Show that only integral value of $frac{a}{b} + frac{b}{a}$ is $2$ when $a,b in Bbb{Z}$
I am given the equation
$$
m=frac{a}{b} + frac{b}{a}
$$
where $a$ and $b$ are positive integers. I have to show that the only integral value $m$ can take is $2$.
I have no idea how to solve it. I tried rewriting the equation as
$$
a^2 + b^2 - mab = 0
$$
But that tells me nothing.
Im stuck, please help!
elementary-number-theory quadratics
edited yesterday
Daniel R
2,45832035
asked yesterday
Funny guyFunny guy
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Funny guy is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |
I am given the equation
$$
m=frac{a}{b} + frac{b}{a}
$$
where $a$ and $b$ are positive integers. I have to show that the only integral value $m$ can take is $2$.
I have no idea how to solve it. I tried rewriting the equation as
$$
a^2 + b^2 - mab = 0
$$
But that tells me nothing.
Im stuck, please help!
elementary-number-theory quadratics
edited yesterday
Daniel R
2,45832035
asked yesterday
Funny guyFunny guy
236
New contributor
Funny guy is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Are $a$ and $b$ integers?
– 5xum
yesterday
Yes, a and b are positive integers
– Funny guy
yesterday
Then your question title should say $a,b in mathbb{Z}^+$ instead of $a,b in mathbb{Z}$
– gandalf61
yesterday
add a comment |
I am given the equation
$$
m=frac{a}{b} + frac{b}{a}
$$
where $a$ and $b$ are positive integers. I have to show that the only integral value $m$ can take is $2$.
I have no idea how to solve it. I tried rewriting the equation as
$$
a^2 + b^2 - mab = 0
$$
But that tells me nothing.
Im stuck, please help!
elementary-number-theory quadratics
edited yesterday
Daniel R
2,45832035
asked yesterday
Funny guyFunny guy
236
New contributor
Funny guy is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
I am given the equation
$$
m=frac{a}{b} + frac{b}{a}
$$
where $a$ and $b$ are positive integers. I have to show that the only integral value $m$ can take is $2$.
I have no idea how to solve it. I tried rewriting the equation as
$$
a^2 + b^2 - mab = 0
$$
But that tells me nothing.
Im stuck, please help!
elementary-number-theory quadratics
elementary-number-theory quadratics
edited yesterday
Daniel R
2,45832035
asked yesterday
Funny guyFunny guy
236
New contributor
Funny guy is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited yesterday
Daniel R
2,45832035
asked yesterday
Funny guyFunny guy
236
New contributor
Funny guy is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited yesterday
Daniel R
2,45832035
edited yesterday
Daniel R
2,45832035
edited yesterday
Daniel R
2,45832035
2,45832035
asked yesterday
Funny guyFunny guy
236
New contributor
Funny guy is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked yesterday
Funny guyFunny guy
236
asked yesterday
Funny guyFunny guy
236
236
New contributor
Funny guy is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Funny guy is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Funny guy is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Are $a$ and $b$ integers?
– 5xum
yesterday
Yes, a and b are positive integers
– Funny guy
yesterday
Then your question title should say $a,b in mathbb{Z}^+$ instead of $a,b in mathbb{Z}$
– gandalf61
yesterday
add a comment |
Are $a$ and $b$ integers?
– 5xum
yesterday
Yes, a and b are positive integers
– Funny guy
yesterday
Then your question title should say $a,b in mathbb{Z}^+$ instead of $a,b in mathbb{Z}$
– gandalf61
yesterday
Are $a$ and $b$ integers?
– 5xum
yesterday
Are $a$ and $b$ integers?
– 5xum
yesterday
Yes, a and b are positive integers
– Funny guy
yesterday
Yes, a and b are positive integers
– Funny guy
yesterday
Then your question title should say $a,b in mathbb{Z}^+$ instead of $a,b in mathbb{Z}$
– gandalf61
yesterday
Then your question title should say $a,b in mathbb{Z}^+$ instead of $a,b in mathbb{Z}$
– gandalf61
yesterday
add a comment |
2 Answers
2
active
oldest
votes
Suppose we have factored out all common factors of $a$ and $b$ so that $a,b$ are relatively prime. Then
$$a^2+b^2-mab = 0$$
implies that $bmid a$ and $amid b$. Do you see how? Thus both are $1$.
answered yesterday
Olof RubinOlof Rubin
1,212316
What do you mean by a|b and b|a?
– Funny guy
yesterday
1
$amid b$ meaning that $a$ divides $b$
– Olof Rubin
yesterday
I don't see how you can conclude a divides b and vice versa
– Funny guy
yesterday
2
For instance we can write $a^2+b^2-mab = 0Leftrightarrow a^2 = b(ma-b)$ and since $b$ divides the right-hand side it must also divide the left-hand side
– Olof Rubin
yesterday
1
Good answer +1.
– Prakhar Nagpal
yesterday
|
show 5 more comments
With
$$a^2 - mab + b^2 = 0 tag{1}label{eq1}$$
consider $a$ to be a variable, with $b$ and $m$ being constant, so you have a quadratic equation in $a$. Using the quadratic formula to solve for $a$ gives that
$$a = cfrac{mb pm sqrt{m^2 b^2 - 4b^2}}{2} tag{2}label{eq2}$$
This simplifies to
$$a = bleft(cfrac{m pm sqrt{m^2 - 4}}{2}right) tag{3}label{eq3}$$
As $a$ and $b$ are positive integers, for $m$ to also be an integer means that $m^2 - 4$ must be a perfect square. For positive integers $n$, note that $left(n + 1right)^2 - n^2 = 2n + 1$ is $gt 4$ for $n ge 2$, meaning the difference of just $2$ consecutive squares is then too large, so $m le 2$, with $m = 2$ working. This gives that $a = b$.
answered yesterday
John OmielanJohn Omielan
1,19918
Why must m^2-4 be a perfect square? Secondly what's the point of doing (n+1)^2-n^2?
– Funny guy
yesterday
@Funnyguy Since $a$, $b$ and $m$ are stated to be integers, the equation (3) can only hold if $sqrt{m^2 - 4}$ is also an integer, i.e., a integer, which means that $m^2 - 4$ must be the square of an integer, which is called a "perfect square". My point for checking $left(n + 1right)^2 - n^2$ is to show that even for as small as $n = 3$, the difference from it squared and the integer just below it is more than $4$, showing that it can't be the value as the difference of squares increase, going from $5$ to $7$, then $9$, etc. Only $m = 2$ works, with the result following from this.
– John Omielan
yesterday
@Funnyguy For $m^2 - 4$ to be a perfect square means that $m^2 - 4 = k^2$, for some integer $k$. Thus, $m^2 = k^2 + 4$, and $m ge k + 1$. However, as the difference of even just $1$ more squared less the value below it is $5$ or more for $n = 3$, this can only be possible if $k = 0$. I'm sorry if this is somewhat confusing (it's still early morning for me), with it perhaps being easier for you to just try a few numbers to see for yourself. It would have been simpler, and perhaps better, to just stated that only $m = 2$ works.
– John Omielan
yesterday
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
Suppose we have factored out all common factors of $a$ and $b$ so that $a,b$ are relatively prime. Then
$$a^2+b^2-mab = 0$$
implies that $bmid a$ and $amid b$. Do you see how? Thus both are $1$.
answered yesterday
Olof RubinOlof Rubin
1,212316
What do you mean by a|b and b|a?
– Funny guy
yesterday
1
$amid b$ meaning that $a$ divides $b$
– Olof Rubin
yesterday
I don't see how you can conclude a divides b and vice versa
– Funny guy
yesterday
2
For instance we can write $a^2+b^2-mab = 0Leftrightarrow a^2 = b(ma-b)$ and since $b$ divides the right-hand side it must also divide the left-hand side
– Olof Rubin
yesterday
1
Good answer +1.
– Prakhar Nagpal
yesterday
|
show 5 more comments
Suppose we have factored out all common factors of $a$ and $b$ so that $a,b$ are relatively prime. Then
$$a^2+b^2-mab = 0$$
implies that $bmid a$ and $amid b$. Do you see how? Thus both are $1$.
answered yesterday
Olof RubinOlof Rubin
1,212316
What do you mean by a|b and b|a?
– Funny guy
yesterday
1
$amid b$ meaning that $a$ divides $b$
– Olof Rubin
yesterday
I don't see how you can conclude a divides b and vice versa
– Funny guy
yesterday
2
For instance we can write $a^2+b^2-mab = 0Leftrightarrow a^2 = b(ma-b)$ and since $b$ divides the right-hand side it must also divide the left-hand side
– Olof Rubin
yesterday
1
Good answer +1.
– Prakhar Nagpal
yesterday
|
show 5 more comments
Suppose we have factored out all common factors of $a$ and $b$ so that $a,b$ are relatively prime. Then
$$a^2+b^2-mab = 0$$
implies that $bmid a$ and $amid b$. Do you see how? Thus both are $1$.
answered yesterday
Olof RubinOlof Rubin
1,212316
Suppose we have factored out all common factors of $a$ and $b$ so that $a,b$ are relatively prime. Then
$$a^2+b^2-mab = 0$$
implies that $bmid a$ and $amid b$. Do you see how? Thus both are $1$.
answered yesterday
Olof RubinOlof Rubin
1,212316
answered yesterday
Olof RubinOlof Rubin
1,212316
answered yesterday
Olof RubinOlof Rubin
1,212316
answered yesterday
Olof RubinOlof Rubin
1,212316
1,212316
What do you mean by a|b and b|a?
– Funny guy
yesterday
1
$amid b$ meaning that $a$ divides $b$
– Olof Rubin
yesterday
I don't see how you can conclude a divides b and vice versa
– Funny guy
yesterday
2
For instance we can write $a^2+b^2-mab = 0Leftrightarrow a^2 = b(ma-b)$ and since $b$ divides the right-hand side it must also divide the left-hand side
– Olof Rubin
yesterday
1
Good answer +1.
– Prakhar Nagpal
yesterday
|
show 5 more comments
What do you mean by a|b and b|a?
– Funny guy
yesterday
1
$amid b$ meaning that $a$ divides $b$
– Olof Rubin
yesterday
I don't see how you can conclude a divides b and vice versa
– Funny guy
yesterday
2
For instance we can write $a^2+b^2-mab = 0Leftrightarrow a^2 = b(ma-b)$ and since $b$ divides the right-hand side it must also divide the left-hand side
– Olof Rubin
yesterday
1
Good answer +1.
– Prakhar Nagpal
yesterday
What do you mean by a|b and b|a?
– Funny guy
yesterday
What do you mean by a|b and b|a?
– Funny guy
yesterday
1
1
$amid b$ meaning that $a$ divides $b$
– Olof Rubin
yesterday
$amid b$ meaning that $a$ divides $b$
– Olof Rubin
yesterday
I don't see how you can conclude a divides b and vice versa
– Funny guy
yesterday
I don't see how you can conclude a divides b and vice versa
– Funny guy
yesterday
2
2
For instance we can write $a^2+b^2-mab = 0Leftrightarrow a^2 = b(ma-b)$ and since $b$ divides the right-hand side it must also divide the left-hand side
– Olof Rubin
yesterday
For instance we can write $a^2+b^2-mab = 0Leftrightarrow a^2 = b(ma-b)$ and since $b$ divides the right-hand side it must also divide the left-hand side
– Olof Rubin
yesterday
1
1
Good answer +1.
– Prakhar Nagpal
yesterday
Good answer +1.
– Prakhar Nagpal
yesterday
|
show 5 more comments
With
$$a^2 - mab + b^2 = 0 tag{1}label{eq1}$$
consider $a$ to be a variable, with $b$ and $m$ being constant, so you have a quadratic equation in $a$. Using the quadratic formula to solve for $a$ gives that
$$a = cfrac{mb pm sqrt{m^2 b^2 - 4b^2}}{2} tag{2}label{eq2}$$
This simplifies to
$$a = bleft(cfrac{m pm sqrt{m^2 - 4}}{2}right) tag{3}label{eq3}$$
As $a$ and $b$ are positive integers, for $m$ to also be an integer means that $m^2 - 4$ must be a perfect square. For positive integers $n$, note that $left(n + 1right)^2 - n^2 = 2n + 1$ is $gt 4$ for $n ge 2$, meaning the difference of just $2$ consecutive squares is then too large, so $m le 2$, with $m = 2$ working. This gives that $a = b$.
answered yesterday
John OmielanJohn Omielan
1,19918
Why must m^2-4 be a perfect square? Secondly what's the point of doing (n+1)^2-n^2?
– Funny guy
yesterday
@Funnyguy Since $a$, $b$ and $m$ are stated to be integers, the equation (3) can only hold if $sqrt{m^2 - 4}$ is also an integer, i.e., a integer, which means that $m^2 - 4$ must be the square of an integer, which is called a "perfect square". My point for checking $left(n + 1right)^2 - n^2$ is to show that even for as small as $n = 3$, the difference from it squared and the integer just below it is more than $4$, showing that it can't be the value as the difference of squares increase, going from $5$ to $7$, then $9$, etc. Only $m = 2$ works, with the result following from this.
– John Omielan
yesterday
@Funnyguy For $m^2 - 4$ to be a perfect square means that $m^2 - 4 = k^2$, for some integer $k$. Thus, $m^2 = k^2 + 4$, and $m ge k + 1$. However, as the difference of even just $1$ more squared less the value below it is $5$ or more for $n = 3$, this can only be possible if $k = 0$. I'm sorry if this is somewhat confusing (it's still early morning for me), with it perhaps being easier for you to just try a few numbers to see for yourself. It would have been simpler, and perhaps better, to just stated that only $m = 2$ works.
– John Omielan
yesterday
add a comment |
With
$$a^2 - mab + b^2 = 0 tag{1}label{eq1}$$
consider $a$ to be a variable, with $b$ and $m$ being constant, so you have a quadratic equation in $a$. Using the quadratic formula to solve for $a$ gives that
$$a = cfrac{mb pm sqrt{m^2 b^2 - 4b^2}}{2} tag{2}label{eq2}$$
This simplifies to
$$a = bleft(cfrac{m pm sqrt{m^2 - 4}}{2}right) tag{3}label{eq3}$$
As $a$ and $b$ are positive integers, for $m$ to also be an integer means that $m^2 - 4$ must be a perfect square. For positive integers $n$, note that $left(n + 1right)^2 - n^2 = 2n + 1$ is $gt 4$ for $n ge 2$, meaning the difference of just $2$ consecutive squares is then too large, so $m le 2$, with $m = 2$ working. This gives that $a = b$.
answered yesterday
John OmielanJohn Omielan
1,19918
Why must m^2-4 be a perfect square? Secondly what's the point of doing (n+1)^2-n^2?
– Funny guy
yesterday
@Funnyguy Since $a$, $b$ and $m$ are stated to be integers, the equation (3) can only hold if $sqrt{m^2 - 4}$ is also an integer, i.e., a integer, which means that $m^2 - 4$ must be the square of an integer, which is called a "perfect square". My point for checking $left(n + 1right)^2 - n^2$ is to show that even for as small as $n = 3$, the difference from it squared and the integer just below it is more than $4$, showing that it can't be the value as the difference of squares increase, going from $5$ to $7$, then $9$, etc. Only $m = 2$ works, with the result following from this.
– John Omielan
yesterday
@Funnyguy For $m^2 - 4$ to be a perfect square means that $m^2 - 4 = k^2$, for some integer $k$. Thus, $m^2 = k^2 + 4$, and $m ge k + 1$. However, as the difference of even just $1$ more squared less the value below it is $5$ or more for $n = 3$, this can only be possible if $k = 0$. I'm sorry if this is somewhat confusing (it's still early morning for me), with it perhaps being easier for you to just try a few numbers to see for yourself. It would have been simpler, and perhaps better, to just stated that only $m = 2$ works.
– John Omielan
yesterday
add a comment |
With
$$a^2 - mab + b^2 = 0 tag{1}label{eq1}$$
consider $a$ to be a variable, with $b$ and $m$ being constant, so you have a quadratic equation in $a$. Using the quadratic formula to solve for $a$ gives that
$$a = cfrac{mb pm sqrt{m^2 b^2 - 4b^2}}{2} tag{2}label{eq2}$$
This simplifies to
$$a = bleft(cfrac{m pm sqrt{m^2 - 4}}{2}right) tag{3}label{eq3}$$
As $a$ and $b$ are positive integers, for $m$ to also be an integer means that $m^2 - 4$ must be a perfect square. For positive integers $n$, note that $left(n + 1right)^2 - n^2 = 2n + 1$ is $gt 4$ for $n ge 2$, meaning the difference of just $2$ consecutive squares is then too large, so $m le 2$, with $m = 2$ working. This gives that $a = b$.
answered yesterday
John OmielanJohn Omielan
1,19918
With
$$a^2 - mab + b^2 = 0 tag{1}label{eq1}$$
consider $a$ to be a variable, with $b$ and $m$ being constant, so you have a quadratic equation in $a$. Using the quadratic formula to solve for $a$ gives that
$$a = cfrac{mb pm sqrt{m^2 b^2 - 4b^2}}{2} tag{2}label{eq2}$$
This simplifies to
$$a = bleft(cfrac{m pm sqrt{m^2 - 4}}{2}right) tag{3}label{eq3}$$
As $a$ and $b$ are positive integers, for $m$ to also be an integer means that $m^2 - 4$ must be a perfect square. For positive integers $n$, note that $left(n + 1right)^2 - n^2 = 2n + 1$ is $gt 4$ for $n ge 2$, meaning the difference of just $2$ consecutive squares is then too large, so $m le 2$, with $m = 2$ working. This gives that $a = b$.
answered yesterday
John OmielanJohn Omielan
1,19918
edited yesterday
answered yesterday
John OmielanJohn Omielan
1,19918
answered yesterday
John OmielanJohn Omielan
1,19918
answered yesterday
John OmielanJohn Omielan
1,19918
1,19918
Why must m^2-4 be a perfect square? Secondly what's the point of doing (n+1)^2-n^2?
– Funny guy
yesterday
@Funnyguy Since $a$, $b$ and $m$ are stated to be integers, the equation (3) can only hold if $sqrt{m^2 - 4}$ is also an integer, i.e., a integer, which means that $m^2 - 4$ must be the square of an integer, which is called a "perfect square". My point for checking $left(n + 1right)^2 - n^2$ is to show that even for as small as $n = 3$, the difference from it squared and the integer just below it is more than $4$, showing that it can't be the value as the difference of squares increase, going from $5$ to $7$, then $9$, etc. Only $m = 2$ works, with the result following from this.
– John Omielan
yesterday
@Funnyguy For $m^2 - 4$ to be a perfect square means that $m^2 - 4 = k^2$, for some integer $k$. Thus, $m^2 = k^2 + 4$, and $m ge k + 1$. However, as the difference of even just $1$ more squared less the value below it is $5$ or more for $n = 3$, this can only be possible if $k = 0$. I'm sorry if this is somewhat confusing (it's still early morning for me), with it perhaps being easier for you to just try a few numbers to see for yourself. It would have been simpler, and perhaps better, to just stated that only $m = 2$ works.
– John Omielan
yesterday
add a comment |
Why must m^2-4 be a perfect square? Secondly what's the point of doing (n+1)^2-n^2?
– Funny guy
yesterday
@Funnyguy Since $a$, $b$ and $m$ are stated to be integers, the equation (3) can only hold if $sqrt{m^2 - 4}$ is also an integer, i.e., a integer, which means that $m^2 - 4$ must be the square of an integer, which is called a "perfect square". My point for checking $left(n + 1right)^2 - n^2$ is to show that even for as small as $n = 3$, the difference from it squared and the integer just below it is more than $4$, showing that it can't be the value as the difference of squares increase, going from $5$ to $7$, then $9$, etc. Only $m = 2$ works, with the result following from this.
– John Omielan
yesterday
@Funnyguy For $m^2 - 4$ to be a perfect square means that $m^2 - 4 = k^2$, for some integer $k$. Thus, $m^2 = k^2 + 4$, and $m ge k + 1$. However, as the difference of even just $1$ more squared less the value below it is $5$ or more for $n = 3$, this can only be possible if $k = 0$. I'm sorry if this is somewhat confusing (it's still early morning for me), with it perhaps being easier for you to just try a few numbers to see for yourself. It would have been simpler, and perhaps better, to just stated that only $m = 2$ works.
– John Omielan
yesterday
Why must m^2-4 be a perfect square? Secondly what's the point of doing (n+1)^2-n^2?
– Funny guy
yesterday
Why must m^2-4 be a perfect square? Secondly what's the point of doing (n+1)^2-n^2?
– Funny guy
yesterday
@Funnyguy Since $a$, $b$ and $m$ are stated to be integers, the equation (3) can only hold if $sqrt{m^2 - 4}$ is also an integer, i.e., a integer, which means that $m^2 - 4$ must be the square of an integer, which is called a "perfect square". My point for checking $left(n + 1right)^2 - n^2$ is to show that even for as small as $n = 3$, the difference from it squared and the integer just below it is more than $4$, showing that it can't be the value as the difference of squares increase, going from $5$ to $7$, then $9$, etc. Only $m = 2$ works, with the result following from this.
– John Omielan
yesterday
@Funnyguy Since $a$, $b$ and $m$ are stated to be integers, the equation (3) can only hold if $sqrt{m^2 - 4}$ is also an integer, i.e., a integer, which means that $m^2 - 4$ must be the square of an integer, which is called a "perfect square". My point for checking $left(n + 1right)^2 - n^2$ is to show that even for as small as $n = 3$, the difference from it squared and the integer just below it is more than $4$, showing that it can't be the value as the difference of squares increase, going from $5$ to $7$, then $9$, etc. Only $m = 2$ works, with the result following from this.
– John Omielan
yesterday
@Funnyguy For $m^2 - 4$ to be a perfect square means that $m^2 - 4 = k^2$, for some integer $k$. Thus, $m^2 = k^2 + 4$, and $m ge k + 1$. However, as the difference of even just $1$ more squared less the value below it is $5$ or more for $n = 3$, this can only be possible if $k = 0$. I'm sorry if this is somewhat confusing (it's still early morning for me), with it perhaps being easier for you to just try a few numbers to see for yourself. It would have been simpler, and perhaps better, to just stated that only $m = 2$ works.
– John Omielan
yesterday
@Funnyguy For $m^2 - 4$ to be a perfect square means that $m^2 - 4 = k^2$, for some integer $k$. Thus, $m^2 = k^2 + 4$, and $m ge k + 1$. However, as the difference of even just $1$ more squared less the value below it is $5$ or more for $n = 3$, this can only be possible if $k = 0$. I'm sorry if this is somewhat confusing (it's still early morning for me), with it perhaps being easier for you to just try a few numbers to see for yourself. It would have been simpler, and perhaps better, to just stated that only $m = 2$ works.
– John Omielan
yesterday
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Funny guy is a new contributor. Be nice, and check out our Code of Conduct.
Funny guy is a new contributor. Be nice, and check out our Code of Conduct.
Funny guy is a new contributor. Be nice, and check out our Code of Conduct.
Funny guy is a new contributor. Be nice, and check out our Code of Conduct.
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Are $a$ and $b$ integers?
– 5xum
yesterday
Yes, a and b are positive integers
– Funny guy
yesterday
Then your question title should say $a,b in mathbb{Z}^+$ instead of $a,b in mathbb{Z}$
– gandalf61
yesterday