Simple solution to Question 6 from the 1988 Math Olympiad [duplicate]











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  • Alternative proof that $(a^2+b^2)/(ab+1)$ is a square when it's an integer

    2 answers




Recall Question 6 of the 1988 Math Olympiad



Question 6 is as follows:




Let $a$ and $b$ be positive integers such that $ab + 1$ divides $a^2 +
b^2$. Show that $$frac{a^2+b^2}{ab+1}$$ is the square of an integer.




My proof:




Proof:



Let us denote the square of an integer by $k^2$. We have to show that
$$frac{a^2+b^2}{ab+1}=k^2$$ for some $a > 0$ and $b > 0$ where $a, b,
> k in Bbb Z$.



Rewriting the equation:



$$a^2+b^2=k^2(ab+1)=k^2ab+k^2$$



Let $k = b$:



$$a^2+b^2=b^3a+b^2 Rightarrow a^2=b^3a Rightarrow a=b^3$$



To check, let's substitute these relations back in the original
expression:



$$frac{a^2+b^2}{ab+1}=frac{(k^3)^2+(k)^2}{(k^3)(k)+1}=frac{k^6+k^2}{k^4+1}=frac{k^2(k^4+1)}{k^4+1}=k^2$$




Two of things I would like to know about this Olympiad Question:




  1. Is this proof correct?

  2. Is there a reliable way to check if a certain solution to a proof already exist?


Background



Every now and then I visit Youtube for educational purposes. I've been a great fan of Kurzgesagt, Numberphile, Computerphile, and other educational channels. For anyone not having heard of these channels I would greatly recommended to check them out!



Today I was watching this video on the channel of Numberphile: The World's Best Mathematician (*) - Numberphile. In this video a great mathematician Professor Tao tells about his passion for mathematics from a young age. At some point in this video it is mentioned that Tao only scored 1 out of 7 points for the infamous Question 6 from the 1988 Math Olympiad. In the next couple of frames this video shows the details of the question. At that moment I paused the video to try and solve this question by myself, with only pen and paper.



After 10 minutes or so I figured out the solution to this problem and questioned myself why the question is called infamous. I came across some solutions at this site that are, in my opinion, mindblowing in terms of length and mathematical notation, for example this solution. Then I read something about Vieta-jumping that I'm not familiar with and how it's used to solve this question (or not). A rather simple solution I came across is this one. I think the elegance and quality of a solution is greatly influenced by its length and level of simplicity in terms of notation, terminology, and construction.



As far as I did my silly online research I couldn't find a solution that was equivalent to mine. Therefore I thought it would be a good idea to share it and let other people, with a much broader understandig of existing mathematical proofs, decide if this proof is correct (and a known solution).










share|cite|improve this question















marked as duplicate by Jyrki Lahtonen, Lord Shark the Unknown, 5xum, Christopher, Jens Nov 13 at 16:03


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.















  • I mentioned that thread in my post. First, it is not a duplicate, since my question differs a lot from that one. Second, the solution mentioned in that thread also differs from my solution.
    – Pascal
    Mar 14 '17 at 18:13






  • 1




    Its always best to get quickly to your question, and then give us details about how you found it. The question is what volunteers can help you with, not your biography, and if people want to figure out quickly if they can help you with the question. Remember: You are asking people to give you their time, don't request more of it than you need.
    – Thomas Andrews
    Mar 14 '17 at 18:13






  • 1




    @Pascal You assumed that $k=b$ without any reason or justification. This reduced the problem from arbitrary integers $a,b$ to the much smaller subset of integers $a,b$ with $a=b^3$. It's (almost) as if I said "look, let's just assume $a=b=1,$, and in that case the fraction is $1$ i.e. a perfect square QED". That's not a proof, and doesn't answer the original question.
    – dxiv
    Mar 14 '17 at 18:23








  • 2




    @Pascal And I took $a=b=1$ because that gave me a square right away. Think at why my "proof" isn't valid, then you'll realize why yours isn't, either. You can't simply postulate that $k=b$ just because it simplifies your calculations.
    – dxiv
    Mar 14 '17 at 18:27






  • 1




    $a=8,b=30$ is a solution, for example, with $k=2$.
    – Thomas Andrews
    Mar 14 '17 at 18:29















up vote
1
down vote

favorite
1













This question already has an answer here:




  • Alternative proof that $(a^2+b^2)/(ab+1)$ is a square when it's an integer

    2 answers




Recall Question 6 of the 1988 Math Olympiad



Question 6 is as follows:




Let $a$ and $b$ be positive integers such that $ab + 1$ divides $a^2 +
b^2$. Show that $$frac{a^2+b^2}{ab+1}$$ is the square of an integer.




My proof:




Proof:



Let us denote the square of an integer by $k^2$. We have to show that
$$frac{a^2+b^2}{ab+1}=k^2$$ for some $a > 0$ and $b > 0$ where $a, b,
> k in Bbb Z$.



Rewriting the equation:



$$a^2+b^2=k^2(ab+1)=k^2ab+k^2$$



Let $k = b$:



$$a^2+b^2=b^3a+b^2 Rightarrow a^2=b^3a Rightarrow a=b^3$$



To check, let's substitute these relations back in the original
expression:



$$frac{a^2+b^2}{ab+1}=frac{(k^3)^2+(k)^2}{(k^3)(k)+1}=frac{k^6+k^2}{k^4+1}=frac{k^2(k^4+1)}{k^4+1}=k^2$$




Two of things I would like to know about this Olympiad Question:




  1. Is this proof correct?

  2. Is there a reliable way to check if a certain solution to a proof already exist?


Background



Every now and then I visit Youtube for educational purposes. I've been a great fan of Kurzgesagt, Numberphile, Computerphile, and other educational channels. For anyone not having heard of these channels I would greatly recommended to check them out!



Today I was watching this video on the channel of Numberphile: The World's Best Mathematician (*) - Numberphile. In this video a great mathematician Professor Tao tells about his passion for mathematics from a young age. At some point in this video it is mentioned that Tao only scored 1 out of 7 points for the infamous Question 6 from the 1988 Math Olympiad. In the next couple of frames this video shows the details of the question. At that moment I paused the video to try and solve this question by myself, with only pen and paper.



After 10 minutes or so I figured out the solution to this problem and questioned myself why the question is called infamous. I came across some solutions at this site that are, in my opinion, mindblowing in terms of length and mathematical notation, for example this solution. Then I read something about Vieta-jumping that I'm not familiar with and how it's used to solve this question (or not). A rather simple solution I came across is this one. I think the elegance and quality of a solution is greatly influenced by its length and level of simplicity in terms of notation, terminology, and construction.



As far as I did my silly online research I couldn't find a solution that was equivalent to mine. Therefore I thought it would be a good idea to share it and let other people, with a much broader understandig of existing mathematical proofs, decide if this proof is correct (and a known solution).










share|cite|improve this question















marked as duplicate by Jyrki Lahtonen, Lord Shark the Unknown, 5xum, Christopher, Jens Nov 13 at 16:03


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.















  • I mentioned that thread in my post. First, it is not a duplicate, since my question differs a lot from that one. Second, the solution mentioned in that thread also differs from my solution.
    – Pascal
    Mar 14 '17 at 18:13






  • 1




    Its always best to get quickly to your question, and then give us details about how you found it. The question is what volunteers can help you with, not your biography, and if people want to figure out quickly if they can help you with the question. Remember: You are asking people to give you their time, don't request more of it than you need.
    – Thomas Andrews
    Mar 14 '17 at 18:13






  • 1




    @Pascal You assumed that $k=b$ without any reason or justification. This reduced the problem from arbitrary integers $a,b$ to the much smaller subset of integers $a,b$ with $a=b^3$. It's (almost) as if I said "look, let's just assume $a=b=1,$, and in that case the fraction is $1$ i.e. a perfect square QED". That's not a proof, and doesn't answer the original question.
    – dxiv
    Mar 14 '17 at 18:23








  • 2




    @Pascal And I took $a=b=1$ because that gave me a square right away. Think at why my "proof" isn't valid, then you'll realize why yours isn't, either. You can't simply postulate that $k=b$ just because it simplifies your calculations.
    – dxiv
    Mar 14 '17 at 18:27






  • 1




    $a=8,b=30$ is a solution, for example, with $k=2$.
    – Thomas Andrews
    Mar 14 '17 at 18:29













up vote
1
down vote

favorite
1









up vote
1
down vote

favorite
1






1






This question already has an answer here:




  • Alternative proof that $(a^2+b^2)/(ab+1)$ is a square when it's an integer

    2 answers




Recall Question 6 of the 1988 Math Olympiad



Question 6 is as follows:




Let $a$ and $b$ be positive integers such that $ab + 1$ divides $a^2 +
b^2$. Show that $$frac{a^2+b^2}{ab+1}$$ is the square of an integer.




My proof:




Proof:



Let us denote the square of an integer by $k^2$. We have to show that
$$frac{a^2+b^2}{ab+1}=k^2$$ for some $a > 0$ and $b > 0$ where $a, b,
> k in Bbb Z$.



Rewriting the equation:



$$a^2+b^2=k^2(ab+1)=k^2ab+k^2$$



Let $k = b$:



$$a^2+b^2=b^3a+b^2 Rightarrow a^2=b^3a Rightarrow a=b^3$$



To check, let's substitute these relations back in the original
expression:



$$frac{a^2+b^2}{ab+1}=frac{(k^3)^2+(k)^2}{(k^3)(k)+1}=frac{k^6+k^2}{k^4+1}=frac{k^2(k^4+1)}{k^4+1}=k^2$$




Two of things I would like to know about this Olympiad Question:




  1. Is this proof correct?

  2. Is there a reliable way to check if a certain solution to a proof already exist?


Background



Every now and then I visit Youtube for educational purposes. I've been a great fan of Kurzgesagt, Numberphile, Computerphile, and other educational channels. For anyone not having heard of these channels I would greatly recommended to check them out!



Today I was watching this video on the channel of Numberphile: The World's Best Mathematician (*) - Numberphile. In this video a great mathematician Professor Tao tells about his passion for mathematics from a young age. At some point in this video it is mentioned that Tao only scored 1 out of 7 points for the infamous Question 6 from the 1988 Math Olympiad. In the next couple of frames this video shows the details of the question. At that moment I paused the video to try and solve this question by myself, with only pen and paper.



After 10 minutes or so I figured out the solution to this problem and questioned myself why the question is called infamous. I came across some solutions at this site that are, in my opinion, mindblowing in terms of length and mathematical notation, for example this solution. Then I read something about Vieta-jumping that I'm not familiar with and how it's used to solve this question (or not). A rather simple solution I came across is this one. I think the elegance and quality of a solution is greatly influenced by its length and level of simplicity in terms of notation, terminology, and construction.



As far as I did my silly online research I couldn't find a solution that was equivalent to mine. Therefore I thought it would be a good idea to share it and let other people, with a much broader understandig of existing mathematical proofs, decide if this proof is correct (and a known solution).










share|cite|improve this question
















This question already has an answer here:




  • Alternative proof that $(a^2+b^2)/(ab+1)$ is a square when it's an integer

    2 answers




Recall Question 6 of the 1988 Math Olympiad



Question 6 is as follows:




Let $a$ and $b$ be positive integers such that $ab + 1$ divides $a^2 +
b^2$. Show that $$frac{a^2+b^2}{ab+1}$$ is the square of an integer.




My proof:




Proof:



Let us denote the square of an integer by $k^2$. We have to show that
$$frac{a^2+b^2}{ab+1}=k^2$$ for some $a > 0$ and $b > 0$ where $a, b,
> k in Bbb Z$.



Rewriting the equation:



$$a^2+b^2=k^2(ab+1)=k^2ab+k^2$$



Let $k = b$:



$$a^2+b^2=b^3a+b^2 Rightarrow a^2=b^3a Rightarrow a=b^3$$



To check, let's substitute these relations back in the original
expression:



$$frac{a^2+b^2}{ab+1}=frac{(k^3)^2+(k)^2}{(k^3)(k)+1}=frac{k^6+k^2}{k^4+1}=frac{k^2(k^4+1)}{k^4+1}=k^2$$




Two of things I would like to know about this Olympiad Question:




  1. Is this proof correct?

  2. Is there a reliable way to check if a certain solution to a proof already exist?


Background



Every now and then I visit Youtube for educational purposes. I've been a great fan of Kurzgesagt, Numberphile, Computerphile, and other educational channels. For anyone not having heard of these channels I would greatly recommended to check them out!



Today I was watching this video on the channel of Numberphile: The World's Best Mathematician (*) - Numberphile. In this video a great mathematician Professor Tao tells about his passion for mathematics from a young age. At some point in this video it is mentioned that Tao only scored 1 out of 7 points for the infamous Question 6 from the 1988 Math Olympiad. In the next couple of frames this video shows the details of the question. At that moment I paused the video to try and solve this question by myself, with only pen and paper.



After 10 minutes or so I figured out the solution to this problem and questioned myself why the question is called infamous. I came across some solutions at this site that are, in my opinion, mindblowing in terms of length and mathematical notation, for example this solution. Then I read something about Vieta-jumping that I'm not familiar with and how it's used to solve this question (or not). A rather simple solution I came across is this one. I think the elegance and quality of a solution is greatly influenced by its length and level of simplicity in terms of notation, terminology, and construction.



As far as I did my silly online research I couldn't find a solution that was equivalent to mine. Therefore I thought it would be a good idea to share it and let other people, with a much broader understandig of existing mathematical proofs, decide if this proof is correct (and a known solution).





This question already has an answer here:




  • Alternative proof that $(a^2+b^2)/(ab+1)$ is a square when it's an integer

    2 answers








contest-math






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share|cite|improve this question













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edited Apr 13 '17 at 12:20









Community

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asked Mar 14 '17 at 18:08









Pascal

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marked as duplicate by Jyrki Lahtonen, Lord Shark the Unknown, 5xum, Christopher, Jens Nov 13 at 16:03


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 Jyrki Lahtonen, Lord Shark the Unknown, 5xum, Christopher, Jens Nov 13 at 16:03


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.














  • I mentioned that thread in my post. First, it is not a duplicate, since my question differs a lot from that one. Second, the solution mentioned in that thread also differs from my solution.
    – Pascal
    Mar 14 '17 at 18:13






  • 1




    Its always best to get quickly to your question, and then give us details about how you found it. The question is what volunteers can help you with, not your biography, and if people want to figure out quickly if they can help you with the question. Remember: You are asking people to give you their time, don't request more of it than you need.
    – Thomas Andrews
    Mar 14 '17 at 18:13






  • 1




    @Pascal You assumed that $k=b$ without any reason or justification. This reduced the problem from arbitrary integers $a,b$ to the much smaller subset of integers $a,b$ with $a=b^3$. It's (almost) as if I said "look, let's just assume $a=b=1,$, and in that case the fraction is $1$ i.e. a perfect square QED". That's not a proof, and doesn't answer the original question.
    – dxiv
    Mar 14 '17 at 18:23








  • 2




    @Pascal And I took $a=b=1$ because that gave me a square right away. Think at why my "proof" isn't valid, then you'll realize why yours isn't, either. You can't simply postulate that $k=b$ just because it simplifies your calculations.
    – dxiv
    Mar 14 '17 at 18:27






  • 1




    $a=8,b=30$ is a solution, for example, with $k=2$.
    – Thomas Andrews
    Mar 14 '17 at 18:29


















  • I mentioned that thread in my post. First, it is not a duplicate, since my question differs a lot from that one. Second, the solution mentioned in that thread also differs from my solution.
    – Pascal
    Mar 14 '17 at 18:13






  • 1




    Its always best to get quickly to your question, and then give us details about how you found it. The question is what volunteers can help you with, not your biography, and if people want to figure out quickly if they can help you with the question. Remember: You are asking people to give you their time, don't request more of it than you need.
    – Thomas Andrews
    Mar 14 '17 at 18:13






  • 1




    @Pascal You assumed that $k=b$ without any reason or justification. This reduced the problem from arbitrary integers $a,b$ to the much smaller subset of integers $a,b$ with $a=b^3$. It's (almost) as if I said "look, let's just assume $a=b=1,$, and in that case the fraction is $1$ i.e. a perfect square QED". That's not a proof, and doesn't answer the original question.
    – dxiv
    Mar 14 '17 at 18:23








  • 2




    @Pascal And I took $a=b=1$ because that gave me a square right away. Think at why my "proof" isn't valid, then you'll realize why yours isn't, either. You can't simply postulate that $k=b$ just because it simplifies your calculations.
    – dxiv
    Mar 14 '17 at 18:27






  • 1




    $a=8,b=30$ is a solution, for example, with $k=2$.
    – Thomas Andrews
    Mar 14 '17 at 18:29
















I mentioned that thread in my post. First, it is not a duplicate, since my question differs a lot from that one. Second, the solution mentioned in that thread also differs from my solution.
– Pascal
Mar 14 '17 at 18:13




I mentioned that thread in my post. First, it is not a duplicate, since my question differs a lot from that one. Second, the solution mentioned in that thread also differs from my solution.
– Pascal
Mar 14 '17 at 18:13




1




1




Its always best to get quickly to your question, and then give us details about how you found it. The question is what volunteers can help you with, not your biography, and if people want to figure out quickly if they can help you with the question. Remember: You are asking people to give you their time, don't request more of it than you need.
– Thomas Andrews
Mar 14 '17 at 18:13




Its always best to get quickly to your question, and then give us details about how you found it. The question is what volunteers can help you with, not your biography, and if people want to figure out quickly if they can help you with the question. Remember: You are asking people to give you their time, don't request more of it than you need.
– Thomas Andrews
Mar 14 '17 at 18:13




1




1




@Pascal You assumed that $k=b$ without any reason or justification. This reduced the problem from arbitrary integers $a,b$ to the much smaller subset of integers $a,b$ with $a=b^3$. It's (almost) as if I said "look, let's just assume $a=b=1,$, and in that case the fraction is $1$ i.e. a perfect square QED". That's not a proof, and doesn't answer the original question.
– dxiv
Mar 14 '17 at 18:23






@Pascal You assumed that $k=b$ without any reason or justification. This reduced the problem from arbitrary integers $a,b$ to the much smaller subset of integers $a,b$ with $a=b^3$. It's (almost) as if I said "look, let's just assume $a=b=1,$, and in that case the fraction is $1$ i.e. a perfect square QED". That's not a proof, and doesn't answer the original question.
– dxiv
Mar 14 '17 at 18:23






2




2




@Pascal And I took $a=b=1$ because that gave me a square right away. Think at why my "proof" isn't valid, then you'll realize why yours isn't, either. You can't simply postulate that $k=b$ just because it simplifies your calculations.
– dxiv
Mar 14 '17 at 18:27




@Pascal And I took $a=b=1$ because that gave me a square right away. Think at why my "proof" isn't valid, then you'll realize why yours isn't, either. You can't simply postulate that $k=b$ just because it simplifies your calculations.
– dxiv
Mar 14 '17 at 18:27




1




1




$a=8,b=30$ is a solution, for example, with $k=2$.
– Thomas Andrews
Mar 14 '17 at 18:29




$a=8,b=30$ is a solution, for example, with $k=2$.
– Thomas Andrews
Mar 14 '17 at 18:29










2 Answers
2






active

oldest

votes

















up vote
6
down vote



accepted










$a=8,b=30$ gives $k=2$. So you haven't gotten all answers when you've assumed $k=b$.



You are trying to prove it for all cases. You've only proven it for some cases.






share|cite|improve this answer




























    up vote
    0
    down vote













    Dividing by the first principle
    $$
    ab+1)a^2+b^2(frac{a}{b}
    $$

    $$
    a^2+frac{a}{b}
    $$





    $$
    b^2-frac{a}{b}
    $$

    Since ab+1 completely divides $$a^2+b^2$$
    $$b^2-frac{a}{b}=0\
    therefore a=b^3
    $$

    The quotient is $$ aover{b}$$
    Putting the value of a
    $$frac{a}{b}=b^2$$
    Hence it is a perfect square






    share|cite|improve this answer








    New contributor




    Arnav Kundu is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
    Check out our Code of Conduct.


















    • Check out the first paragraph, please. I don't have a clue as to what you are doing there. Also, the solution $a=30, b=8$ proves that your claim $a=b^3$ is simply false.
      – Jyrki Lahtonen
      Nov 13 at 5:37


















    2 Answers
    2






    active

    oldest

    votes








    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes








    up vote
    6
    down vote



    accepted










    $a=8,b=30$ gives $k=2$. So you haven't gotten all answers when you've assumed $k=b$.



    You are trying to prove it for all cases. You've only proven it for some cases.






    share|cite|improve this answer

























      up vote
      6
      down vote



      accepted










      $a=8,b=30$ gives $k=2$. So you haven't gotten all answers when you've assumed $k=b$.



      You are trying to prove it for all cases. You've only proven it for some cases.






      share|cite|improve this answer























        up vote
        6
        down vote



        accepted







        up vote
        6
        down vote



        accepted






        $a=8,b=30$ gives $k=2$. So you haven't gotten all answers when you've assumed $k=b$.



        You are trying to prove it for all cases. You've only proven it for some cases.






        share|cite|improve this answer












        $a=8,b=30$ gives $k=2$. So you haven't gotten all answers when you've assumed $k=b$.



        You are trying to prove it for all cases. You've only proven it for some cases.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Mar 14 '17 at 18:30









        Thomas Andrews

        129k10145293




        129k10145293






















            up vote
            0
            down vote













            Dividing by the first principle
            $$
            ab+1)a^2+b^2(frac{a}{b}
            $$

            $$
            a^2+frac{a}{b}
            $$





            $$
            b^2-frac{a}{b}
            $$

            Since ab+1 completely divides $$a^2+b^2$$
            $$b^2-frac{a}{b}=0\
            therefore a=b^3
            $$

            The quotient is $$ aover{b}$$
            Putting the value of a
            $$frac{a}{b}=b^2$$
            Hence it is a perfect square






            share|cite|improve this answer








            New contributor




            Arnav Kundu is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
            Check out our Code of Conduct.


















            • Check out the first paragraph, please. I don't have a clue as to what you are doing there. Also, the solution $a=30, b=8$ proves that your claim $a=b^3$ is simply false.
              – Jyrki Lahtonen
              Nov 13 at 5:37















            up vote
            0
            down vote













            Dividing by the first principle
            $$
            ab+1)a^2+b^2(frac{a}{b}
            $$

            $$
            a^2+frac{a}{b}
            $$





            $$
            b^2-frac{a}{b}
            $$

            Since ab+1 completely divides $$a^2+b^2$$
            $$b^2-frac{a}{b}=0\
            therefore a=b^3
            $$

            The quotient is $$ aover{b}$$
            Putting the value of a
            $$frac{a}{b}=b^2$$
            Hence it is a perfect square






            share|cite|improve this answer








            New contributor




            Arnav Kundu is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
            Check out our Code of Conduct.


















            • Check out the first paragraph, please. I don't have a clue as to what you are doing there. Also, the solution $a=30, b=8$ proves that your claim $a=b^3$ is simply false.
              – Jyrki Lahtonen
              Nov 13 at 5:37













            up vote
            0
            down vote










            up vote
            0
            down vote









            Dividing by the first principle
            $$
            ab+1)a^2+b^2(frac{a}{b}
            $$

            $$
            a^2+frac{a}{b}
            $$





            $$
            b^2-frac{a}{b}
            $$

            Since ab+1 completely divides $$a^2+b^2$$
            $$b^2-frac{a}{b}=0\
            therefore a=b^3
            $$

            The quotient is $$ aover{b}$$
            Putting the value of a
            $$frac{a}{b}=b^2$$
            Hence it is a perfect square






            share|cite|improve this answer








            New contributor




            Arnav Kundu is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
            Check out our Code of Conduct.









            Dividing by the first principle
            $$
            ab+1)a^2+b^2(frac{a}{b}
            $$

            $$
            a^2+frac{a}{b}
            $$





            $$
            b^2-frac{a}{b}
            $$

            Since ab+1 completely divides $$a^2+b^2$$
            $$b^2-frac{a}{b}=0\
            therefore a=b^3
            $$

            The quotient is $$ aover{b}$$
            Putting the value of a
            $$frac{a}{b}=b^2$$
            Hence it is a perfect square







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            answered Nov 13 at 5:13









            Arnav Kundu

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            New contributor





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            • Check out the first paragraph, please. I don't have a clue as to what you are doing there. Also, the solution $a=30, b=8$ proves that your claim $a=b^3$ is simply false.
              – Jyrki Lahtonen
              Nov 13 at 5:37


















            • Check out the first paragraph, please. I don't have a clue as to what you are doing there. Also, the solution $a=30, b=8$ proves that your claim $a=b^3$ is simply false.
              – Jyrki Lahtonen
              Nov 13 at 5:37
















            Check out the first paragraph, please. I don't have a clue as to what you are doing there. Also, the solution $a=30, b=8$ proves that your claim $a=b^3$ is simply false.
            – Jyrki Lahtonen
            Nov 13 at 5:37




            Check out the first paragraph, please. I don't have a clue as to what you are doing there. Also, the solution $a=30, b=8$ proves that your claim $a=b^3$ is simply false.
            – Jyrki Lahtonen
            Nov 13 at 5:37



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