Mixing
Determine all triples $(m,n,p)$ of positive rational numbers.
$begingroup$
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
edited Jan 13 at 5:04
Haran
1,076323
asked Jan 12 at 16:39
YellowYellow
16011
$endgroup$
add a comment |
$begingroup$
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
edited Jan 13 at 5:04
Haran
1,076323
asked Jan 12 at 16:39
YellowYellow
16011
$endgroup$
$begingroup$
$m+frac{1}{n}p$ 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
add a comment |
$begingroup$
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
edited Jan 13 at 5:04
Haran
1,076323
asked Jan 12 at 16:39
YellowYellow
16011
$endgroup$
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
edited Jan 13 at 5:04
Haran
1,076323
asked Jan 12 at 16:39
YellowYellow
16011
edited Jan 13 at 5:04
Haran
1,076323
asked Jan 12 at 16:39
YellowYellow
16011
edited Jan 13 at 5:04
Haran
1,076323
edited Jan 13 at 5:04
Haran
1,076323
edited Jan 13 at 5:04
Haran
1,076323
1,076323
asked Jan 12 at 16:39
YellowYellow
16011
asked Jan 12 at 16:39
YellowYellow
16011
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
add a comment |
$begingroup$
$m+frac{1}{n}p$ 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
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
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.
answered Jan 12 at 16:56
HaranHaran
1,076323
$endgroup$
$begingroup$
@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
add a comment |
Your Answer
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
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.
answered Jan 12 at 16:56
HaranHaran
1,076323
$endgroup$
$begingroup$
@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
add a comment |
$begingroup$
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.
answered Jan 12 at 16:56
HaranHaran
1,076323
$endgroup$
$begingroup$
@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
add a comment |
$begingroup$
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.
answered Jan 12 at 16:56
HaranHaran
1,076323
$endgroup$
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.
answered Jan 12 at 16:56
HaranHaran
1,076323
answered Jan 12 at 16:56
HaranHaran
1,076323
answered Jan 12 at 16:56
HaranHaran
1,076323
answered Jan 12 at 16:56
HaranHaran
1,076323
1,076323
$begingroup$
@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
add a comment |
$begingroup$
@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
add a comment |
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$begingroup$
$m+frac{1}{n}p$ 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