Two answers $(1;0)$ and $(0;1)$

You're wrong, there are 8 solutions

$ (0;1)(0;-1)(1;0)(-1;0)(3;4)(4;3)(-3;-4)(-4;-3)$

rmtf1111 wrote: Find all solutions for (x,y) , both integers such that: $xy=3(\sqrt{x^2+y^2}-1)$ If $xy \leq 0$ then obviously we obtain solutions $\{(0,1),(1,0),(-1,0),(0,-1)\}$. Else $WLOG$ $x,y > 0$. Now $x^2 + y^2 = z^2$ so by pytagorean triples we obtain that:$WLOG$ $x = k^2 - l^2$,$y = 2kl$,$z = k^2 + l^2$ and $k,l > 0$. We have that $2kl(k-l)(k+l) = 3(k^2+l^2 - 1)$,note with $d = k - l$ and get that $2ld(2l^2+3ld+d^2) = 3(2l^2+d^2+2ld-1)$. If $ld \ge 2$ then $6l^2 + 3d^2 + 6ld - 3 \ge 8l^2 + 12ld + 4d^2$ ,contradiction $LHS > RHS$ so $ld = 1$ and becasue $l,d > 0$ we obtain that $l = d = 1$,$(x,y) \in \{ (3,4) , (4,3) , (-3,-4) , (-4, -3) \}$.