<s>[INST] 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.
 [/INST] $x^2+3y=a^2$
$y^2+3x=b^2$
$a,b \in \mathbb{N}$
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!!!!
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