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.
 A: Answered at the source
question.  Briefly: There are no such $(x,y,z)$, because then we'd have
a nonconstant arithmetic progression of four squares:
$$
1 = 1^2, \
\frac43(x^2+x)+1 = y^2, \
\frac83(x^2+x)+1 = z^2, \
4(x^2+x)+1 = (2x+1)^2.
$$
The impossibility of such a progression is a theorem of Euler
(1780, answering a question "first raised by Fermat in 1640" according to
Keith
Conrad's exposition).  This even proves that there are no
rational solutions $(x,y,z)$ other than the obvious ones with
$x=0$ or $x=-1$.
A: We consider only the general question. It was proved by Matijasevich that there is no algorithm which, on input any Diophantine equation $P(x_1,x_2,\dots,x_m)=0$, where $P$ is a polynomial with integer coefficients, will determine whether the equation has an integer solution.
Using a little trick that goes back to Skolem, given any Diophantine equation $P(x_1,x_2,\dots,x_m)=0$, we can algorithmically produce a system $Q_i(y_1,y_2, \dots, y_n)=0$, $i=1,\dots, s$ of quadratic Diophantine equations such that the system has a solution in integers if and only if $P(x_1,x_2,\dots,x_m)=0$ has a solution in integers.
It follows that there is no algorithm which, given any system of quadratic Diophantine equations, will determine whether the system has a solution in integers. 
A: Incomplete... First take $w = 2 x + 1$ 
From $z^2 - 2 y^2 = -1$ we get the sequence of $y$ values 
$$ y_0 = 1, \; \; y_1 = 5, \; \; y_{n+2} = 6 y_{n+1} - y_n,  $$ or
$$ 1,5,29,169,985,5741,33461, 195025 \ldots  $$
From $w^2 - 3 y^2 = -2$ we get the sequence of $y$ values 
$$ Y_0 = 1, \; \; Y_1 = 3, \; \; Y_{n+2} = 4 Y_{n+1} - Y_n,  $$ or
$$ 1,3,11,41,153,571,2131, 7953, 29681, 110771 \ldots  $$ 
The question can then be written about the two sequences. They grow at different rates, so the indices would be different anyway; but, are there indices $i,j$ with $$ y_i = Y_j? $$
At the least, the matter can be experimented with further this way: both sequences have explicit recipes involving square roots and exponents, same as the Fibonacci numbers and the golden ratio. The two roots of the characteristic polynomials are real, and after a relatively short while you can ignore the power of the smaller roots, and compare $y_i$ and $Y_j$ using logarithms to find good pairs $i,j.$ Finally, as mentioned, I imagine this can be finished with some algebraic number theory. But you could also finish it by showing $\log y_m$ and $\log Y_n$ never get quite close enough.   
For what it may be worth,
$$  y_n = \left( \frac{\sqrt 2 + 1}{\sqrt 8}  \right)  \left( 3 + \sqrt 8   \right)^n +  \left( \frac{\sqrt 2 - 1}{\sqrt 8}  \right)  \left( 3 - \sqrt 8   \right)^n   $$  and
$$  Y_n = \left( \frac{\sqrt 3 + 1}{\sqrt {12}}  \right)  \left( 2 + \sqrt 3   \right)^n +  \left( \frac{\sqrt 3 - 1}{\sqrt {12}}  \right)  \left( 2 - \sqrt 3   \right)^n   $$ 
A: 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.
