$y=1 : (x-6)^2 |x^2+1 \to (x-6) |37 \to x=5,7$ $y=2:x=2$ is solution Let $xy>6$ and $x \geq y \geq 2$ If $ y \geq 3$ then $2x^2 \geq x^2+y^2 \geq (xy-6)^2 \geq (3x-6)^2 \to x=3$ If $y=2$ then $(2x-6)^2 | x^2+4 \to x=2t,t>1, (2t-3)^2|t^2+1 \to t \leq 3 \to t=2 \to x=4$ Answer: $(5,1),(7,1),(2,2),(4,2),(3,3)$

Inequality. $$(xy - 6)^2 | x^2 + y^2$$$$14x^2 \ge x^2+y^2+12xy \ge x^2 y^2+36$$$$y=1,2,3$$Remaining easy