Showing that the Diophantine equation $3x^2 + 6y^6 + 1 = 8xy^3$ has no solutions $x,y in mathbb{Q}$












0












$begingroup$



I want to show that the Diophantine equation $3x^2 + 6y^6 + 1 = 8xy^3$ has no solutions $x,y in mathbb{Q}$.




I tried factoring, but didn't manage (but I'm not good at factoring). Then I tried reducing $mod{7}$, but this didn't gave decisive results.






number-theory elementary-number-theory diophantine-equations







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  • 3




    $begingroup$
    I think your problem statement is missing something... what do you want to show about the Diophantine equation?
    $endgroup$
    – RandomMathGuy
    Dec 7 '18 at 20:36










  • $begingroup$
    I fixed the question with what I thought would be the correct question. Please immediately edit it if my guess was wrong.
    $endgroup$
    – Batominovski
    Dec 7 '18 at 20:46










  • $begingroup$
    You were right in the modification, thanks! Indeed, I wanted to show that there are no solutions.
    $endgroup$
    – Jens Wagemaker
    Dec 7 '18 at 20:47
















0












$begingroup$



I want to show that the Diophantine equation $3x^2 + 6y^6 + 1 = 8xy^3$ has no solutions $x,y in mathbb{Q}$.




I tried factoring, but didn't manage (but I'm not good at factoring). Then I tried reducing $mod{7}$, but this didn't gave decisive results.






number-theory elementary-number-theory diophantine-equations







share|cite|improve this question











$endgroup$








  • 3




    $begingroup$
    I think your problem statement is missing something... what do you want to show about the Diophantine equation?
    $endgroup$
    – RandomMathGuy
    Dec 7 '18 at 20:36










  • $begingroup$
    I fixed the question with what I thought would be the correct question. Please immediately edit it if my guess was wrong.
    $endgroup$
    – Batominovski
    Dec 7 '18 at 20:46










  • $begingroup$
    You were right in the modification, thanks! Indeed, I wanted to show that there are no solutions.
    $endgroup$
    – Jens Wagemaker
    Dec 7 '18 at 20:47














0












0








0





$begingroup$



I want to show that the Diophantine equation $3x^2 + 6y^6 + 1 = 8xy^3$ has no solutions $x,y in mathbb{Q}$.




I tried factoring, but didn't manage (but I'm not good at factoring). Then I tried reducing $mod{7}$, but this didn't gave decisive results.






number-theory elementary-number-theory diophantine-equations







share|cite|improve this question











$endgroup$





I want to show that the Diophantine equation $3x^2 + 6y^6 + 1 = 8xy^3$ has no solutions $x,y in mathbb{Q}$.




I tried factoring, but didn't manage (but I'm not good at factoring). Then I tried reducing $mod{7}$, but this didn't gave decisive results.






number-theory elementary-number-theory diophantine-equations




number-theory elementary-number-theory diophantine-equations






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 7 '18 at 20:45









Batominovski

1




1










asked Dec 7 '18 at 20:30









Jens WagemakerJens Wagemaker

550312




550312








  • 3




    $begingroup$
    I think your problem statement is missing something... what do you want to show about the Diophantine equation?
    $endgroup$
    – RandomMathGuy
    Dec 7 '18 at 20:36










  • $begingroup$
    I fixed the question with what I thought would be the correct question. Please immediately edit it if my guess was wrong.
    $endgroup$
    – Batominovski
    Dec 7 '18 at 20:46










  • $begingroup$
    You were right in the modification, thanks! Indeed, I wanted to show that there are no solutions.
    $endgroup$
    – Jens Wagemaker
    Dec 7 '18 at 20:47














  • 3




    $begingroup$
    I think your problem statement is missing something... what do you want to show about the Diophantine equation?
    $endgroup$
    – RandomMathGuy
    Dec 7 '18 at 20:36










  • $begingroup$
    I fixed the question with what I thought would be the correct question. Please immediately edit it if my guess was wrong.
    $endgroup$
    – Batominovski
    Dec 7 '18 at 20:46










  • $begingroup$
    You were right in the modification, thanks! Indeed, I wanted to show that there are no solutions.
    $endgroup$
    – Jens Wagemaker
    Dec 7 '18 at 20:47








3




3




$begingroup$
I think your problem statement is missing something... what do you want to show about the Diophantine equation?
$endgroup$
– RandomMathGuy
Dec 7 '18 at 20:36




$begingroup$
I think your problem statement is missing something... what do you want to show about the Diophantine equation?
$endgroup$
– RandomMathGuy
Dec 7 '18 at 20:36












$begingroup$
I fixed the question with what I thought would be the correct question. Please immediately edit it if my guess was wrong.
$endgroup$
– Batominovski
Dec 7 '18 at 20:46




$begingroup$
I fixed the question with what I thought would be the correct question. Please immediately edit it if my guess was wrong.
$endgroup$
– Batominovski
Dec 7 '18 at 20:46












$begingroup$
You were right in the modification, thanks! Indeed, I wanted to show that there are no solutions.
$endgroup$
– Jens Wagemaker
Dec 7 '18 at 20:47




$begingroup$
You were right in the modification, thanks! Indeed, I wanted to show that there are no solutions.
$endgroup$
– Jens Wagemaker
Dec 7 '18 at 20:47










3 Answers
3






active

oldest

votes


















1












$begingroup$

Rewriting it as $6y^6-8xy^3+3x^2+1=0$, we have a quadratic equation in $y^3$, so that



$$y^3={4xpmsqrt{(4x)^2-6(3x^2+1)}over6}={4xpmsqrt{-(2x^2+6)}over6}$$



The square root is imaginary for all real $x$, so there are not only no rational solutions, there aren't any real ones either.



Alternatively, it's a quadratic in $x$, with solution



$$x={4y^3pmsqrt{(4y^3)^2-3(6y^6+1)}over3}={4y^3pmsqrt{-(2y^6+3)}over3}$$



for which the square root is imaginary for real $y$. (For some reason I noticed the equation as a quadratic in $y^3$ first!)






share|cite|improve this answer











$endgroup$





















    4












    $begingroup$

    $$ 3 u^2 - 8uv + 6 v^2 = frac{1}{3} (3u-4v)^2 + frac{2}{3} v^2 $$ is positive definite for real $u,v$



    $$ Q^T D Q = H $$
    $$left(
    begin{array}{rr}
    1 & 0 \
    - frac{ 4 }{ 3 } & 1 \
    end{array}
    right)
    left(
    begin{array}{rr}
    3 & 0 \
    0 & frac{ 2 }{ 3 } \
    end{array}
    right)
    left(
    begin{array}{rr}
    1 & - frac{ 4 }{ 3 } \
    0 & 1 \
    end{array}
    right)
    = left(
    begin{array}{rr}
    3 & - 4 \
    - 4 & 6 \
    end{array}
    right)
    $$






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      The answer looks nice, but I'm a bit puzzled. Can you explain your factorization technique?
      $endgroup$
      – Jens Wagemaker
      Dec 7 '18 at 20:50










    • $begingroup$
      @JensWagemaker see math.stackexchange.com/questions/1388421/…
      $endgroup$
      – Will Jagy
      Dec 7 '18 at 20:51



















    1












    $begingroup$

    Let the pair $(a, b)$ soluation for that equation
    $$
    \6sqrt{2}>8=>
    \8ab^3=3a^2+6b^6+1ge2sqrt{18a^2b^6}+1>6sqrt{2}|ab^3|ge8|ab^3|ge8ab^3=>
    \8ab^3>8ab^3
    $$
    A contradiction.






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      What is $AM$ and $GM$?
      $endgroup$
      – Jens Wagemaker
      Dec 7 '18 at 22:05










    • $begingroup$
      Arifmetic and Geometric mean.
      $endgroup$
      – Samvel Safaryan
      Dec 7 '18 at 22:05










    • $begingroup$
      How did you come up with these estimates?
      $endgroup$
      – Jens Wagemaker
      Dec 7 '18 at 22:07










    • $begingroup$
      if $a, bge 0$ then $a+bge2sqrt{ab}$
      $endgroup$
      – Samvel Safaryan
      Dec 7 '18 at 22:09











    Your Answer





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    3 Answers
    3






    active

    oldest

    votes








    3 Answers
    3






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    1












    $begingroup$

    Rewriting it as $6y^6-8xy^3+3x^2+1=0$, we have a quadratic equation in $y^3$, so that



    $$y^3={4xpmsqrt{(4x)^2-6(3x^2+1)}over6}={4xpmsqrt{-(2x^2+6)}over6}$$



    The square root is imaginary for all real $x$, so there are not only no rational solutions, there aren't any real ones either.



    Alternatively, it's a quadratic in $x$, with solution



    $$x={4y^3pmsqrt{(4y^3)^2-3(6y^6+1)}over3}={4y^3pmsqrt{-(2y^6+3)}over3}$$



    for which the square root is imaginary for real $y$. (For some reason I noticed the equation as a quadratic in $y^3$ first!)






    share|cite|improve this answer











    $endgroup$


















      1












      $begingroup$

      Rewriting it as $6y^6-8xy^3+3x^2+1=0$, we have a quadratic equation in $y^3$, so that



      $$y^3={4xpmsqrt{(4x)^2-6(3x^2+1)}over6}={4xpmsqrt{-(2x^2+6)}over6}$$



      The square root is imaginary for all real $x$, so there are not only no rational solutions, there aren't any real ones either.



      Alternatively, it's a quadratic in $x$, with solution



      $$x={4y^3pmsqrt{(4y^3)^2-3(6y^6+1)}over3}={4y^3pmsqrt{-(2y^6+3)}over3}$$



      for which the square root is imaginary for real $y$. (For some reason I noticed the equation as a quadratic in $y^3$ first!)






      share|cite|improve this answer











      $endgroup$
















        1












        1








        1





        $begingroup$

        Rewriting it as $6y^6-8xy^3+3x^2+1=0$, we have a quadratic equation in $y^3$, so that



        $$y^3={4xpmsqrt{(4x)^2-6(3x^2+1)}over6}={4xpmsqrt{-(2x^2+6)}over6}$$



        The square root is imaginary for all real $x$, so there are not only no rational solutions, there aren't any real ones either.



        Alternatively, it's a quadratic in $x$, with solution



        $$x={4y^3pmsqrt{(4y^3)^2-3(6y^6+1)}over3}={4y^3pmsqrt{-(2y^6+3)}over3}$$



        for which the square root is imaginary for real $y$. (For some reason I noticed the equation as a quadratic in $y^3$ first!)






        share|cite|improve this answer











        $endgroup$



        Rewriting it as $6y^6-8xy^3+3x^2+1=0$, we have a quadratic equation in $y^3$, so that



        $$y^3={4xpmsqrt{(4x)^2-6(3x^2+1)}over6}={4xpmsqrt{-(2x^2+6)}over6}$$



        The square root is imaginary for all real $x$, so there are not only no rational solutions, there aren't any real ones either.



        Alternatively, it's a quadratic in $x$, with solution



        $$x={4y^3pmsqrt{(4y^3)^2-3(6y^6+1)}over3}={4y^3pmsqrt{-(2y^6+3)}over3}$$



        for which the square root is imaginary for real $y$. (For some reason I noticed the equation as a quadratic in $y^3$ first!)







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Dec 7 '18 at 22:18

























        answered Dec 7 '18 at 22:13









        Barry CipraBarry Cipra

        59.4k653126




        59.4k653126























            4












            $begingroup$

            $$ 3 u^2 - 8uv + 6 v^2 = frac{1}{3} (3u-4v)^2 + frac{2}{3} v^2 $$ is positive definite for real $u,v$



            $$ Q^T D Q = H $$
            $$left(
            begin{array}{rr}
            1 & 0 \
            - frac{ 4 }{ 3 } & 1 \
            end{array}
            right)
            left(
            begin{array}{rr}
            3 & 0 \
            0 & frac{ 2 }{ 3 } \
            end{array}
            right)
            left(
            begin{array}{rr}
            1 & - frac{ 4 }{ 3 } \
            0 & 1 \
            end{array}
            right)
            = left(
            begin{array}{rr}
            3 & - 4 \
            - 4 & 6 \
            end{array}
            right)
            $$






            share|cite|improve this answer











            $endgroup$













            • $begingroup$
              The answer looks nice, but I'm a bit puzzled. Can you explain your factorization technique?
              $endgroup$
              – Jens Wagemaker
              Dec 7 '18 at 20:50










            • $begingroup$
              @JensWagemaker see math.stackexchange.com/questions/1388421/…
              $endgroup$
              – Will Jagy
              Dec 7 '18 at 20:51
















            4












            $begingroup$

            $$ 3 u^2 - 8uv + 6 v^2 = frac{1}{3} (3u-4v)^2 + frac{2}{3} v^2 $$ is positive definite for real $u,v$



            $$ Q^T D Q = H $$
            $$left(
            begin{array}{rr}
            1 & 0 \
            - frac{ 4 }{ 3 } & 1 \
            end{array}
            right)
            left(
            begin{array}{rr}
            3 & 0 \
            0 & frac{ 2 }{ 3 } \
            end{array}
            right)
            left(
            begin{array}{rr}
            1 & - frac{ 4 }{ 3 } \
            0 & 1 \
            end{array}
            right)
            = left(
            begin{array}{rr}
            3 & - 4 \
            - 4 & 6 \
            end{array}
            right)
            $$






            share|cite|improve this answer











            $endgroup$













            • $begingroup$
              The answer looks nice, but I'm a bit puzzled. Can you explain your factorization technique?
              $endgroup$
              – Jens Wagemaker
              Dec 7 '18 at 20:50










            • $begingroup$
              @JensWagemaker see math.stackexchange.com/questions/1388421/…
              $endgroup$
              – Will Jagy
              Dec 7 '18 at 20:51














            4












            4








            4





            $begingroup$

            $$ 3 u^2 - 8uv + 6 v^2 = frac{1}{3} (3u-4v)^2 + frac{2}{3} v^2 $$ is positive definite for real $u,v$



            $$ Q^T D Q = H $$
            $$left(
            begin{array}{rr}
            1 & 0 \
            - frac{ 4 }{ 3 } & 1 \
            end{array}
            right)
            left(
            begin{array}{rr}
            3 & 0 \
            0 & frac{ 2 }{ 3 } \
            end{array}
            right)
            left(
            begin{array}{rr}
            1 & - frac{ 4 }{ 3 } \
            0 & 1 \
            end{array}
            right)
            = left(
            begin{array}{rr}
            3 & - 4 \
            - 4 & 6 \
            end{array}
            right)
            $$






            share|cite|improve this answer











            $endgroup$



            $$ 3 u^2 - 8uv + 6 v^2 = frac{1}{3} (3u-4v)^2 + frac{2}{3} v^2 $$ is positive definite for real $u,v$



            $$ Q^T D Q = H $$
            $$left(
            begin{array}{rr}
            1 & 0 \
            - frac{ 4 }{ 3 } & 1 \
            end{array}
            right)
            left(
            begin{array}{rr}
            3 & 0 \
            0 & frac{ 2 }{ 3 } \
            end{array}
            right)
            left(
            begin{array}{rr}
            1 & - frac{ 4 }{ 3 } \
            0 & 1 \
            end{array}
            right)
            = left(
            begin{array}{rr}
            3 & - 4 \
            - 4 & 6 \
            end{array}
            right)
            $$







            share|cite|improve this answer














            share|cite|improve this answer



            share|cite|improve this answer








            edited Dec 7 '18 at 20:47

























            answered Dec 7 '18 at 20:34









            Will JagyWill Jagy

            103k5101200




            103k5101200












            • $begingroup$
              The answer looks nice, but I'm a bit puzzled. Can you explain your factorization technique?
              $endgroup$
              – Jens Wagemaker
              Dec 7 '18 at 20:50










            • $begingroup$
              @JensWagemaker see math.stackexchange.com/questions/1388421/…
              $endgroup$
              – Will Jagy
              Dec 7 '18 at 20:51


















            • $begingroup$
              The answer looks nice, but I'm a bit puzzled. Can you explain your factorization technique?
              $endgroup$
              – Jens Wagemaker
              Dec 7 '18 at 20:50










            • $begingroup$
              @JensWagemaker see math.stackexchange.com/questions/1388421/…
              $endgroup$
              – Will Jagy
              Dec 7 '18 at 20:51
















            $begingroup$
            The answer looks nice, but I'm a bit puzzled. Can you explain your factorization technique?
            $endgroup$
            – Jens Wagemaker
            Dec 7 '18 at 20:50




            $begingroup$
            The answer looks nice, but I'm a bit puzzled. Can you explain your factorization technique?
            $endgroup$
            – Jens Wagemaker
            Dec 7 '18 at 20:50












            $begingroup$
            @JensWagemaker see math.stackexchange.com/questions/1388421/…
            $endgroup$
            – Will Jagy
            Dec 7 '18 at 20:51




            $begingroup$
            @JensWagemaker see math.stackexchange.com/questions/1388421/…
            $endgroup$
            – Will Jagy
            Dec 7 '18 at 20:51











            1












            $begingroup$

            Let the pair $(a, b)$ soluation for that equation
            $$
            \6sqrt{2}>8=>
            \8ab^3=3a^2+6b^6+1ge2sqrt{18a^2b^6}+1>6sqrt{2}|ab^3|ge8|ab^3|ge8ab^3=>
            \8ab^3>8ab^3
            $$
            A contradiction.






            share|cite|improve this answer











            $endgroup$













            • $begingroup$
              What is $AM$ and $GM$?
              $endgroup$
              – Jens Wagemaker
              Dec 7 '18 at 22:05










            • $begingroup$
              Arifmetic and Geometric mean.
              $endgroup$
              – Samvel Safaryan
              Dec 7 '18 at 22:05










            • $begingroup$
              How did you come up with these estimates?
              $endgroup$
              – Jens Wagemaker
              Dec 7 '18 at 22:07










            • $begingroup$
              if $a, bge 0$ then $a+bge2sqrt{ab}$
              $endgroup$
              – Samvel Safaryan
              Dec 7 '18 at 22:09
















            1












            $begingroup$

            Let the pair $(a, b)$ soluation for that equation
            $$
            \6sqrt{2}>8=>
            \8ab^3=3a^2+6b^6+1ge2sqrt{18a^2b^6}+1>6sqrt{2}|ab^3|ge8|ab^3|ge8ab^3=>
            \8ab^3>8ab^3
            $$
            A contradiction.






            share|cite|improve this answer











            $endgroup$













            • $begingroup$
              What is $AM$ and $GM$?
              $endgroup$
              – Jens Wagemaker
              Dec 7 '18 at 22:05










            • $begingroup$
              Arifmetic and Geometric mean.
              $endgroup$
              – Samvel Safaryan
              Dec 7 '18 at 22:05










            • $begingroup$
              How did you come up with these estimates?
              $endgroup$
              – Jens Wagemaker
              Dec 7 '18 at 22:07










            • $begingroup$
              if $a, bge 0$ then $a+bge2sqrt{ab}$
              $endgroup$
              – Samvel Safaryan
              Dec 7 '18 at 22:09














            1












            1








            1





            $begingroup$

            Let the pair $(a, b)$ soluation for that equation
            $$
            \6sqrt{2}>8=>
            \8ab^3=3a^2+6b^6+1ge2sqrt{18a^2b^6}+1>6sqrt{2}|ab^3|ge8|ab^3|ge8ab^3=>
            \8ab^3>8ab^3
            $$
            A contradiction.






            share|cite|improve this answer











            $endgroup$



            Let the pair $(a, b)$ soluation for that equation
            $$
            \6sqrt{2}>8=>
            \8ab^3=3a^2+6b^6+1ge2sqrt{18a^2b^6}+1>6sqrt{2}|ab^3|ge8|ab^3|ge8ab^3=>
            \8ab^3>8ab^3
            $$
            A contradiction.







            share|cite|improve this answer














            share|cite|improve this answer



            share|cite|improve this answer








            edited Dec 7 '18 at 22:06

























            answered Dec 7 '18 at 22:03









            Samvel SafaryanSamvel Safaryan

            493111




            493111












            • $begingroup$
              What is $AM$ and $GM$?
              $endgroup$
              – Jens Wagemaker
              Dec 7 '18 at 22:05










            • $begingroup$
              Arifmetic and Geometric mean.
              $endgroup$
              – Samvel Safaryan
              Dec 7 '18 at 22:05










            • $begingroup$
              How did you come up with these estimates?
              $endgroup$
              – Jens Wagemaker
              Dec 7 '18 at 22:07










            • $begingroup$
              if $a, bge 0$ then $a+bge2sqrt{ab}$
              $endgroup$
              – Samvel Safaryan
              Dec 7 '18 at 22:09


















            • $begingroup$
              What is $AM$ and $GM$?
              $endgroup$
              – Jens Wagemaker
              Dec 7 '18 at 22:05










            • $begingroup$
              Arifmetic and Geometric mean.
              $endgroup$
              – Samvel Safaryan
              Dec 7 '18 at 22:05










            • $begingroup$
              How did you come up with these estimates?
              $endgroup$
              – Jens Wagemaker
              Dec 7 '18 at 22:07










            • $begingroup$
              if $a, bge 0$ then $a+bge2sqrt{ab}$
              $endgroup$
              – Samvel Safaryan
              Dec 7 '18 at 22:09
















            $begingroup$
            What is $AM$ and $GM$?
            $endgroup$
            – Jens Wagemaker
            Dec 7 '18 at 22:05




            $begingroup$
            What is $AM$ and $GM$?
            $endgroup$
            – Jens Wagemaker
            Dec 7 '18 at 22:05












            $begingroup$
            Arifmetic and Geometric mean.
            $endgroup$
            – Samvel Safaryan
            Dec 7 '18 at 22:05




            $begingroup$
            Arifmetic and Geometric mean.
            $endgroup$
            – Samvel Safaryan
            Dec 7 '18 at 22:05












            $begingroup$
            How did you come up with these estimates?
            $endgroup$
            – Jens Wagemaker
            Dec 7 '18 at 22:07




            $begingroup$
            How did you come up with these estimates?
            $endgroup$
            – Jens Wagemaker
            Dec 7 '18 at 22:07












            $begingroup$
            if $a, bge 0$ then $a+bge2sqrt{ab}$
            $endgroup$
            – Samvel Safaryan
            Dec 7 '18 at 22:09




            $begingroup$
            if $a, bge 0$ then $a+bge2sqrt{ab}$
            $endgroup$
            – Samvel Safaryan
            Dec 7 '18 at 22:09


















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