Case 1 $(a+b)^2+4|2(a^2+b^2)$ $2(a^2+b^2) < 2((a+b)^2+4)$ so $2(a^2+b^2)=(a+b)^2+4$ or $(a-b)^2=4$ Case 2: $2(a^2+b^2)|(a+b)^2+4$ and $(a+b)^2+4 \geq 4(a^2+b^2)$ because equality case already done $2ab+4 \geq 3(a^2+b^2)$ $2 \geq (a^2+b^2) \to (a,b)=(1,1)$ So $(a,b)=(1,1), (t,t+2),(t+2,t)$