System of quadratic Diophantine equations Is there a method for determining if a system of quadratic diophantine equations has any solutions?
My specific example (which comes from this question) is:
$$\frac{4}{3}x^2 + \frac{4}{3}x + 1 = y^2$$
$$\frac{8}{3}x^2 + \frac{8}{3}x + 1 = z^2$$
I want to know if there are any positive integer triples $(x,y,z)$ which satisfy both equations.

Your specific example can be formalized in the following way:
$$
\begin{eqnarray}
4x^2+4x+3=3y^2\\
8x^2+8x+3=3z^2\\
4x^2+4x=3(z^2-y^2)\\
(z^2-y^2)=4x(x+1)/3
\end{eqnarray}
$$
let $x+1=3n$
$$
\begin{eqnarray}
(z^2-y^2)=4(3n-1)n
\end{eqnarray}
$$
Let $z=4n-1$ and $y=2n-1$ (more generally if $3n^2-n=uv$, then $z=u+v$ and $y=v-u$).  So I would have thought there are tons of integer solutions, but there's some other constraint on the go. Let's punch $x+1=3n$ into the original expressions:
$$
\begin{eqnarray}
4x(x+1)+3&=&12n(3n-1)+3&=&3y^2\\
y^2&=&12n^2-4n+1&&\\
8x(x+1)+3&=&24n(3n-1)+3&=&3z^2\\
z^2&=&24n^2-8n+1&&\\
z^2-y^2&=&12n^2-4n&=&y^2-1\\
(z+y)(z-y)&=&(y+1)(y-1)&&
\end{eqnarray}
$$
The only solution I can come up with here is $z=y$, $y=1$, $x=0$.  One other approach works too:
let $x=3n$
$$
\begin{eqnarray}
(z^2-y^2)=4(3n+1)n
\end{eqnarray}
$$
Let $z=4n+1$ and $y=2n+1$. Let's punch $x=3n$ into the original expressions:
$$
\begin{eqnarray}
4x(x+1)+3&=&12n(3n+1)+3&=&3y^2\\
y^2&=&12n^2+4n+1&&\\
8x(x+1)+3&=&24n(3n+1)+3&=&3z^2\\
z^2&=&24n^2+8n+1&&\\
z^2-y^2&=&12n^2+4n&=&y^2-1\\
(z+y)(z-y)&=&(y+1)(y-1)&&
\end{eqnarray}
$$
The end result is the same.
