When does $sqrt{4x^3+4y^3+1}$ a natural number?












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If $x$ and $y$ is a natural number, how can we find $y$ in terms of $x$ so the $sqrt{4x^3+4y^3+1}$ is a natural number?






elementary-number-theory







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  • 1




    $begingroup$
    Is $yinBbb N$ too?
    $endgroup$
    – Lord Shark the Unknown
    Jan 17 at 7:22








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    $begingroup$
    You could take $y=-x$.
    $endgroup$
    – Lord Shark the Unknown
    Jan 17 at 7:24






  • 1




    $begingroup$
    I want to know when does 4x^3 + 4y^3 +1 is perfect square
    $endgroup$
    – Jerome Aligan
    Jan 17 at 7:25










  • $begingroup$
    What have you tried so far?
    $endgroup$
    – Klangen
    Jan 17 at 8:30
















0












$begingroup$


If $x$ and $y$ is a natural number, how can we find $y$ in terms of $x$ so the $sqrt{4x^3+4y^3+1}$ is a natural number?






elementary-number-theory







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    Is $yinBbb N$ too?
    $endgroup$
    – Lord Shark the Unknown
    Jan 17 at 7:22








  • 5




    $begingroup$
    You could take $y=-x$.
    $endgroup$
    – Lord Shark the Unknown
    Jan 17 at 7:24






  • 1




    $begingroup$
    I want to know when does 4x^3 + 4y^3 +1 is perfect square
    $endgroup$
    – Jerome Aligan
    Jan 17 at 7:25










  • $begingroup$
    What have you tried so far?
    $endgroup$
    – Klangen
    Jan 17 at 8:30














0












0








0


2



$begingroup$


If $x$ and $y$ is a natural number, how can we find $y$ in terms of $x$ so the $sqrt{4x^3+4y^3+1}$ is a natural number?






elementary-number-theory







share|cite|improve this question











$endgroup$




If $x$ and $y$ is a natural number, how can we find $y$ in terms of $x$ so the $sqrt{4x^3+4y^3+1}$ is a natural number?






elementary-number-theory




elementary-number-theory






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share|cite|improve this question













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share|cite|improve this question








edited Jan 17 at 8:22







Jerome Aligan

















asked Jan 17 at 7:20









Jerome AliganJerome Aligan

42




42








  • 1




    $begingroup$
    Is $yinBbb N$ too?
    $endgroup$
    – Lord Shark the Unknown
    Jan 17 at 7:22








  • 5




    $begingroup$
    You could take $y=-x$.
    $endgroup$
    – Lord Shark the Unknown
    Jan 17 at 7:24






  • 1




    $begingroup$
    I want to know when does 4x^3 + 4y^3 +1 is perfect square
    $endgroup$
    – Jerome Aligan
    Jan 17 at 7:25










  • $begingroup$
    What have you tried so far?
    $endgroup$
    – Klangen
    Jan 17 at 8:30














  • 1




    $begingroup$
    Is $yinBbb N$ too?
    $endgroup$
    – Lord Shark the Unknown
    Jan 17 at 7:22








  • 5




    $begingroup$
    You could take $y=-x$.
    $endgroup$
    – Lord Shark the Unknown
    Jan 17 at 7:24






  • 1




    $begingroup$
    I want to know when does 4x^3 + 4y^3 +1 is perfect square
    $endgroup$
    – Jerome Aligan
    Jan 17 at 7:25










  • $begingroup$
    What have you tried so far?
    $endgroup$
    – Klangen
    Jan 17 at 8:30








1




1




$begingroup$
Is $yinBbb N$ too?
$endgroup$
– Lord Shark the Unknown
Jan 17 at 7:22






$begingroup$
Is $yinBbb N$ too?
$endgroup$
– Lord Shark the Unknown
Jan 17 at 7:22






5




5




$begingroup$
You could take $y=-x$.
$endgroup$
– Lord Shark the Unknown
Jan 17 at 7:24




$begingroup$
You could take $y=-x$.
$endgroup$
– Lord Shark the Unknown
Jan 17 at 7:24




1




1




$begingroup$
I want to know when does 4x^3 + 4y^3 +1 is perfect square
$endgroup$
– Jerome Aligan
Jan 17 at 7:25




$begingroup$
I want to know when does 4x^3 + 4y^3 +1 is perfect square
$endgroup$
– Jerome Aligan
Jan 17 at 7:25












$begingroup$
What have you tried so far?
$endgroup$
– Klangen
Jan 17 at 8:30




$begingroup$
What have you tried so far?
$endgroup$
– Klangen
Jan 17 at 8:30










2 Answers
2






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oldest

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1












$begingroup$

For one family of solutions set $y=x^2$. Then



$z=sqrt{4x^3+4y^3+1}=sqrt{4x^3+4x^6+1}=sqrt{(2x^3+1)^2}=2x^3+1$



This gives:



$(x,y,z) = (1,1,3), space (2,4,17), space (3,9,55), dots$



There may, of course, be other solutions.






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    1












    $begingroup$

    $4x^3+4y^3=k^2-1=(k-1)(k+1)$



    Since LHS is even RHS is a product of two consecutive even numbers, let $k=2m+1$ we have:



    $4x^3+4y^3=2m(2m+2)=4m(m+1)$



    $x^3+y^3=m(m+1)$



    So whenever $x^3+y^3=m(m+1)$ and $k=2m+1$,then $4x^3+4y^3+1=k^2$



    For example:



    $m=1$ ⇒ $1(1+1)=2=1^3+1^3$⇒$k=2+1=3$



    $m=8$ ⇒ $8(8+1)=72=4^3+2^3$ ⇒ $k=2.8+1=17$



    $m=90$ ⇒ $90(90+1)=8190=19^3+11^3$ ⇒ $k=2.90+1=181$.






    share|cite|improve this answer









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      2 Answers
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      active

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      2 Answers
      2






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      active

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      1












      $begingroup$

      For one family of solutions set $y=x^2$. Then



      $z=sqrt{4x^3+4y^3+1}=sqrt{4x^3+4x^6+1}=sqrt{(2x^3+1)^2}=2x^3+1$



      This gives:



      $(x,y,z) = (1,1,3), space (2,4,17), space (3,9,55), dots$



      There may, of course, be other solutions.






      share|cite|improve this answer









      $endgroup$


















        1












        $begingroup$

        For one family of solutions set $y=x^2$. Then



        $z=sqrt{4x^3+4y^3+1}=sqrt{4x^3+4x^6+1}=sqrt{(2x^3+1)^2}=2x^3+1$



        This gives:



        $(x,y,z) = (1,1,3), space (2,4,17), space (3,9,55), dots$



        There may, of course, be other solutions.






        share|cite|improve this answer









        $endgroup$
















          1












          1








          1





          $begingroup$

          For one family of solutions set $y=x^2$. Then



          $z=sqrt{4x^3+4y^3+1}=sqrt{4x^3+4x^6+1}=sqrt{(2x^3+1)^2}=2x^3+1$



          This gives:



          $(x,y,z) = (1,1,3), space (2,4,17), space (3,9,55), dots$



          There may, of course, be other solutions.






          share|cite|improve this answer









          $endgroup$



          For one family of solutions set $y=x^2$. Then



          $z=sqrt{4x^3+4y^3+1}=sqrt{4x^3+4x^6+1}=sqrt{(2x^3+1)^2}=2x^3+1$



          This gives:



          $(x,y,z) = (1,1,3), space (2,4,17), space (3,9,55), dots$



          There may, of course, be other solutions.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Jan 17 at 11:31









          gandalf61gandalf61

          9,263825




          9,263825























              1












              $begingroup$

              $4x^3+4y^3=k^2-1=(k-1)(k+1)$



              Since LHS is even RHS is a product of two consecutive even numbers, let $k=2m+1$ we have:



              $4x^3+4y^3=2m(2m+2)=4m(m+1)$



              $x^3+y^3=m(m+1)$



              So whenever $x^3+y^3=m(m+1)$ and $k=2m+1$,then $4x^3+4y^3+1=k^2$



              For example:



              $m=1$ ⇒ $1(1+1)=2=1^3+1^3$⇒$k=2+1=3$



              $m=8$ ⇒ $8(8+1)=72=4^3+2^3$ ⇒ $k=2.8+1=17$



              $m=90$ ⇒ $90(90+1)=8190=19^3+11^3$ ⇒ $k=2.90+1=181$.






              share|cite|improve this answer









              $endgroup$


















                1












                $begingroup$

                $4x^3+4y^3=k^2-1=(k-1)(k+1)$



                Since LHS is even RHS is a product of two consecutive even numbers, let $k=2m+1$ we have:



                $4x^3+4y^3=2m(2m+2)=4m(m+1)$



                $x^3+y^3=m(m+1)$



                So whenever $x^3+y^3=m(m+1)$ and $k=2m+1$,then $4x^3+4y^3+1=k^2$



                For example:



                $m=1$ ⇒ $1(1+1)=2=1^3+1^3$⇒$k=2+1=3$



                $m=8$ ⇒ $8(8+1)=72=4^3+2^3$ ⇒ $k=2.8+1=17$



                $m=90$ ⇒ $90(90+1)=8190=19^3+11^3$ ⇒ $k=2.90+1=181$.






                share|cite|improve this answer









                $endgroup$
















                  1












                  1








                  1





                  $begingroup$

                  $4x^3+4y^3=k^2-1=(k-1)(k+1)$



                  Since LHS is even RHS is a product of two consecutive even numbers, let $k=2m+1$ we have:



                  $4x^3+4y^3=2m(2m+2)=4m(m+1)$



                  $x^3+y^3=m(m+1)$



                  So whenever $x^3+y^3=m(m+1)$ and $k=2m+1$,then $4x^3+4y^3+1=k^2$



                  For example:



                  $m=1$ ⇒ $1(1+1)=2=1^3+1^3$⇒$k=2+1=3$



                  $m=8$ ⇒ $8(8+1)=72=4^3+2^3$ ⇒ $k=2.8+1=17$



                  $m=90$ ⇒ $90(90+1)=8190=19^3+11^3$ ⇒ $k=2.90+1=181$.






                  share|cite|improve this answer









                  $endgroup$



                  $4x^3+4y^3=k^2-1=(k-1)(k+1)$



                  Since LHS is even RHS is a product of two consecutive even numbers, let $k=2m+1$ we have:



                  $4x^3+4y^3=2m(2m+2)=4m(m+1)$



                  $x^3+y^3=m(m+1)$



                  So whenever $x^3+y^3=m(m+1)$ and $k=2m+1$,then $4x^3+4y^3+1=k^2$



                  For example:



                  $m=1$ ⇒ $1(1+1)=2=1^3+1^3$⇒$k=2+1=3$



                  $m=8$ ⇒ $8(8+1)=72=4^3+2^3$ ⇒ $k=2.8+1=17$



                  $m=90$ ⇒ $90(90+1)=8190=19^3+11^3$ ⇒ $k=2.90+1=181$.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Jan 21 at 5:08









                  siroussirous

                  1,6981514




                  1,6981514






























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