Show that the curve $2Y^2 = X^4 - 17$ has no points in $mathbb{Q}$












6












$begingroup$


There is a hint to show that if there were points in $mathbb{Q}$, then there would exist $r,s, t in mathbb{Z}$ with $gcd(r, t) = 1$ such that $2s^2 = t^4 - 17r^4$
, and then show that any
prime dividing
$s$
is a quadratic residue modulo $17$.



I'm stuck on the first bit - if you suppose $left(frac{a}{b}, frac{c}{d}right)$ is a point on the curve in $mathbb{Q}$ (so $a, b, c, d in mathbb{Z}$) then you get to $2a^2d^4 = b^2c^4 - 17b^2d^4$, but I'm not sure where to go from here since you don't know that $b$ is a square?






number-theory elliptic-curves







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$endgroup$

















    6












    $begingroup$


    There is a hint to show that if there were points in $mathbb{Q}$, then there would exist $r,s, t in mathbb{Z}$ with $gcd(r, t) = 1$ such that $2s^2 = t^4 - 17r^4$
    , and then show that any
    prime dividing
    $s$
    is a quadratic residue modulo $17$.



    I'm stuck on the first bit - if you suppose $left(frac{a}{b}, frac{c}{d}right)$ is a point on the curve in $mathbb{Q}$ (so $a, b, c, d in mathbb{Z}$) then you get to $2a^2d^4 = b^2c^4 - 17b^2d^4$, but I'm not sure where to go from here since you don't know that $b$ is a square?






    number-theory elliptic-curves







    share|cite|improve this question









    $endgroup$















      6












      6








      6


      1



      $begingroup$


      There is a hint to show that if there were points in $mathbb{Q}$, then there would exist $r,s, t in mathbb{Z}$ with $gcd(r, t) = 1$ such that $2s^2 = t^4 - 17r^4$
      , and then show that any
      prime dividing
      $s$
      is a quadratic residue modulo $17$.



      I'm stuck on the first bit - if you suppose $left(frac{a}{b}, frac{c}{d}right)$ is a point on the curve in $mathbb{Q}$ (so $a, b, c, d in mathbb{Z}$) then you get to $2a^2d^4 = b^2c^4 - 17b^2d^4$, but I'm not sure where to go from here since you don't know that $b$ is a square?






      number-theory elliptic-curves







      share|cite|improve this question









      $endgroup$




      There is a hint to show that if there were points in $mathbb{Q}$, then there would exist $r,s, t in mathbb{Z}$ with $gcd(r, t) = 1$ such that $2s^2 = t^4 - 17r^4$
      , and then show that any
      prime dividing
      $s$
      is a quadratic residue modulo $17$.



      I'm stuck on the first bit - if you suppose $left(frac{a}{b}, frac{c}{d}right)$ is a point on the curve in $mathbb{Q}$ (so $a, b, c, d in mathbb{Z}$) then you get to $2a^2d^4 = b^2c^4 - 17b^2d^4$, but I'm not sure where to go from here since you don't know that $b$ is a square?






      number-theory elliptic-curves




      number-theory elliptic-curves






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked Feb 24 '14 at 9:48









      jemimajemima

      49929




      49929






















          1 Answer
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          $begingroup$

          $b$ need not be a square. Instead, the trick is to show that you can actually divide the equation by $b^2$.



          Let $p$ be a prime that divides $b$ and is not 2. Let $p^n$ be the highest power of $p$ that divides $b$.



          Since $gcd(a,b) = 1$, and we know that $ 2a^2 d^4 equiv 0 pmod{p^{2n}}$, thus $ d^4 equiv 0 pmod{p^{2n}}$ and this implies that $ p^n mid d^2 $.



          If $ 2^n mid mid b$ (highest power of 2 that divides $b$), then we have $ 2 not mid a$. Similarly, we get $ 2 d^4 equiv 0 pmod{2^{2n}}$. Let $2^m mid mid d^2$, then we know that $ 1 + 2m geq 2n$. Since $m$ and $n$ are integers, we can conclude that $m geq n$. Thus, we have $ 2^n mid d^2$.



          Hence, $b mid d^2$, and so we may divide throughout by $b^2$, to obtain $ 2 left( a timesfrac{d^2}{b} right)^2 = c^4 - 17 d^4$, and $gcd (c, d) = 1 $ by assumption.




          In general, you see that this will work, as long as the coefficient (2) is square-free.






          share|cite|improve this answer











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            $begingroup$

            $b$ need not be a square. Instead, the trick is to show that you can actually divide the equation by $b^2$.



            Let $p$ be a prime that divides $b$ and is not 2. Let $p^n$ be the highest power of $p$ that divides $b$.



            Since $gcd(a,b) = 1$, and we know that $ 2a^2 d^4 equiv 0 pmod{p^{2n}}$, thus $ d^4 equiv 0 pmod{p^{2n}}$ and this implies that $ p^n mid d^2 $.



            If $ 2^n mid mid b$ (highest power of 2 that divides $b$), then we have $ 2 not mid a$. Similarly, we get $ 2 d^4 equiv 0 pmod{2^{2n}}$. Let $2^m mid mid d^2$, then we know that $ 1 + 2m geq 2n$. Since $m$ and $n$ are integers, we can conclude that $m geq n$. Thus, we have $ 2^n mid d^2$.



            Hence, $b mid d^2$, and so we may divide throughout by $b^2$, to obtain $ 2 left( a timesfrac{d^2}{b} right)^2 = c^4 - 17 d^4$, and $gcd (c, d) = 1 $ by assumption.




            In general, you see that this will work, as long as the coefficient (2) is square-free.






            share|cite|improve this answer











            $endgroup$


















              5












              $begingroup$

              $b$ need not be a square. Instead, the trick is to show that you can actually divide the equation by $b^2$.



              Let $p$ be a prime that divides $b$ and is not 2. Let $p^n$ be the highest power of $p$ that divides $b$.



              Since $gcd(a,b) = 1$, and we know that $ 2a^2 d^4 equiv 0 pmod{p^{2n}}$, thus $ d^4 equiv 0 pmod{p^{2n}}$ and this implies that $ p^n mid d^2 $.



              If $ 2^n mid mid b$ (highest power of 2 that divides $b$), then we have $ 2 not mid a$. Similarly, we get $ 2 d^4 equiv 0 pmod{2^{2n}}$. Let $2^m mid mid d^2$, then we know that $ 1 + 2m geq 2n$. Since $m$ and $n$ are integers, we can conclude that $m geq n$. Thus, we have $ 2^n mid d^2$.



              Hence, $b mid d^2$, and so we may divide throughout by $b^2$, to obtain $ 2 left( a timesfrac{d^2}{b} right)^2 = c^4 - 17 d^4$, and $gcd (c, d) = 1 $ by assumption.




              In general, you see that this will work, as long as the coefficient (2) is square-free.






              share|cite|improve this answer











              $endgroup$
















                5












                5








                5





                $begingroup$

                $b$ need not be a square. Instead, the trick is to show that you can actually divide the equation by $b^2$.



                Let $p$ be a prime that divides $b$ and is not 2. Let $p^n$ be the highest power of $p$ that divides $b$.



                Since $gcd(a,b) = 1$, and we know that $ 2a^2 d^4 equiv 0 pmod{p^{2n}}$, thus $ d^4 equiv 0 pmod{p^{2n}}$ and this implies that $ p^n mid d^2 $.



                If $ 2^n mid mid b$ (highest power of 2 that divides $b$), then we have $ 2 not mid a$. Similarly, we get $ 2 d^4 equiv 0 pmod{2^{2n}}$. Let $2^m mid mid d^2$, then we know that $ 1 + 2m geq 2n$. Since $m$ and $n$ are integers, we can conclude that $m geq n$. Thus, we have $ 2^n mid d^2$.



                Hence, $b mid d^2$, and so we may divide throughout by $b^2$, to obtain $ 2 left( a timesfrac{d^2}{b} right)^2 = c^4 - 17 d^4$, and $gcd (c, d) = 1 $ by assumption.




                In general, you see that this will work, as long as the coefficient (2) is square-free.






                share|cite|improve this answer











                $endgroup$



                $b$ need not be a square. Instead, the trick is to show that you can actually divide the equation by $b^2$.



                Let $p$ be a prime that divides $b$ and is not 2. Let $p^n$ be the highest power of $p$ that divides $b$.



                Since $gcd(a,b) = 1$, and we know that $ 2a^2 d^4 equiv 0 pmod{p^{2n}}$, thus $ d^4 equiv 0 pmod{p^{2n}}$ and this implies that $ p^n mid d^2 $.



                If $ 2^n mid mid b$ (highest power of 2 that divides $b$), then we have $ 2 not mid a$. Similarly, we get $ 2 d^4 equiv 0 pmod{2^{2n}}$. Let $2^m mid mid d^2$, then we know that $ 1 + 2m geq 2n$. Since $m$ and $n$ are integers, we can conclude that $m geq n$. Thus, we have $ 2^n mid d^2$.



                Hence, $b mid d^2$, and so we may divide throughout by $b^2$, to obtain $ 2 left( a timesfrac{d^2}{b} right)^2 = c^4 - 17 d^4$, and $gcd (c, d) = 1 $ by assumption.




                In general, you see that this will work, as long as the coefficient (2) is square-free.







                share|cite|improve this answer














                share|cite|improve this answer



                share|cite|improve this answer








                edited Feb 24 '14 at 10:14

























                answered Feb 24 '14 at 10:01









                Calvin LinCalvin Lin

                36.3k349114




                36.3k349114






























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