<s>[INST] 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??
 [/INST] $$x^2-y^2-3(x-y)=u^2-v^2$$
$$(x-y)(x+y-3)=(u+v)(u-v)$$
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.
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