Csdrhrt

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?

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.

(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.