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??
 A: 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  $(a,s) - (\pm3;\pm1)$
A: $$x^2-y^2-3(x-y)=u^2-v^2$$
$$(x-y)(x+y-3)=(u+v)(u-v)$$
Assuming $u^2-v^2\neq 0$, we have:
$$\dfrac{x-y}{u-v}=\dfrac{u+v}{x+y-3}=\dfrac{r}{s}$$ where $\gcd(r,s)=1$.
$$x-y=\dfrac{r(u-v)}{s}$$
$$x+y=\dfrac{s(u+v)}{r}+3$$
since $x,y$ are integers, and $u,v$ have the same parity, then there exist $p,q$ such that: $$u-v=ps $$ $$ u+v=qr $$
Hence,
$$x-y=pr $$
$$x+y=qs+3 $$
with $$x=\dfrac{qs+pr+3}{2}$$ $$y=\dfrac{qs-pr+3}{2}$$
$$u=\dfrac{qr+ps}{2} $$ $$v=\dfrac{qr-ps}{2} $$ where either $qr$ and $ps$ have the same parity and $pr$ and $qs$ have opposite parity.
A: We can without (much) loss of generality assume that $y\le x$. Note that $x^2+3y$ is a perfect square  greater than $x^2$. 
Thus we have $x^2+3y\ge (x+1)^2$. But $x^2+3y\lt (x+2)^2$.  It follows that  $3y=2x+1$. 
Since $y^2+3x$ is a perfect square, so is $9y^2+27x$, that is, $4x^2+31x+1$.  This has to be a square, so $4x^2+31x+1$ is equal to one of $(2x+1)^2$ or $(2x+2)^2$ and so on up to $(2x+7)^2$, since $(2x+8)^2$ is clearly too big. 
The case $(2x+1)^2$ does not work, and neither does $(2x+2)^2$, nor $(2x+3)^2$.  The case $(2x+4)^2$ gives $x=1$. The case $(2x+5)^2$ does not work. The case $(2x+6)^2$ gives $x=5$, which does not work because $y$ is not an integer, and the case $(2x+7)^2$ gives $x=16$. 
We conclude that the solutions are $x=1,y=1$, $x=16, y=11$, (and $x=11,y=16$).
