On $2xy+x+y+1 = 2ab+a+b$
$begingroup$
Let us suppose that $2xy+x+y = n$, where $x,y,n$ are positive integers and $x≥2$ and $y≥2$.
Let us choose x and y, so that $n+1 = 2ab+a+b$ , where $a, b$ are also positive integer numbers.
I was wondering what we can say about the relation between $x,y$ and $a,b$.
As $n$ and $n+1$ are very close, and as $xy > x+y$, there must be a maximal difference between $x$ and $a$, also between $y$ and $b$.
Did anyone read a paper about it, or have any suggestion on how much this difference can be?
elementary-number-theory
asked Nov 25 '18 at 12:22
TilsightTilsight
195
$endgroup$
add a comment |
$begingroup$
Let us suppose that $2xy+x+y = n$, where $x,y,n$ are positive integers and $x≥2$ and $y≥2$.
Let us choose x and y, so that $n+1 = 2ab+a+b$ , where $a, b$ are also positive integer numbers.
I was wondering what we can say about the relation between $x,y$ and $a,b$.
As $n$ and $n+1$ are very close, and as $xy > x+y$, there must be a maximal difference between $x$ and $a$, also between $y$ and $b$.
Did anyone read a paper about it, or have any suggestion on how much this difference can be?
elementary-number-theory
asked Nov 25 '18 at 12:22
TilsightTilsight
195
$endgroup$
add a comment |
$begingroup$
Let us suppose that $2xy+x+y = n$, where $x,y,n$ are positive integers and $x≥2$ and $y≥2$.
Let us choose x and y, so that $n+1 = 2ab+a+b$ , where $a, b$ are also positive integer numbers.
I was wondering what we can say about the relation between $x,y$ and $a,b$.
As $n$ and $n+1$ are very close, and as $xy > x+y$, there must be a maximal difference between $x$ and $a$, also between $y$ and $b$.
Did anyone read a paper about it, or have any suggestion on how much this difference can be?
elementary-number-theory
asked Nov 25 '18 at 12:22
TilsightTilsight
195
$endgroup$
Let us suppose that $2xy+x+y = n$, where $x,y,n$ are positive integers and $x≥2$ and $y≥2$.
Let us choose x and y, so that $n+1 = 2ab+a+b$ , where $a, b$ are also positive integer numbers.
I was wondering what we can say about the relation between $x,y$ and $a,b$.
As $n$ and $n+1$ are very close, and as $xy > x+y$, there must be a maximal difference between $x$ and $a$, also between $y$ and $b$.
Did anyone read a paper about it, or have any suggestion on how much this difference can be?
elementary-number-theory
elementary-number-theory
asked Nov 25 '18 at 12:22
TilsightTilsight
195
asked Nov 25 '18 at 12:22
TilsightTilsight
195
asked Nov 25 '18 at 12:22
TilsightTilsight
195
asked Nov 25 '18 at 12:22
TilsightTilsight
195
asked Nov 25 '18 at 12:22
TilsightTilsight
195
195
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
If you multiply by $2$ and add $1$ you get:$$(2x+1)(2y+1)+2=(2a+1)(2b+1)$$ so you are looking at the factorisation of successive odd numbers with the factors on the left both at least $5$ and the factors on the right both at least $3$.
If we say $x=y=2$ we get $27$ and to get the maximum difference (in fact this is the only option here) we choose $a=1, b=4$ corresponding to $3times 9$.
This should be enough clue to allow you to explore further.
There are other cases to consider and you might want $x$ and $y$ to be different. But $x=y=3m$ and $a=1, b=6m^2+2m$ are enough to show that for $x=y$ the differences can be arbitrarily large.
answered Nov 25 '18 at 12:42
Mark BennetMark Bennet
80.8k981179
$endgroup$
$begingroup$
Thanks. I need to think your answer over. But I was just thinking that ab can not be more than xy because then the difference between n and n+1 would me more than 1.This would limit the possible values, wouldn't it?
$endgroup$
– Tilsight
Nov 25 '18 at 14:08
$begingroup$
@Tilsight $441times 1=21times 21$ you have potentially large differences between pairs of factors even when the difference is zero.
$endgroup$
– Mark Bennet
Nov 25 '18 at 14:33
$begingroup$
Yes, but it can't be, because then n = 2*441*1+441+1, and n+1 is not 2*21*21+21+21. As the difference between n and n+1 is 1, my gut says that the difference between x and a (and y and b) has to be limited.
$endgroup$
– Tilsight
Nov 25 '18 at 16:01
$begingroup$
Using your example x=2, y=2, so n = 12. Here n+1=13, and 13 = 2*4*1+4+1, so a=4 and b=1. See that xy=ab here. My question is how much larger $a*b$ can be compared to xy maximum, in order to find an a,b that fulfils the conditions in the description. My feeling is that ab is either xy or x*y+1, if it were bigger, then it will be bigger than n+1.
$endgroup$
– Tilsight
Nov 25 '18 at 16:07
$begingroup$
@Tilsight Your question does not ask for the difference between the two products, but the difference between the individual numbers. If you want to compare $ab$ and $xy$ you are really asking how different the sums can be for similar products. Try something like $x=y=300$, and $a=1, b=60200$. All I have been trying to do is give you clues so you can explore this yourself - you really need to try some things out - you will learn next to nothing until you do.
$endgroup$
– Mark Bennet
Nov 25 '18 at 16:42
|
show 2 more comments
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1 Answer
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$begingroup$
If you multiply by $2$ and add $1$ you get:$$(2x+1)(2y+1)+2=(2a+1)(2b+1)$$ so you are looking at the factorisation of successive odd numbers with the factors on the left both at least $5$ and the factors on the right both at least $3$.
If we say $x=y=2$ we get $27$ and to get the maximum difference (in fact this is the only option here) we choose $a=1, b=4$ corresponding to $3times 9$.
This should be enough clue to allow you to explore further.
There are other cases to consider and you might want $x$ and $y$ to be different. But $x=y=3m$ and $a=1, b=6m^2+2m$ are enough to show that for $x=y$ the differences can be arbitrarily large.
answered Nov 25 '18 at 12:42
Mark BennetMark Bennet
80.8k981179
$endgroup$
$begingroup$
Thanks. I need to think your answer over. But I was just thinking that ab can not be more than xy because then the difference between n and n+1 would me more than 1.This would limit the possible values, wouldn't it?
$endgroup$
– Tilsight
Nov 25 '18 at 14:08
$begingroup$
@Tilsight $441times 1=21times 21$ you have potentially large differences between pairs of factors even when the difference is zero.
$endgroup$
– Mark Bennet
Nov 25 '18 at 14:33
$begingroup$
Yes, but it can't be, because then n = 2*441*1+441+1, and n+1 is not 2*21*21+21+21. As the difference between n and n+1 is 1, my gut says that the difference between x and a (and y and b) has to be limited.
$endgroup$
– Tilsight
Nov 25 '18 at 16:01
$begingroup$
Using your example x=2, y=2, so n = 12. Here n+1=13, and 13 = 2*4*1+4+1, so a=4 and b=1. See that xy=ab here. My question is how much larger $a*b$ can be compared to xy maximum, in order to find an a,b that fulfils the conditions in the description. My feeling is that ab is either xy or x*y+1, if it were bigger, then it will be bigger than n+1.
$endgroup$
– Tilsight
Nov 25 '18 at 16:07
$begingroup$
@Tilsight Your question does not ask for the difference between the two products, but the difference between the individual numbers. If you want to compare $ab$ and $xy$ you are really asking how different the sums can be for similar products. Try something like $x=y=300$, and $a=1, b=60200$. All I have been trying to do is give you clues so you can explore this yourself - you really need to try some things out - you will learn next to nothing until you do.
$endgroup$
– Mark Bennet
Nov 25 '18 at 16:42
|
show 2 more comments
$begingroup$
If you multiply by $2$ and add $1$ you get:$$(2x+1)(2y+1)+2=(2a+1)(2b+1)$$ so you are looking at the factorisation of successive odd numbers with the factors on the left both at least $5$ and the factors on the right both at least $3$.
If we say $x=y=2$ we get $27$ and to get the maximum difference (in fact this is the only option here) we choose $a=1, b=4$ corresponding to $3times 9$.
This should be enough clue to allow you to explore further.
There are other cases to consider and you might want $x$ and $y$ to be different. But $x=y=3m$ and $a=1, b=6m^2+2m$ are enough to show that for $x=y$ the differences can be arbitrarily large.
answered Nov 25 '18 at 12:42
Mark BennetMark Bennet
80.8k981179
$endgroup$
$begingroup$
Thanks. I need to think your answer over. But I was just thinking that ab can not be more than xy because then the difference between n and n+1 would me more than 1.This would limit the possible values, wouldn't it?
$endgroup$
– Tilsight
Nov 25 '18 at 14:08
$begingroup$
@Tilsight $441times 1=21times 21$ you have potentially large differences between pairs of factors even when the difference is zero.
$endgroup$
– Mark Bennet
Nov 25 '18 at 14:33
$begingroup$
Yes, but it can't be, because then n = 2*441*1+441+1, and n+1 is not 2*21*21+21+21. As the difference between n and n+1 is 1, my gut says that the difference between x and a (and y and b) has to be limited.
$endgroup$
– Tilsight
Nov 25 '18 at 16:01
$begingroup$
Using your example x=2, y=2, so n = 12. Here n+1=13, and 13 = 2*4*1+4+1, so a=4 and b=1. See that xy=ab here. My question is how much larger $a*b$ can be compared to xy maximum, in order to find an a,b that fulfils the conditions in the description. My feeling is that ab is either xy or x*y+1, if it were bigger, then it will be bigger than n+1.
$endgroup$
– Tilsight
Nov 25 '18 at 16:07
$begingroup$
@Tilsight Your question does not ask for the difference between the two products, but the difference between the individual numbers. If you want to compare $ab$ and $xy$ you are really asking how different the sums can be for similar products. Try something like $x=y=300$, and $a=1, b=60200$. All I have been trying to do is give you clues so you can explore this yourself - you really need to try some things out - you will learn next to nothing until you do.
$endgroup$
– Mark Bennet
Nov 25 '18 at 16:42
|
show 2 more comments
$begingroup$
If you multiply by $2$ and add $1$ you get:$$(2x+1)(2y+1)+2=(2a+1)(2b+1)$$ so you are looking at the factorisation of successive odd numbers with the factors on the left both at least $5$ and the factors on the right both at least $3$.
If we say $x=y=2$ we get $27$ and to get the maximum difference (in fact this is the only option here) we choose $a=1, b=4$ corresponding to $3times 9$.
This should be enough clue to allow you to explore further.
There are other cases to consider and you might want $x$ and $y$ to be different. But $x=y=3m$ and $a=1, b=6m^2+2m$ are enough to show that for $x=y$ the differences can be arbitrarily large.
answered Nov 25 '18 at 12:42
Mark BennetMark Bennet
80.8k981179
$endgroup$
If you multiply by $2$ and add $1$ you get:$$(2x+1)(2y+1)+2=(2a+1)(2b+1)$$ so you are looking at the factorisation of successive odd numbers with the factors on the left both at least $5$ and the factors on the right both at least $3$.
If we say $x=y=2$ we get $27$ and to get the maximum difference (in fact this is the only option here) we choose $a=1, b=4$ corresponding to $3times 9$.
This should be enough clue to allow you to explore further.
There are other cases to consider and you might want $x$ and $y$ to be different. But $x=y=3m$ and $a=1, b=6m^2+2m$ are enough to show that for $x=y$ the differences can be arbitrarily large.
answered Nov 25 '18 at 12:42
Mark BennetMark Bennet
80.8k981179
edited Nov 25 '18 at 14:26
answered Nov 25 '18 at 12:42
Mark BennetMark Bennet
80.8k981179
answered Nov 25 '18 at 12:42
Mark BennetMark Bennet
80.8k981179
answered Nov 25 '18 at 12:42
Mark BennetMark Bennet
80.8k981179
80.8k981179
$begingroup$
Thanks. I need to think your answer over. But I was just thinking that ab can not be more than xy because then the difference between n and n+1 would me more than 1.This would limit the possible values, wouldn't it?
$endgroup$
– Tilsight
Nov 25 '18 at 14:08
$begingroup$
@Tilsight $441times 1=21times 21$ you have potentially large differences between pairs of factors even when the difference is zero.
$endgroup$
– Mark Bennet
Nov 25 '18 at 14:33
$begingroup$
Yes, but it can't be, because then n = 2*441*1+441+1, and n+1 is not 2*21*21+21+21. As the difference between n and n+1 is 1, my gut says that the difference between x and a (and y and b) has to be limited.
$endgroup$
– Tilsight
Nov 25 '18 at 16:01
$begingroup$
Using your example x=2, y=2, so n = 12. Here n+1=13, and 13 = 2*4*1+4+1, so a=4 and b=1. See that xy=ab here. My question is how much larger $a*b$ can be compared to xy maximum, in order to find an a,b that fulfils the conditions in the description. My feeling is that ab is either xy or x*y+1, if it were bigger, then it will be bigger than n+1.
$endgroup$
– Tilsight
Nov 25 '18 at 16:07
$begingroup$
@Tilsight Your question does not ask for the difference between the two products, but the difference between the individual numbers. If you want to compare $ab$ and $xy$ you are really asking how different the sums can be for similar products. Try something like $x=y=300$, and $a=1, b=60200$. All I have been trying to do is give you clues so you can explore this yourself - you really need to try some things out - you will learn next to nothing until you do.
$endgroup$
– Mark Bennet
Nov 25 '18 at 16:42
|
show 2 more comments
$begingroup$
Thanks. I need to think your answer over. But I was just thinking that ab can not be more than xy because then the difference between n and n+1 would me more than 1.This would limit the possible values, wouldn't it?
$endgroup$
– Tilsight
Nov 25 '18 at 14:08
$begingroup$
@Tilsight $441times 1=21times 21$ you have potentially large differences between pairs of factors even when the difference is zero.
$endgroup$
– Mark Bennet
Nov 25 '18 at 14:33
$begingroup$
Yes, but it can't be, because then n = 2*441*1+441+1, and n+1 is not 2*21*21+21+21. As the difference between n and n+1 is 1, my gut says that the difference between x and a (and y and b) has to be limited.
$endgroup$
– Tilsight
Nov 25 '18 at 16:01
$begingroup$
Using your example x=2, y=2, so n = 12. Here n+1=13, and 13 = 2*4*1+4+1, so a=4 and b=1. See that xy=ab here. My question is how much larger $a*b$ can be compared to xy maximum, in order to find an a,b that fulfils the conditions in the description. My feeling is that ab is either xy or x*y+1, if it were bigger, then it will be bigger than n+1.
$endgroup$
– Tilsight
Nov 25 '18 at 16:07
$begingroup$
@Tilsight Your question does not ask for the difference between the two products, but the difference between the individual numbers. If you want to compare $ab$ and $xy$ you are really asking how different the sums can be for similar products. Try something like $x=y=300$, and $a=1, b=60200$. All I have been trying to do is give you clues so you can explore this yourself - you really need to try some things out - you will learn next to nothing until you do.
$endgroup$
– Mark Bennet
Nov 25 '18 at 16:42
$begingroup$
Thanks. I need to think your answer over. But I was just thinking that ab can not be more than xy because then the difference between n and n+1 would me more than 1.This would limit the possible values, wouldn't it?
$endgroup$
– Tilsight
Nov 25 '18 at 14:08
$begingroup$
Thanks. I need to think your answer over. But I was just thinking that ab can not be more than xy because then the difference between n and n+1 would me more than 1.This would limit the possible values, wouldn't it?
$endgroup$
– Tilsight
Nov 25 '18 at 14:08
$begingroup$
@Tilsight $441times 1=21times 21$ you have potentially large differences between pairs of factors even when the difference is zero.
$endgroup$
– Mark Bennet
Nov 25 '18 at 14:33
$begingroup$
@Tilsight $441times 1=21times 21$ you have potentially large differences between pairs of factors even when the difference is zero.
$endgroup$
– Mark Bennet
Nov 25 '18 at 14:33
$begingroup$
Yes, but it can't be, because then n = 2*441*1+441+1, and n+1 is not 2*21*21+21+21. As the difference between n and n+1 is 1, my gut says that the difference between x and a (and y and b) has to be limited.
$endgroup$
– Tilsight
Nov 25 '18 at 16:01
$begingroup$
Yes, but it can't be, because then n = 2*441*1+441+1, and n+1 is not 2*21*21+21+21. As the difference between n and n+1 is 1, my gut says that the difference between x and a (and y and b) has to be limited.
$endgroup$
– Tilsight
Nov 25 '18 at 16:01
$begingroup$
Using your example x=2, y=2, so n = 12. Here n+1=13, and 13 = 2*4*1+4+1, so a=4 and b=1. See that xy=ab here. My question is how much larger $a*b$ can be compared to xy maximum, in order to find an a,b that fulfils the conditions in the description. My feeling is that ab is either xy or x*y+1, if it were bigger, then it will be bigger than n+1.
$endgroup$
– Tilsight
Nov 25 '18 at 16:07
$begingroup$
Using your example x=2, y=2, so n = 12. Here n+1=13, and 13 = 2*4*1+4+1, so a=4 and b=1. See that xy=ab here. My question is how much larger $a*b$ can be compared to xy maximum, in order to find an a,b that fulfils the conditions in the description. My feeling is that ab is either xy or x*y+1, if it were bigger, then it will be bigger than n+1.
$endgroup$
– Tilsight
Nov 25 '18 at 16:07
$begingroup$
@Tilsight Your question does not ask for the difference between the two products, but the difference between the individual numbers. If you want to compare $ab$ and $xy$ you are really asking how different the sums can be for similar products. Try something like $x=y=300$, and $a=1, b=60200$. All I have been trying to do is give you clues so you can explore this yourself - you really need to try some things out - you will learn next to nothing until you do.
$endgroup$
– Mark Bennet
Nov 25 '18 at 16:42
$begingroup$
@Tilsight Your question does not ask for the difference between the two products, but the difference between the individual numbers. If you want to compare $ab$ and $xy$ you are really asking how different the sums can be for similar products. Try something like $x=y=300$, and $a=1, b=60200$. All I have been trying to do is give you clues so you can explore this yourself - you really need to try some things out - you will learn next to nothing until you do.
$endgroup$
– Mark Bennet
Nov 25 '18 at 16:42
|
show 2 more comments
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