Birational Equivalence of Diophantine Equations and Elliptic Curves

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+2\frac{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.

(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,

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

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

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

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,

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

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