=frac{xz}{wy}$, we can see using the three statements with divisibility solving that the only solutions are $(1,1,1)$ , $(2,1,frac{1}{2})$ and $(4,frac{1}{2},frac{1}{2})$, and all its symmetric permutations.



The statements we get are $x mid 2w$, $z mid 2y$ and $wy mid 2xz$. You can take casework to solve these.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    @Yellow can we speak in private chat?
    $endgroup$
    – Haran
    Jan 12 at 16:58






  • 2




    $begingroup$
    To add, this problem is from 2006 Balkan Mathematical Olympiad, proposed by Valentin Vornicu, and the solution outlined above, as far as I remember, goes back to Gabriel Dospinescu (aka harazi in AoPS forums).
    $endgroup$
    – Aaron
    Jan 12 at 17:45










  • $begingroup$
    Haran, in nowhere I wrote, I did say it is not yours. I said this outline is quite old, goes back to Harazi.
    $endgroup$
    – Aaron
    Jan 12 at 17:54










  • $begingroup$
    See AoPS (www.mathlinks.ro) forums, he's a member there, and had many questions in Romanian and USAMO olympiads. In fact, if you are into olympiad mathematics, I highly recommend that you get an account from there, it has a rich collection of many olympiad problems from all across the world.
    $endgroup$
    – Aaron
    Jan 12 at 18:04












  • $begingroup$
    I already have an account Aaron. Thanks for the recommendation though!
    $endgroup$
    – Haran
    Jan 12 at 18:05











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1 Answer
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active

oldest

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1 Answer
1






active

oldest

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active

oldest

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active

oldest

votes









2












$begingroup$

First, by multiplying, we can see that $frac{(mnp+1)^3}{(mnp)^2}$ is an integer. Consider $mnp = frac{X}{Y}$, where $gcd(X,Y)=1$. Then, you can see that $X=Y=1$. This gives you $mnp=1$. Replace $m=frac{w}{x}$, $n=frac{y}{z}$, $p=frac{xz}{wy}$, we can see using the three statements with divisibility solving that the only solutions are $(1,1,1)$ , $(2,1,frac{1}{2})$ and $(4,frac{1}{2},frac{1}{2})$, and all its symmetric permutations.



The statements we get are $x mid 2w$, $z mid 2y$ and $wy mid 2xz$. You can take casework to solve these.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    @Yellow can we speak in private chat?
    $endgroup$
    – Haran
    Jan 12 at 16:58






  • 2




    $begingroup$
    To add, this problem is from 2006 Balkan Mathematical Olympiad, proposed by Valentin Vornicu, and the solution outlined above, as far as I remember, goes back to Gabriel Dospinescu (aka harazi in AoPS forums).
    $endgroup$
    – Aaron
    Jan 12 at 17:45










  • $begingroup$
    Haran, in nowhere I wrote, I did say it is not yours. I said this outline is quite old, goes back to Harazi.
    $endgroup$
    – Aaron
    Jan 12 at 17:54










  • $begingroup$
    See AoPS (www.mathlinks.ro) forums, he's a member there, and had many questions in Romanian and USAMO olympiads. In fact, if you are into olympiad mathematics, I highly recommend that you get an account from there, it has a rich collection of many olympiad problems from all across the world.
    $endgroup$
    – Aaron
    Jan 12 at 18:04












  • $begingroup$
    I already have an account Aaron. Thanks for the recommendation though!
    $endgroup$
    – Haran
    Jan 12 at 18:05
















2












$begingroup$

First, by multiplying, we can see that $frac{(mnp+1)^3}{(mnp)^2}$ is an integer. Consider $mnp = frac{X}{Y}$, where $gcd(X,Y)=1$. Then, you can see that $X=Y=1$. This gives you $mnp=1$. Replace $m=frac{w}{x}$, $n=frac{y}{z}$, $p=frac{xz}{wy}$, we can see using the three statements with divisibility solving that the only solutions are $(1,1,1)$ , $(2,1,frac{1}{2})$ and $(4,frac{1}{2},frac{1}{2})$, and all its symmetric permutations.



The statements we get are $x mid 2w$, $z mid 2y$ and $wy mid 2xz$. You can take casework to solve these.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    @Yellow can we speak in private chat?
    $endgroup$
    – Haran
    Jan 12 at 16:58






  • 2




    $begingroup$
    To add, this problem is from 2006 Balkan Mathematical Olympiad, proposed by Valentin Vornicu, and the solution outlined above, as far as I remember, goes back to Gabriel Dospinescu (aka harazi in AoPS forums).
    $endgroup$
    – Aaron
    Jan 12 at 17:45










  • $begingroup$
    Haran, in nowhere I wrote, I did say it is not yours. I said this outline is quite old, goes back to Harazi.
    $endgroup$
    – Aaron
    Jan 12 at 17:54










  • $begingroup$
    See AoPS (www.mathlinks.ro) forums, he's a member there, and had many questions in Romanian and USAMO olympiads. In fact, if you are into olympiad mathematics, I highly recommend that you get an account from there, it has a rich collection of many olympiad problems from all across the world.
    $endgroup$
    – Aaron
    Jan 12 at 18:04












  • $begingroup$
    I already have an account Aaron. Thanks for the recommendation though!
    $endgroup$
    – Haran
    Jan 12 at 18:05














2












2








2





$begingroup$

First, by multiplying, we can see that $frac{(mnp+1)^3}{(mnp)^2}$ is an integer. Consider $mnp = frac{X}{Y}$, where $gcd(X,Y)=1$. Then, you can see that $X=Y=1$. This gives you $mnp=1$. Replace $m=frac{w}{x}$, $n=frac{y}{z}$, $p=frac{xz}{wy}$, we can see using the three statements with divisibility solving that the only solutions are $(1,1,1)$ , $(2,1,frac{1}{2})$ and $(4,frac{1}{2},frac{1}{2})$, and all its symmetric permutations.



The statements we get are $x mid 2w$, $z mid 2y$ and $wy mid 2xz$. You can take casework to solve these.






share|cite|improve this answer









$endgroup$



First, by multiplying, we can see that $frac{(mnp+1)^3}{(mnp)^2}$ is an integer. Consider $mnp = frac{X}{Y}$, where $gcd(X,Y)=1$. Then, you can see that $X=Y=1$. This gives you $mnp=1$. Replace $m=frac{w}{x}$, $n=frac{y}{z}$, $p=frac{xz}{wy}$, we can see using the three statements with divisibility solving that the only solutions are $(1,1,1)$ , $(2,1,frac{1}{2})$ and $(4,frac{1}{2},frac{1}{2})$, and all its symmetric permutations.



The statements we get are $x mid 2w$, $z mid 2y$ and $wy mid 2xz$. You can take casework to solve these.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Jan 12 at 16:56









HaranHaran

1,144424




1,144424












  • $begingroup$
    @Yellow can we speak in private chat?
    $endgroup$
    – Haran
    Jan 12 at 16:58






  • 2




    $begingroup$
    To add, this problem is from 2006 Balkan Mathematical Olympiad, proposed by Valentin Vornicu, and the solution outlined above, as far as I remember, goes back to Gabriel Dospinescu (aka harazi in AoPS forums).
    $endgroup$
    – Aaron
    Jan 12 at 17:45










  • $begingroup$
    Haran, in nowhere I wrote, I did say it is not yours. I said this outline is quite old, goes back to Harazi.
    $endgroup$
    – Aaron
    Jan 12 at 17:54










  • $begingroup$
    See AoPS (www.mathlinks.ro) forums, he's a member there, and had many questions in Romanian and USAMO olympiads. In fact, if you are into olympiad mathematics, I highly recommend that you get an account from there, it has a rich collection of many olympiad problems from all across the world.
    $endgroup$
    – Aaron
    Jan 12 at 18:04












  • $begingroup$
    I already have an account Aaron. Thanks for the recommendation though!
    $endgroup$
    – Haran
    Jan 12 at 18:05


















  • $begingroup$
    @Yellow can we speak in private chat?
    $endgroup$
    – Haran
    Jan 12 at 16:58






  • 2




    $begingroup$
    To add, this problem is from 2006 Balkan Mathematical Olympiad, proposed by Valentin Vornicu, and the solution outlined above, as far as I remember, goes back to Gabriel Dospinescu (aka harazi in AoPS forums).
    $endgroup$
    – Aaron
    Jan 12 at 17:45










  • $begingroup$
    Haran, in nowhere I wrote, I did say it is not yours. I said this outline is quite old, goes back to Harazi.
    $endgroup$
    – Aaron
    Jan 12 at 17:54










  • $begingroup$
    See AoPS (www.mathlinks.ro) forums, he's a member there, and had many questions in Romanian and USAMO olympiads. In fact, if you are into olympiad mathematics, I highly recommend that you get an account from there, it has a rich collection of many olympiad problems from all across the world.
    $endgroup$
    – Aaron
    Jan 12 at 18:04












  • $begingroup$
    I already have an account Aaron. Thanks for the recommendation though!
    $endgroup$
    – Haran
    Jan 12 at 18:05
















$begingroup$
@Yellow can we speak in private chat?
$endgroup$
– Haran
Jan 12 at 16:58




$begingroup$
@Yellow can we speak in private chat?
$endgroup$
– Haran
Jan 12 at 16:58




2




2




$begingroup$
To add, this problem is from 2006 Balkan Mathematical Olympiad, proposed by Valentin Vornicu, and the solution outlined above, as far as I remember, goes back to Gabriel Dospinescu (aka harazi in AoPS forums).
$endgroup$
– Aaron
Jan 12 at 17:45




$begingroup$
To add, this problem is from 2006 Balkan Mathematical Olympiad, proposed by Valentin Vornicu, and the solution outlined above, as far as I remember, goes back to Gabriel Dospinescu (aka harazi in AoPS forums).
$endgroup$
– Aaron
Jan 12 at 17:45












$begingroup$
Haran, in nowhere I wrote, I did say it is not yours. I said this outline is quite old, goes back to Harazi.
$endgroup$
– Aaron
Jan 12 at 17:54




$begingroup$
Haran, in nowhere I wrote, I did say it is not yours. I said this outline is quite old, goes back to Harazi.
$endgroup$
– Aaron
Jan 12 at 17:54












$begingroup$
See AoPS (www.mathlinks.ro) forums, he's a member there, and had many questions in Romanian and USAMO olympiads. In fact, if you are into olympiad mathematics, I highly recommend that you get an account from there, it has a rich collection of many olympiad problems from all across the world.
$endgroup$
– Aaron
Jan 12 at 18:04






$begingroup$
See AoPS (www.mathlinks.ro) forums, he's a member there, and had many questions in Romanian and USAMO olympiads. In fact, if you are into olympiad mathematics, I highly recommend that you get an account from there, it has a rich collection of many olympiad problems from all across the world.
$endgroup$
– Aaron
Jan 12 at 18:04














$begingroup$
I already have an account Aaron. Thanks for the recommendation though!
$endgroup$
– Haran
Jan 12 at 18:05




$begingroup$
I already have an account Aaron. Thanks for the recommendation though!
$endgroup$
– Haran
Jan 12 at 18:05


















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