Solutions of diophantine eq: $x^4-2x^3y+2xy^3+y^4=2s^2$ I'm examining solutions of this diophantine equation:
$$x^4-2x^3y+2xy^3+y^4=2s^2$$
It looks like all the solutions are of the form $(x,y,s) = (t,\pm t, \pm t^2)$ where $t$ is any integer. But how do I prove that these are all the solutions? Thanks.
 A: It depends on whether you want an elementary proof or not.
The following argument uses properties of elliptic curves to show the result.
Define $z=s/y^2$ and $w=x/y$ giving
\begin{equation*}
2z^2=w^4-2w^3+2w+1
\end{equation*}
This quartic has a rational solution $w=1, z=1$, so is birationally equivalent to an elliptic curve.
Define $u=1/(w-1)$, so that $w=1+1/u$ giving
\begin{equation*}
2z^2u^4=2u^4+2u+1
\end{equation*}
and set $4zu^2=g$ and $2u=h$ so that
\begin{equation*}
g^2=h^4+8h+8
\end{equation*}
This equation can be transformed to the elliptic curve
\begin{equation*}
c^2=d^3+3d^2+d
\end{equation*}
with
\begin{equation*}
w=\frac{x}{y}=\frac{c+2d+1}{c-1}
\end{equation*}
Using Pari-GP, we have that the elliptic curve has $3$ finite torsion points at $(0,0)$, $(-1,1)$ and $(-1,-1)$. Denis Simon's ellrank package tells us that the rank of the elliptic curve is zero, so the afore-mentioned points are the only rational points on the curve.
These give $w= \pm 1$ only and the partial homogeneity of the original equation gives the solutions $(t, \pm t, \pm t^2)$ as stated.
