Finding pairs of integers such that $x^2+3y$ and $y^2+3x$ are both perfect squares Can we find pairs $(x,y)$ of positive integers such that $x^2+3y$ and $y^2+3x$ are simultaneously perfect squares? Thanks a lot in advance. My progress is minimal.
 A: Not a complete solution but an approach that seems like it will work.
Assume $y \gt x$
Then we have that
$(y+2)^2 \gt y^2 + 3y \gt y^2+3x \gt y^2$
If $y^2 + 3x$ was a perfect square, then we have that $y^2 + 3x = (y+1)^2$. 
This gives us $3x = 2y+1$.
Substitute in the other expression, and form similar inequalities. This will narrow down to few small cases to consider.
To elaborate, given $3x = 2y+1$
$x^2 + 3y = x^2 + \frac{9x-3}{2} \lt (x+3)^2$
Thus $x^2 + 3y$ is either $(x+1)^2$ or $(x+2)^2$.
Thus we have that
$3y = 2x+1$ or $3y = 4x + 4$.
Substitute $y = \frac{3x -1}{2}$ and solve the linear equation in $x$ and compute $y$. In the end, don't forget to verify that both the expressions are indeed perfect squares.
The other case $y=x$ can be treated similarly.
A: $x^2+3y=a^2$
$y^2+3x=b^2$
$a,b \in \mathbb{N}$
assume wlog $x\ge y$
$a^2> x^2$
$a > x$ but since $x\ge y$ we have $x+2>a$
so $a=x+1 $
$3y=2x+1$
$x=3k+1 ,y=2k+1$
$4k^2+4k+1+9k+3=b^2$
solving for k we get
$16b^2+105=t^2$ is a perfect square
$b=1,2,4,13$
so $b=2,4,13$ and $x=y=1$ or  $x=16 $, $y=11 $ and if $y\ge x$ we can get  $x=11 $, $y=16$
Done!!!!
