Birational Equivalence of Diophantine Equations and Elliptic Curves












7












$begingroup$


A while ago I saw this question Quartic diophantine equation: $16r^4+112r^3+200r^2-112r+16=s^2$ which was very relevant to a undergraduate research paper I am currently working on. The answer given for this problem describes a method for determining solutions to diophantine equations by finding a "birational equivalence" to the solution set of an elliptic curve.



The result I am seeking depends on the solution set to a quartic diophantine equation of two variables and I believe that there is only one solution and no more (which was indicated by a program which checked possible values up to a high number).



So my question is, is there a general method for determining such a birational equivalence between solutions sets of diophantine equations and elliptic curves? And where is a good source for an undergraduate to learn to use these tools?










share|cite|improve this question











$endgroup$












  • $begingroup$
    I'm not quite sure what "a birational equivalence" between diophantine equations and elliptic curves should mean. When we want to find integer points on these "elliptic curves", it is a diophantine problem. Generally speaking it is much easier to identify rational points on the curves (and birational parameterization is one way to answer that), and it can be AFAIK a difficult problem to say whether integer points exist even given some rational solutions.
    $endgroup$
    – hardmath
    Dec 28 '15 at 20:30










  • $begingroup$
    Perhaps what you have in mind is the subject discussed here, at Ask SageMath.
    $endgroup$
    – hardmath
    Dec 28 '15 at 20:33






  • 1




    $begingroup$
    Certain curves are birationally equivalent to elliptic curves, so in this sense Diophantine equations give rise to elliptic curves. I think this is what the OP meant when he referred to birational equivalence between Diophantine equations and elliptic curves.
    $endgroup$
    – Jose Arnaldo Bebita-Dris
    Jan 2 '16 at 18:27












  • $begingroup$
    Yes that is exactly it :)
    $endgroup$
    – Brandon Thomas Van Over
    Jan 2 '16 at 20:32
















7












$begingroup$


A while ago I saw this question Quartic diophantine equation: $16r^4+112r^3+200r^2-112r+16=s^2$ which was very relevant to a undergraduate research paper I am currently working on. The answer given for this problem describes a method for determining solutions to diophantine equations by finding a "birational equivalence" to the solution set of an elliptic curve.



The result I am seeking depends on the solution set to a quartic diophantine equation of two variables and I believe that there is only one solution and no more (which was indicated by a program which checked possible values up to a high number).



So my question is, is there a general method for determining such a birational equivalence between solutions sets of diophantine equations and elliptic curves? And where is a good source for an undergraduate to learn to use these tools?










share|cite|improve this question











$endgroup$












  • $begingroup$
    I'm not quite sure what "a birational equivalence" between diophantine equations and elliptic curves should mean. When we want to find integer points on these "elliptic curves", it is a diophantine problem. Generally speaking it is much easier to identify rational points on the curves (and birational parameterization is one way to answer that), and it can be AFAIK a difficult problem to say whether integer points exist even given some rational solutions.
    $endgroup$
    – hardmath
    Dec 28 '15 at 20:30










  • $begingroup$
    Perhaps what you have in mind is the subject discussed here, at Ask SageMath.
    $endgroup$
    – hardmath
    Dec 28 '15 at 20:33






  • 1




    $begingroup$
    Certain curves are birationally equivalent to elliptic curves, so in this sense Diophantine equations give rise to elliptic curves. I think this is what the OP meant when he referred to birational equivalence between Diophantine equations and elliptic curves.
    $endgroup$
    – Jose Arnaldo Bebita-Dris
    Jan 2 '16 at 18:27












  • $begingroup$
    Yes that is exactly it :)
    $endgroup$
    – Brandon Thomas Van Over
    Jan 2 '16 at 20:32














7












7








7


4



$begingroup$


A while ago I saw this question Quartic diophantine equation: $16r^4+112r^3+200r^2-112r+16=s^2$ which was very relevant to a undergraduate research paper I am currently working on. The answer given for this problem describes a method for determining solutions to diophantine equations by finding a "birational equivalence" to the solution set of an elliptic curve.



The result I am seeking depends on the solution set to a quartic diophantine equation of two variables and I believe that there is only one solution and no more (which was indicated by a program which checked possible values up to a high number).



So my question is, is there a general method for determining such a birational equivalence between solutions sets of diophantine equations and elliptic curves? And where is a good source for an undergraduate to learn to use these tools?










share|cite|improve this question











$endgroup$




A while ago I saw this question Quartic diophantine equation: $16r^4+112r^3+200r^2-112r+16=s^2$ which was very relevant to a undergraduate research paper I am currently working on. The answer given for this problem describes a method for determining solutions to diophantine equations by finding a "birational equivalence" to the solution set of an elliptic curve.



The result I am seeking depends on the solution set to a quartic diophantine equation of two variables and I believe that there is only one solution and no more (which was indicated by a program which checked possible values up to a high number).



So my question is, is there a general method for determining such a birational equivalence between solutions sets of diophantine equations and elliptic curves? And where is a good source for an undergraduate to learn to use these tools?







diophantine-equations elliptic-curves book-recommendation birational-geometry






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Apr 13 '17 at 12:21









Community

1




1










asked Dec 28 '15 at 20:23









Brandon Thomas Van OverBrandon Thomas Van Over

3,83221128




3,83221128












  • $begingroup$
    I'm not quite sure what "a birational equivalence" between diophantine equations and elliptic curves should mean. When we want to find integer points on these "elliptic curves", it is a diophantine problem. Generally speaking it is much easier to identify rational points on the curves (and birational parameterization is one way to answer that), and it can be AFAIK a difficult problem to say whether integer points exist even given some rational solutions.
    $endgroup$
    – hardmath
    Dec 28 '15 at 20:30










  • $begingroup$
    Perhaps what you have in mind is the subject discussed here, at Ask SageMath.
    $endgroup$
    – hardmath
    Dec 28 '15 at 20:33






  • 1




    $begingroup$
    Certain curves are birationally equivalent to elliptic curves, so in this sense Diophantine equations give rise to elliptic curves. I think this is what the OP meant when he referred to birational equivalence between Diophantine equations and elliptic curves.
    $endgroup$
    – Jose Arnaldo Bebita-Dris
    Jan 2 '16 at 18:27












  • $begingroup$
    Yes that is exactly it :)
    $endgroup$
    – Brandon Thomas Van Over
    Jan 2 '16 at 20:32


















  • $begingroup$
    I'm not quite sure what "a birational equivalence" between diophantine equations and elliptic curves should mean. When we want to find integer points on these "elliptic curves", it is a diophantine problem. Generally speaking it is much easier to identify rational points on the curves (and birational parameterization is one way to answer that), and it can be AFAIK a difficult problem to say whether integer points exist even given some rational solutions.
    $endgroup$
    – hardmath
    Dec 28 '15 at 20:30










  • $begingroup$
    Perhaps what you have in mind is the subject discussed here, at Ask SageMath.
    $endgroup$
    – hardmath
    Dec 28 '15 at 20:33






  • 1




    $begingroup$
    Certain curves are birationally equivalent to elliptic curves, so in this sense Diophantine equations give rise to elliptic curves. I think this is what the OP meant when he referred to birational equivalence between Diophantine equations and elliptic curves.
    $endgroup$
    – Jose Arnaldo Bebita-Dris
    Jan 2 '16 at 18:27












  • $begingroup$
    Yes that is exactly it :)
    $endgroup$
    – Brandon Thomas Van Over
    Jan 2 '16 at 20:32
















$begingroup$
I'm not quite sure what "a birational equivalence" between diophantine equations and elliptic curves should mean. When we want to find integer points on these "elliptic curves", it is a diophantine problem. Generally speaking it is much easier to identify rational points on the curves (and birational parameterization is one way to answer that), and it can be AFAIK a difficult problem to say whether integer points exist even given some rational solutions.
$endgroup$
– hardmath
Dec 28 '15 at 20:30




$begingroup$
I'm not quite sure what "a birational equivalence" between diophantine equations and elliptic curves should mean. When we want to find integer points on these "elliptic curves", it is a diophantine problem. Generally speaking it is much easier to identify rational points on the curves (and birational parameterization is one way to answer that), and it can be AFAIK a difficult problem to say whether integer points exist even given some rational solutions.
$endgroup$
– hardmath
Dec 28 '15 at 20:30












$begingroup$
Perhaps what you have in mind is the subject discussed here, at Ask SageMath.
$endgroup$
– hardmath
Dec 28 '15 at 20:33




$begingroup$
Perhaps what you have in mind is the subject discussed here, at Ask SageMath.
$endgroup$
– hardmath
Dec 28 '15 at 20:33




1




1




$begingroup$
Certain curves are birationally equivalent to elliptic curves, so in this sense Diophantine equations give rise to elliptic curves. I think this is what the OP meant when he referred to birational equivalence between Diophantine equations and elliptic curves.
$endgroup$
– Jose Arnaldo Bebita-Dris
Jan 2 '16 at 18:27






$begingroup$
Certain curves are birationally equivalent to elliptic curves, so in this sense Diophantine equations give rise to elliptic curves. I think this is what the OP meant when he referred to birational equivalence between Diophantine equations and elliptic curves.
$endgroup$
– Jose Arnaldo Bebita-Dris
Jan 2 '16 at 18:27














$begingroup$
Yes that is exactly it :)
$endgroup$
– Brandon Thomas Van Over
Jan 2 '16 at 20:32




$begingroup$
Yes that is exactly it :)
$endgroup$
– Brandon Thomas Van Over
Jan 2 '16 at 20:32










2 Answers
2






active

oldest

votes


















10












$begingroup$

The following shows how to transform the quartic
begin{equation}
y^2=ax^4+bx^3+cx^2+dx+e ,
end{equation}
with $a,b,c,d,e in mathbb{Q}$, into an equivalent elliptic curve.




Case 1:




We first consider the case when $a=1$, which is very common. If $a=alpha^2$ for some rational $alpha$, we
substitute $y=Y/alpha$ and $x=X/alpha$, giving
begin{equation*}
Y^2=X^4+frac{b}{alpha}X^3+cX^2+alpha d X+alpha^2 e
end{equation*} Thus, suppose
begin{equation}
y^2=x^4+bx^3+cx^2+dx+e
end{equation}



We describe the method given by Mordell on page $77$ of his book Diophantine Equations, with some minor modifications.



We first get rid of the cubic term by making the standard substitution $x=z-b/4$ giving
begin{equation*}
y^2=z^4+fz^2+gz+h
end{equation*}
where
begin{equation*}
f=frac{8c-3b^2}{8} hspace{1cm} g=frac{b^3-4bc+8d}{8} hspace{1cm}
h=frac{-(3b^4-16b^2c+64bd-256e)}{256}
end{equation*}



We now get rid off the quartic term by defining $y=z^2+u+k$, where $u$ is a new variable and $k$ is a constant to be
determined. This gives the quadratic in $z$
begin{equation*}
(f-2(k+u))z^2+gz+h-k^2-u(2k+u)=0
end{equation*}



For $x$ to be rational, then $z$ should be rational, so the discriminant of this quadratic would have to be a rational square.
The discriminant is a cubic in $u$. We do not get a term in $u^2$ if we make $k=f/6$, giving
begin{equation*}
D^2=-8u^3+2frac{f^2+12h}{3}u+frac{2f^3-72fh+27g^2}{27}
end{equation*}
and, if we substitute the formulae for $f,g,h$, and clear denominators we have
begin{equation}
G^2=H^3+27(3bd-c^2-12e)H+27(27b^2e-9bcd+2c^3-72ce+27d^2)
end{equation}
with
begin{equation}
x=frac{2G-3bH+9(bc-6d)}{12H-9(3b^2-8c)}
end{equation}
and
begin{equation}
y= pm frac{18x^2+9bx+3c-H}{18}
end{equation}




Case 2:




If $a ne alpha^2$, we need a rational point $(p,q)$ lying on the quartic.



Let $z=1/(x-p)$, so that $x=p+1/z$ giving
begin{equation*}
y^2z^4=(ap^4+bp^3+cp^2+dp+e)z^4+(4ap^3+3bp^2+2cp+d)z^3+
end{equation*}
begin{equation*}
(6ap^2+3bp+c)z^2+(4ap+b)z+a
end{equation*}
where the coefficient of $z^4$ is, in fact, $q^2$. Define, $y=w/z^2$, and then
$z=u/q^2, , w=v/q^3$
giving
begin{equation}
v^2=u^4+(4ap^3+3bp^2+2cp+d)u^3+q^2(6ap^2+3bp+c)u^2+
end{equation}
begin{equation*}
q^4(4ap+b)u+aq^6 equiv u^4+fu^3+gu^2+hu+k
end{equation*}



We now, essentially, complete the square. We can write
begin{equation*}
y^2=u^4+fu^3+gu^2+hu+k=(u^2+mu+n)^2+(su+t)
end{equation*}
if we set
begin{equation*}
m=frac{f}{2} hspace{1cm} n=frac{4g-f^2}{8} hspace{1cm} s=frac{f^3-4fg+8h}{8}
end{equation*}
and
begin{equation*}
t=frac{-(f^4-8f^2g+16(g^2-4k))}{64}
end{equation*}



This gives
begin{equation*}
(y+u^2+mu+n)(y-u^2-mu-n)=su+t
end{equation*}
and if we define $y+u^2+mu+n=Z$ we have
begin{equation}
2(u^2+mu+n)=Z-frac{su+t}{Z}
end{equation}



Multiply both sides by $Z^2$, giving
begin{equation*}
2u^2Z^2+2muZ^2+suZ=Z^3-2nZ^2-tZ
end{equation*}
which, on defining $W=uZ$, gives
begin{equation*}
2W^2+2mWZ+sW=Z^3-2nZ^2-tZ
end{equation*}



Define, $Z=X/2$ and $W=Y/4$ giving
begin{equation}
Y^2+2mXY+2sY=X^3-4nX^2-4tX
end{equation}
which is an elliptic curve. If we define $Y=G-s-mX$, we transform to the form
begin{equation}
G^2=X^3+(m^2-4n)X^2+(2ms-4t)X+s^2
end{equation}



All of the above are easy to program in any symbolic package. Hopefully, I have all the formulae correct.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Thank you for taking the time to explain this in detail and provide a reference. I really appreciate it.
    $endgroup$
    – Brandon Thomas Van Over
    Dec 29 '15 at 14:20










  • $begingroup$
    This will seem to solve one of my longstanding problems. I'm going to try programming this one out and report back.
    $endgroup$
    – J. M. is not a mathematician
    May 14 '16 at 15:39



















3












$begingroup$

(This is a long comment re MacLeod's answer.)



We can also combine the two cases together. Assume a quartic polynomial to be made a square,



$$pu^4+qu^3+ru^2+su+t=z_1^2tag1$$



has a known rational point, call it $w$. We substitute $u=v+w$ and collect the new variable $v$,



$$c_4v^4+c_3v^3+c_2v^2+c_1v+color{blue}{c_0^2}=z_2^2tag2$$



where the $c_i$ are polynomials in $w$ and the coefficients of $(1)$. The constant term of $(2)$ turns out to be a square, specifically, $c_0^2:=pw^4+qw^3+rw^2+sw+t$.



Let $v=1/x,,$ $z_2=c_0, z_3/x^2$ and $(2)$ becomes,



$$x^4+d_3x^3+d_2x^2+d_1x+d_0=z_3^2tag3$$



which is the Case 1 and then be solved as explained by MacLeod.






share|cite|improve this answer











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    2 Answers
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    active

    oldest

    votes








    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    10












    $begingroup$

    The following shows how to transform the quartic
    begin{equation}
    y^2=ax^4+bx^3+cx^2+dx+e ,
    end{equation}
    with $a,b,c,d,e in mathbb{Q}$, into an equivalent elliptic curve.




    Case 1:




    We first consider the case when $a=1$, which is very common. If $a=alpha^2$ for some rational $alpha$, we
    substitute $y=Y/alpha$ and $x=X/alpha$, giving
    begin{equation*}
    Y^2=X^4+frac{b}{alpha}X^3+cX^2+alpha d X+alpha^2 e
    end{equation*} Thus, suppose
    begin{equation}
    y^2=x^4+bx^3+cx^2+dx+e
    end{equation}



    We describe the method given by Mordell on page $77$ of his book Diophantine Equations, with some minor modifications.



    We first get rid of the cubic term by making the standard substitution $x=z-b/4$ giving
    begin{equation*}
    y^2=z^4+fz^2+gz+h
    end{equation*}
    where
    begin{equation*}
    f=frac{8c-3b^2}{8} hspace{1cm} g=frac{b^3-4bc+8d}{8} hspace{1cm}
    h=frac{-(3b^4-16b^2c+64bd-256e)}{256}
    end{equation*}



    We now get rid off the quartic term by defining $y=z^2+u+k$, where $u$ is a new variable and $k$ is a constant to be
    determined. This gives the quadratic in $z$
    begin{equation*}
    (f-2(k+u))z^2+gz+h-k^2-u(2k+u)=0
    end{equation*}



    For $x$ to be rational, then $z$ should be rational, so the discriminant of this quadratic would have to be a rational square.
    The discriminant is a cubic in $u$. We do not get a term in $u^2$ if we make $k=f/6$, giving
    begin{equation*}
    D^2=-8u^3+2frac{f^2+12h}{3}u+frac{2f^3-72fh+27g^2}{27}
    end{equation*}
    and, if we substitute the formulae for $f,g,h$, and clear denominators we have
    begin{equation}
    G^2=H^3+27(3bd-c^2-12e)H+27(27b^2e-9bcd+2c^3-72ce+27d^2)
    end{equation}
    with
    begin{equation}
    x=frac{2G-3bH+9(bc-6d)}{12H-9(3b^2-8c)}
    end{equation}
    and
    begin{equation}
    y= pm frac{18x^2+9bx+3c-H}{18}
    end{equation}




    Case 2:




    If $a ne alpha^2$, we need a rational point $(p,q)$ lying on the quartic.



    Let $z=1/(x-p)$, so that $x=p+1/z$ giving
    begin{equation*}
    y^2z^4=(ap^4+bp^3+cp^2+dp+e)z^4+(4ap^3+3bp^2+2cp+d)z^3+
    end{equation*}
    begin{equation*}
    (6ap^2+3bp+c)z^2+(4ap+b)z+a
    end{equation*}
    where the coefficient of $z^4$ is, in fact, $q^2$. Define, $y=w/z^2$, and then
    $z=u/q^2, , w=v/q^3$
    giving
    begin{equation}
    v^2=u^4+(4ap^3+3bp^2+2cp+d)u^3+q^2(6ap^2+3bp+c)u^2+
    end{equation}
    begin{equation*}
    q^4(4ap+b)u+aq^6 equiv u^4+fu^3+gu^2+hu+k
    end{equation*}



    We now, essentially, complete the square. We can write
    begin{equation*}
    y^2=u^4+fu^3+gu^2+hu+k=(u^2+mu+n)^2+(su+t)
    end{equation*}
    if we set
    begin{equation*}
    m=frac{f}{2} hspace{1cm} n=frac{4g-f^2}{8} hspace{1cm} s=frac{f^3-4fg+8h}{8}
    end{equation*}
    and
    begin{equation*}
    t=frac{-(f^4-8f^2g+16(g^2-4k))}{64}
    end{equation*}



    This gives
    begin{equation*}
    (y+u^2+mu+n)(y-u^2-mu-n)=su+t
    end{equation*}
    and if we define $y+u^2+mu+n=Z$ we have
    begin{equation}
    2(u^2+mu+n)=Z-frac{su+t}{Z}
    end{equation}



    Multiply both sides by $Z^2$, giving
    begin{equation*}
    2u^2Z^2+2muZ^2+suZ=Z^3-2nZ^2-tZ
    end{equation*}
    which, on defining $W=uZ$, gives
    begin{equation*}
    2W^2+2mWZ+sW=Z^3-2nZ^2-tZ
    end{equation*}



    Define, $Z=X/2$ and $W=Y/4$ giving
    begin{equation}
    Y^2+2mXY+2sY=X^3-4nX^2-4tX
    end{equation}
    which is an elliptic curve. If we define $Y=G-s-mX$, we transform to the form
    begin{equation}
    G^2=X^3+(m^2-4n)X^2+(2ms-4t)X+s^2
    end{equation}



    All of the above are easy to program in any symbolic package. Hopefully, I have all the formulae correct.






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      Thank you for taking the time to explain this in detail and provide a reference. I really appreciate it.
      $endgroup$
      – Brandon Thomas Van Over
      Dec 29 '15 at 14:20










    • $begingroup$
      This will seem to solve one of my longstanding problems. I'm going to try programming this one out and report back.
      $endgroup$
      – J. M. is not a mathematician
      May 14 '16 at 15:39
















    10












    $begingroup$

    The following shows how to transform the quartic
    begin{equation}
    y^2=ax^4+bx^3+cx^2+dx+e ,
    end{equation}
    with $a,b,c,d,e in mathbb{Q}$, into an equivalent elliptic curve.




    Case 1:




    We first consider the case when $a=1$, which is very common. If $a=alpha^2$ for some rational $alpha$, we
    substitute $y=Y/alpha$ and $x=X/alpha$, giving
    begin{equation*}
    Y^2=X^4+frac{b}{alpha}X^3+cX^2+alpha d X+alpha^2 e
    end{equation*} Thus, suppose
    begin{equation}
    y^2=x^4+bx^3+cx^2+dx+e
    end{equation}



    We describe the method given by Mordell on page $77$ of his book Diophantine Equations, with some minor modifications.



    We first get rid of the cubic term by making the standard substitution $x=z-b/4$ giving
    begin{equation*}
    y^2=z^4+fz^2+gz+h
    end{equation*}
    where
    begin{equation*}
    f=frac{8c-3b^2}{8} hspace{1cm} g=frac{b^3-4bc+8d}{8} hspace{1cm}
    h=frac{-(3b^4-16b^2c+64bd-256e)}{256}
    end{equation*}



    We now get rid off the quartic term by defining $y=z^2+u+k$, where $u$ is a new variable and $k$ is a constant to be
    determined. This gives the quadratic in $z$
    begin{equation*}
    (f-2(k+u))z^2+gz+h-k^2-u(2k+u)=0
    end{equation*}



    For $x$ to be rational, then $z$ should be rational, so the discriminant of this quadratic would have to be a rational square.
    The discriminant is a cubic in $u$. We do not get a term in $u^2$ if we make $k=f/6$, giving
    begin{equation*}
    D^2=-8u^3+2frac{f^2+12h}{3}u+frac{2f^3-72fh+27g^2}{27}
    end{equation*}
    and, if we substitute the formulae for $f,g,h$, and clear denominators we have
    begin{equation}
    G^2=H^3+27(3bd-c^2-12e)H+27(27b^2e-9bcd+2c^3-72ce+27d^2)
    end{equation}
    with
    begin{equation}
    x=frac{2G-3bH+9(bc-6d)}{12H-9(3b^2-8c)}
    end{equation}
    and
    begin{equation}
    y= pm frac{18x^2+9bx+3c-H}{18}
    end{equation}




    Case 2:




    If $a ne alpha^2$, we need a rational point $(p,q)$ lying on the quartic.



    Let $z=1/(x-p)$, so that $x=p+1/z$ giving
    begin{equation*}
    y^2z^4=(ap^4+bp^3+cp^2+dp+e)z^4+(4ap^3+3bp^2+2cp+d)z^3+
    end{equation*}
    begin{equation*}
    (6ap^2+3bp+c)z^2+(4ap+b)z+a
    end{equation*}
    where the coefficient of $z^4$ is, in fact, $q^2$. Define, $y=w/z^2$, and then
    $z=u/q^2, , w=v/q^3$
    giving
    begin{equation}
    v^2=u^4+(4ap^3+3bp^2+2cp+d)u^3+q^2(6ap^2+3bp+c)u^2+
    end{equation}
    begin{equation*}
    q^4(4ap+b)u+aq^6 equiv u^4+fu^3+gu^2+hu+k
    end{equation*}



    We now, essentially, complete the square. We can write
    begin{equation*}
    y^2=u^4+fu^3+gu^2+hu+k=(u^2+mu+n)^2+(su+t)
    end{equation*}
    if we set
    begin{equation*}
    m=frac{f}{2} hspace{1cm} n=frac{4g-f^2}{8} hspace{1cm} s=frac{f^3-4fg+8h}{8}
    end{equation*}
    and
    begin{equation*}
    t=frac{-(f^4-8f^2g+16(g^2-4k))}{64}
    end{equation*}



    This gives
    begin{equation*}
    (y+u^2+mu+n)(y-u^2-mu-n)=su+t
    end{equation*}
    and if we define $y+u^2+mu+n=Z$ we have
    begin{equation}
    2(u^2+mu+n)=Z-frac{su+t}{Z}
    end{equation}



    Multiply both sides by $Z^2$, giving
    begin{equation*}
    2u^2Z^2+2muZ^2+suZ=Z^3-2nZ^2-tZ
    end{equation*}
    which, on defining $W=uZ$, gives
    begin{equation*}
    2W^2+2mWZ+sW=Z^3-2nZ^2-tZ
    end{equation*}



    Define, $Z=X/2$ and $W=Y/4$ giving
    begin{equation}
    Y^2+2mXY+2sY=X^3-4nX^2-4tX
    end{equation}
    which is an elliptic curve. If we define $Y=G-s-mX$, we transform to the form
    begin{equation}
    G^2=X^3+(m^2-4n)X^2+(2ms-4t)X+s^2
    end{equation}



    All of the above are easy to program in any symbolic package. Hopefully, I have all the formulae correct.






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      Thank you for taking the time to explain this in detail and provide a reference. I really appreciate it.
      $endgroup$
      – Brandon Thomas Van Over
      Dec 29 '15 at 14:20










    • $begingroup$
      This will seem to solve one of my longstanding problems. I'm going to try programming this one out and report back.
      $endgroup$
      – J. M. is not a mathematician
      May 14 '16 at 15:39














    10












    10








    10





    $begingroup$

    The following shows how to transform the quartic
    begin{equation}
    y^2=ax^4+bx^3+cx^2+dx+e ,
    end{equation}
    with $a,b,c,d,e in mathbb{Q}$, into an equivalent elliptic curve.




    Case 1:




    We first consider the case when $a=1$, which is very common. If $a=alpha^2$ for some rational $alpha$, we
    substitute $y=Y/alpha$ and $x=X/alpha$, giving
    begin{equation*}
    Y^2=X^4+frac{b}{alpha}X^3+cX^2+alpha d X+alpha^2 e
    end{equation*} Thus, suppose
    begin{equation}
    y^2=x^4+bx^3+cx^2+dx+e
    end{equation}



    We describe the method given by Mordell on page $77$ of his book Diophantine Equations, with some minor modifications.



    We first get rid of the cubic term by making the standard substitution $x=z-b/4$ giving
    begin{equation*}
    y^2=z^4+fz^2+gz+h
    end{equation*}
    where
    begin{equation*}
    f=frac{8c-3b^2}{8} hspace{1cm} g=frac{b^3-4bc+8d}{8} hspace{1cm}
    h=frac{-(3b^4-16b^2c+64bd-256e)}{256}
    end{equation*}



    We now get rid off the quartic term by defining $y=z^2+u+k$, where $u$ is a new variable and $k$ is a constant to be
    determined. This gives the quadratic in $z$
    begin{equation*}
    (f-2(k+u))z^2+gz+h-k^2-u(2k+u)=0
    end{equation*}



    For $x$ to be rational, then $z$ should be rational, so the discriminant of this quadratic would have to be a rational square.
    The discriminant is a cubic in $u$. We do not get a term in $u^2$ if we make $k=f/6$, giving
    begin{equation*}
    D^2=-8u^3+2frac{f^2+12h}{3}u+frac{2f^3-72fh+27g^2}{27}
    end{equation*}
    and, if we substitute the formulae for $f,g,h$, and clear denominators we have
    begin{equation}
    G^2=H^3+27(3bd-c^2-12e)H+27(27b^2e-9bcd+2c^3-72ce+27d^2)
    end{equation}
    with
    begin{equation}
    x=frac{2G-3bH+9(bc-6d)}{12H-9(3b^2-8c)}
    end{equation}
    and
    begin{equation}
    y= pm frac{18x^2+9bx+3c-H}{18}
    end{equation}




    Case 2:




    If $a ne alpha^2$, we need a rational point $(p,q)$ lying on the quartic.



    Let $z=1/(x-p)$, so that $x=p+1/z$ giving
    begin{equation*}
    y^2z^4=(ap^4+bp^3+cp^2+dp+e)z^4+(4ap^3+3bp^2+2cp+d)z^3+
    end{equation*}
    begin{equation*}
    (6ap^2+3bp+c)z^2+(4ap+b)z+a
    end{equation*}
    where the coefficient of $z^4$ is, in fact, $q^2$. Define, $y=w/z^2$, and then
    $z=u/q^2, , w=v/q^3$
    giving
    begin{equation}
    v^2=u^4+(4ap^3+3bp^2+2cp+d)u^3+q^2(6ap^2+3bp+c)u^2+
    end{equation}
    begin{equation*}
    q^4(4ap+b)u+aq^6 equiv u^4+fu^3+gu^2+hu+k
    end{equation*}



    We now, essentially, complete the square. We can write
    begin{equation*}
    y^2=u^4+fu^3+gu^2+hu+k=(u^2+mu+n)^2+(su+t)
    end{equation*}
    if we set
    begin{equation*}
    m=frac{f}{2} hspace{1cm} n=frac{4g-f^2}{8} hspace{1cm} s=frac{f^3-4fg+8h}{8}
    end{equation*}
    and
    begin{equation*}
    t=frac{-(f^4-8f^2g+16(g^2-4k))}{64}
    end{equation*}



    This gives
    begin{equation*}
    (y+u^2+mu+n)(y-u^2-mu-n)=su+t
    end{equation*}
    and if we define $y+u^2+mu+n=Z$ we have
    begin{equation}
    2(u^2+mu+n)=Z-frac{su+t}{Z}
    end{equation}



    Multiply both sides by $Z^2$, giving
    begin{equation*}
    2u^2Z^2+2muZ^2+suZ=Z^3-2nZ^2-tZ
    end{equation*}
    which, on defining $W=uZ$, gives
    begin{equation*}
    2W^2+2mWZ+sW=Z^3-2nZ^2-tZ
    end{equation*}



    Define, $Z=X/2$ and $W=Y/4$ giving
    begin{equation}
    Y^2+2mXY+2sY=X^3-4nX^2-4tX
    end{equation}
    which is an elliptic curve. If we define $Y=G-s-mX$, we transform to the form
    begin{equation}
    G^2=X^3+(m^2-4n)X^2+(2ms-4t)X+s^2
    end{equation}



    All of the above are easy to program in any symbolic package. Hopefully, I have all the formulae correct.






    share|cite|improve this answer











    $endgroup$



    The following shows how to transform the quartic
    begin{equation}
    y^2=ax^4+bx^3+cx^2+dx+e ,
    end{equation}
    with $a,b,c,d,e in mathbb{Q}$, into an equivalent elliptic curve.




    Case 1:




    We first consider the case when $a=1$, which is very common. If $a=alpha^2$ for some rational $alpha$, we
    substitute $y=Y/alpha$ and $x=X/alpha$, giving
    begin{equation*}
    Y^2=X^4+frac{b}{alpha}X^3+cX^2+alpha d X+alpha^2 e
    end{equation*} Thus, suppose
    begin{equation}
    y^2=x^4+bx^3+cx^2+dx+e
    end{equation}



    We describe the method given by Mordell on page $77$ of his book Diophantine Equations, with some minor modifications.



    We first get rid of the cubic term by making the standard substitution $x=z-b/4$ giving
    begin{equation*}
    y^2=z^4+fz^2+gz+h
    end{equation*}
    where
    begin{equation*}
    f=frac{8c-3b^2}{8} hspace{1cm} g=frac{b^3-4bc+8d}{8} hspace{1cm}
    h=frac{-(3b^4-16b^2c+64bd-256e)}{256}
    end{equation*}



    We now get rid off the quartic term by defining $y=z^2+u+k$, where $u$ is a new variable and $k$ is a constant to be
    determined. This gives the quadratic in $z$
    begin{equation*}
    (f-2(k+u))z^2+gz+h-k^2-u(2k+u)=0
    end{equation*}



    For $x$ to be rational, then $z$ should be rational, so the discriminant of this quadratic would have to be a rational square.
    The discriminant is a cubic in $u$. We do not get a term in $u^2$ if we make $k=f/6$, giving
    begin{equation*}
    D^2=-8u^3+2frac{f^2+12h}{3}u+frac{2f^3-72fh+27g^2}{27}
    end{equation*}
    and, if we substitute the formulae for $f,g,h$, and clear denominators we have
    begin{equation}
    G^2=H^3+27(3bd-c^2-12e)H+27(27b^2e-9bcd+2c^3-72ce+27d^2)
    end{equation}
    with
    begin{equation}
    x=frac{2G-3bH+9(bc-6d)}{12H-9(3b^2-8c)}
    end{equation}
    and
    begin{equation}
    y= pm frac{18x^2+9bx+3c-H}{18}
    end{equation}




    Case 2:




    If $a ne alpha^2$, we need a rational point $(p,q)$ lying on the quartic.



    Let $z=1/(x-p)$, so that $x=p+1/z$ giving
    begin{equation*}
    y^2z^4=(ap^4+bp^3+cp^2+dp+e)z^4+(4ap^3+3bp^2+2cp+d)z^3+
    end{equation*}
    begin{equation*}
    (6ap^2+3bp+c)z^2+(4ap+b)z+a
    end{equation*}
    where the coefficient of $z^4$ is, in fact, $q^2$. Define, $y=w/z^2$, and then
    $z=u/q^2, , w=v/q^3$
    giving
    begin{equation}
    v^2=u^4+(4ap^3+3bp^2+2cp+d)u^3+q^2(6ap^2+3bp+c)u^2+
    end{equation}
    begin{equation*}
    q^4(4ap+b)u+aq^6 equiv u^4+fu^3+gu^2+hu+k
    end{equation*}



    We now, essentially, complete the square. We can write
    begin{equation*}
    y^2=u^4+fu^3+gu^2+hu+k=(u^2+mu+n)^2+(su+t)
    end{equation*}
    if we set
    begin{equation*}
    m=frac{f}{2} hspace{1cm} n=frac{4g-f^2}{8} hspace{1cm} s=frac{f^3-4fg+8h}{8}
    end{equation*}
    and
    begin{equation*}
    t=frac{-(f^4-8f^2g+16(g^2-4k))}{64}
    end{equation*}



    This gives
    begin{equation*}
    (y+u^2+mu+n)(y-u^2-mu-n)=su+t
    end{equation*}
    and if we define $y+u^2+mu+n=Z$ we have
    begin{equation}
    2(u^2+mu+n)=Z-frac{su+t}{Z}
    end{equation}



    Multiply both sides by $Z^2$, giving
    begin{equation*}
    2u^2Z^2+2muZ^2+suZ=Z^3-2nZ^2-tZ
    end{equation*}
    which, on defining $W=uZ$, gives
    begin{equation*}
    2W^2+2mWZ+sW=Z^3-2nZ^2-tZ
    end{equation*}



    Define, $Z=X/2$ and $W=Y/4$ giving
    begin{equation}
    Y^2+2mXY+2sY=X^3-4nX^2-4tX
    end{equation}
    which is an elliptic curve. If we define $Y=G-s-mX$, we transform to the form
    begin{equation}
    G^2=X^3+(m^2-4n)X^2+(2ms-4t)X+s^2
    end{equation}



    All of the above are easy to program in any symbolic package. Hopefully, I have all the formulae correct.







    share|cite|improve this answer














    share|cite|improve this answer



    share|cite|improve this answer








    edited Jul 25 '16 at 1:06









    Tito Piezas III

    27.8k369178




    27.8k369178










    answered Dec 29 '15 at 12:02









    Allan MacLeodAllan MacLeod

    1,24975




    1,24975












    • $begingroup$
      Thank you for taking the time to explain this in detail and provide a reference. I really appreciate it.
      $endgroup$
      – Brandon Thomas Van Over
      Dec 29 '15 at 14:20










    • $begingroup$
      This will seem to solve one of my longstanding problems. I'm going to try programming this one out and report back.
      $endgroup$
      – J. M. is not a mathematician
      May 14 '16 at 15:39


















    • $begingroup$
      Thank you for taking the time to explain this in detail and provide a reference. I really appreciate it.
      $endgroup$
      – Brandon Thomas Van Over
      Dec 29 '15 at 14:20










    • $begingroup$
      This will seem to solve one of my longstanding problems. I'm going to try programming this one out and report back.
      $endgroup$
      – J. M. is not a mathematician
      May 14 '16 at 15:39
















    $begingroup$
    Thank you for taking the time to explain this in detail and provide a reference. I really appreciate it.
    $endgroup$
    – Brandon Thomas Van Over
    Dec 29 '15 at 14:20




    $begingroup$
    Thank you for taking the time to explain this in detail and provide a reference. I really appreciate it.
    $endgroup$
    – Brandon Thomas Van Over
    Dec 29 '15 at 14:20












    $begingroup$
    This will seem to solve one of my longstanding problems. I'm going to try programming this one out and report back.
    $endgroup$
    – J. M. is not a mathematician
    May 14 '16 at 15:39




    $begingroup$
    This will seem to solve one of my longstanding problems. I'm going to try programming this one out and report back.
    $endgroup$
    – J. M. is not a mathematician
    May 14 '16 at 15:39











    3












    $begingroup$

    (This is a long comment re MacLeod's answer.)



    We can also combine the two cases together. Assume a quartic polynomial to be made a square,



    $$pu^4+qu^3+ru^2+su+t=z_1^2tag1$$



    has a known rational point, call it $w$. We substitute $u=v+w$ and collect the new variable $v$,



    $$c_4v^4+c_3v^3+c_2v^2+c_1v+color{blue}{c_0^2}=z_2^2tag2$$



    where the $c_i$ are polynomials in $w$ and the coefficients of $(1)$. The constant term of $(2)$ turns out to be a square, specifically, $c_0^2:=pw^4+qw^3+rw^2+sw+t$.



    Let $v=1/x,,$ $z_2=c_0, z_3/x^2$ and $(2)$ becomes,



    $$x^4+d_3x^3+d_2x^2+d_1x+d_0=z_3^2tag3$$



    which is the Case 1 and then be solved as explained by MacLeod.






    share|cite|improve this answer











    $endgroup$


















      3












      $begingroup$

      (This is a long comment re MacLeod's answer.)



      We can also combine the two cases together. Assume a quartic polynomial to be made a square,



      $$pu^4+qu^3+ru^2+su+t=z_1^2tag1$$



      has a known rational point, call it $w$. We substitute $u=v+w$ and collect the new variable $v$,



      $$c_4v^4+c_3v^3+c_2v^2+c_1v+color{blue}{c_0^2}=z_2^2tag2$$



      where the $c_i$ are polynomials in $w$ and the coefficients of $(1)$. The constant term of $(2)$ turns out to be a square, specifically, $c_0^2:=pw^4+qw^3+rw^2+sw+t$.



      Let $v=1/x,,$ $z_2=c_0, z_3/x^2$ and $(2)$ becomes,



      $$x^4+d_3x^3+d_2x^2+d_1x+d_0=z_3^2tag3$$



      which is the Case 1 and then be solved as explained by MacLeod.






      share|cite|improve this answer











      $endgroup$
















        3












        3








        3





        $begingroup$

        (This is a long comment re MacLeod's answer.)



        We can also combine the two cases together. Assume a quartic polynomial to be made a square,



        $$pu^4+qu^3+ru^2+su+t=z_1^2tag1$$



        has a known rational point, call it $w$. We substitute $u=v+w$ and collect the new variable $v$,



        $$c_4v^4+c_3v^3+c_2v^2+c_1v+color{blue}{c_0^2}=z_2^2tag2$$



        where the $c_i$ are polynomials in $w$ and the coefficients of $(1)$. The constant term of $(2)$ turns out to be a square, specifically, $c_0^2:=pw^4+qw^3+rw^2+sw+t$.



        Let $v=1/x,,$ $z_2=c_0, z_3/x^2$ and $(2)$ becomes,



        $$x^4+d_3x^3+d_2x^2+d_1x+d_0=z_3^2tag3$$



        which is the Case 1 and then be solved as explained by MacLeod.






        share|cite|improve this answer











        $endgroup$



        (This is a long comment re MacLeod's answer.)



        We can also combine the two cases together. Assume a quartic polynomial to be made a square,



        $$pu^4+qu^3+ru^2+su+t=z_1^2tag1$$



        has a known rational point, call it $w$. We substitute $u=v+w$ and collect the new variable $v$,



        $$c_4v^4+c_3v^3+c_2v^2+c_1v+color{blue}{c_0^2}=z_2^2tag2$$



        where the $c_i$ are polynomials in $w$ and the coefficients of $(1)$. The constant term of $(2)$ turns out to be a square, specifically, $c_0^2:=pw^4+qw^3+rw^2+sw+t$.



        Let $v=1/x,,$ $z_2=c_0, z_3/x^2$ and $(2)$ becomes,



        $$x^4+d_3x^3+d_2x^2+d_1x+d_0=z_3^2tag3$$



        which is the Case 1 and then be solved as explained by MacLeod.







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Dec 14 '18 at 3:12

























        answered Jul 25 '16 at 19:10









        Tito Piezas IIITito Piezas III

        27.8k369178




        27.8k369178






























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