System of diophantine equations $x^2+3y=u^2$, $y^2+3x=v^2$ 
Solve the following system of Diophantine equations(the unknowns are positive integers):
  $$
\left\{ 
\begin{array}{c}
x^2+3y=u^2 \\ 
y^2+3x=v^2
\end{array}
\right. 
$$  

I worked as follows:
subtract the two equations to get: $4x^2-4y^2-12(x-y)=9y^2-9x^2\ \implies\ ... (x-y)(13x+13y-12)=0\implies x=y\  or\  13x+13y-12=0$
The first equation has infinite answers and the second has none(since $gcd(13,13)$ does not divide $12$), am I right??

The solution to this system of equations there.   http://www.artofproblemsolving.com/community/c3046h1046718__4
$$\left\{\begin{aligned}&x^2+qy=z^2\\&y^2+qx=v^2\end{aligned}\right.$$
Use this decision.
$$x=2psb^2-a^2p^2$$
$$y=2abp^2-b^2s^2$$
$$q=as(4bp-as)$$
$$z=a^2p^2+2psb^2-abs^2$$
$$v=2abp^2-psa^2+b^2s^2$$
In our case, it is necessary to $b=1$ ; $p=1$ and  $(a,s) - (\pm3;\pm1)$
