Show that only integral value of $frac{a}{b} + frac{b}{a}$ is $2$ when $a,b in Bbb{Z}$ [duplicate]












3















This question already has an answer here:




  • Can sum of a rational number and its reciprocal be an integer?

    8 answers




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







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marked as duplicate by Bill Dubuque elementary-number-theory
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 10 hours ago


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.















  • Are $a$ and $b$ integers?
    – 5xum
    2 days ago










  • Yes, a and b are positive integers
    – Funny guy
    2 days ago










  • Then your question title should say $a,b in mathbb{Z}^+$ instead of $a,b in mathbb{Z}$
    – gandalf61
    2 days ago
















3















This question already has an answer here:




  • Can sum of a rational number and its reciprocal be an integer?

    8 answers




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







share|cite|improve this question









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.











marked as duplicate by Bill Dubuque elementary-number-theory
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 10 hours ago


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.















  • Are $a$ and $b$ integers?
    – 5xum
    2 days ago










  • Yes, a and b are positive integers
    – Funny guy
    2 days ago










  • Then your question title should say $a,b in mathbb{Z}^+$ instead of $a,b in mathbb{Z}$
    – gandalf61
    2 days ago














3












3








3


2






This question already has an answer here:




  • Can sum of a rational number and its reciprocal be an integer?

    8 answers




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







share|cite|improve this question









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.












This question already has an answer here:




  • Can sum of a rational number and its reciprocal be an integer?

    8 answers




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!





This question already has an answer here:




  • Can sum of a rational number and its reciprocal be an integer?

    8 answers







elementary-number-theory quadratics




elementary-number-theory quadratics






share|cite|improve this question









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Funny guy is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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edited 2 days ago









Daniel R

2,45832035




2,45832035






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asked 2 days ago









Funny guyFunny guy

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marked as duplicate by Bill Dubuque elementary-number-theory
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 10 hours ago


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 Bill Dubuque elementary-number-theory
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 10 hours ago


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.














  • Are $a$ and $b$ integers?
    – 5xum
    2 days ago










  • Yes, a and b are positive integers
    – Funny guy
    2 days ago










  • Then your question title should say $a,b in mathbb{Z}^+$ instead of $a,b in mathbb{Z}$
    – gandalf61
    2 days ago


















  • Are $a$ and $b$ integers?
    – 5xum
    2 days ago










  • Yes, a and b are positive integers
    – Funny guy
    2 days ago










  • Then your question title should say $a,b in mathbb{Z}^+$ instead of $a,b in mathbb{Z}$
    – gandalf61
    2 days ago
















Are $a$ and $b$ integers?
– 5xum
2 days ago




Are $a$ and $b$ integers?
– 5xum
2 days ago












Yes, a and b are positive integers
– Funny guy
2 days ago




Yes, a and b are positive integers
– Funny guy
2 days ago












Then your question title should say $a,b in mathbb{Z}^+$ instead of $a,b in mathbb{Z}$
– gandalf61
2 days ago




Then your question title should say $a,b in mathbb{Z}^+$ instead of $a,b in mathbb{Z}$
– gandalf61
2 days ago










2 Answers
2






active

oldest

votes


















7














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 $1$.






share|cite|improve this answer





















  • What do you mean by a|b and b|a?
    – Funny guy
    2 days ago






  • 1




    $amid b$ meaning that $a$ divides $b$
    – Olof Rubin
    2 days ago










  • I don't see how you can conclude a divides b and vice versa
    – Funny guy
    2 days ago








  • 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
    2 days ago








  • 1




    Good answer +1.
    – Prakhar Nagpal
    2 days ago



















3














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 $a = b$.






share|cite|improve this answer























  • Why must m^2-4 be a perfect square? Secondly what's the point of doing (n+1)^2-n^2?
    – Funny guy
    2 days ago












  • @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
    2 days ago










  • @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
    2 days ago




















2 Answers
2






active

oldest

votes








2 Answers
2






active

oldest

votes









active

oldest

votes






active

oldest

votes









7














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 $1$.






share|cite|improve this answer





















  • What do you mean by a|b and b|a?
    – Funny guy
    2 days ago






  • 1




    $amid b$ meaning that $a$ divides $b$
    – Olof Rubin
    2 days ago










  • I don't see how you can conclude a divides b and vice versa
    – Funny guy
    2 days ago








  • 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
    2 days ago








  • 1




    Good answer +1.
    – Prakhar Nagpal
    2 days ago
















7














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 $1$.






share|cite|improve this answer





















  • What do you mean by a|b and b|a?
    – Funny guy
    2 days ago






  • 1




    $amid b$ meaning that $a$ divides $b$
    – Olof Rubin
    2 days ago










  • I don't see how you can conclude a divides b and vice versa
    – Funny guy
    2 days ago








  • 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
    2 days ago








  • 1




    Good answer +1.
    – Prakhar Nagpal
    2 days ago














7












7








7






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 $1$.






share|cite|improve this answer












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 $1$.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered 2 days ago









Olof RubinOlof Rubin

1,202316




1,202316












  • What do you mean by a|b and b|a?
    – Funny guy
    2 days ago






  • 1




    $amid b$ meaning that $a$ divides $b$
    – Olof Rubin
    2 days ago










  • I don't see how you can conclude a divides b and vice versa
    – Funny guy
    2 days ago








  • 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
    2 days ago








  • 1




    Good answer +1.
    – Prakhar Nagpal
    2 days ago


















  • What do you mean by a|b and b|a?
    – Funny guy
    2 days ago






  • 1




    $amid b$ meaning that $a$ divides $b$
    – Olof Rubin
    2 days ago










  • I don't see how you can conclude a divides b and vice versa
    – Funny guy
    2 days ago








  • 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
    2 days ago








  • 1




    Good answer +1.
    – Prakhar Nagpal
    2 days ago
















What do you mean by a|b and b|a?
– Funny guy
2 days ago




What do you mean by a|b and b|a?
– Funny guy
2 days ago




1




1




$amid b$ meaning that $a$ divides $b$
– Olof Rubin
2 days ago




$amid b$ meaning that $a$ divides $b$
– Olof Rubin
2 days ago












I don't see how you can conclude a divides b and vice versa
– Funny guy
2 days ago






I don't see how you can conclude a divides b and vice versa
– Funny guy
2 days ago






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
2 days ago






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
2 days ago






1




1




Good answer +1.
– Prakhar Nagpal
2 days ago




Good answer +1.
– Prakhar Nagpal
2 days ago











3














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 $a = b$.






share|cite|improve this answer























  • Why must m^2-4 be a perfect square? Secondly what's the point of doing (n+1)^2-n^2?
    – Funny guy
    2 days ago












  • @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
    2 days ago










  • @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
    2 days ago


















3














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 $a = b$.






share|cite|improve this answer























  • Why must m^2-4 be a perfect square? Secondly what's the point of doing (n+1)^2-n^2?
    – Funny guy
    2 days ago












  • @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
    2 days ago










  • @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
    2 days ago
















3












3








3






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 $a = b$.






share|cite|improve this answer














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 $a = b$.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited 2 days ago

























answered 2 days ago









John OmielanJohn Omielan

1,27618




1,27618












  • Why must m^2-4 be a perfect square? Secondly what's the point of doing (n+1)^2-n^2?
    – Funny guy
    2 days ago












  • @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
    2 days ago










  • @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
    2 days ago




















  • Why must m^2-4 be a perfect square? Secondly what's the point of doing (n+1)^2-n^2?
    – Funny guy
    2 days ago












  • @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
    2 days ago










  • @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
    2 days ago


















Why must m^2-4 be a perfect square? Secondly what's the point of doing (n+1)^2-n^2?
– Funny guy
2 days ago






Why must m^2-4 be a perfect square? Secondly what's the point of doing (n+1)^2-n^2?
– Funny guy
2 days ago














@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
2 days ago




@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
2 days ago












@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
2 days ago






@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
2 days ago





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