jmc

Since $x+y=2,$ there exists a real number $t$ such that $x=1+t$ and $y=1-t.$ Then $$(1+t{)}^5+(1-t{)}^5=82.$$This simplifies to $10t^4+20t^2-80=0.$ This equation factors as $10(t^2-2)(t^2+4)=0,$ so $t=\pm \sqrt{2}.$

Hence, the solutions are $(1+\sqrt{2},1-\sqrt{2})$ and $(1-\sqrt{2},1+\sqrt{2}),$ and the final answer is $$(1+\sqrt{2}{)}^2+(1-\sqrt{2}{)}^2+(1-\sqrt{2}{)}^2+(1+\sqrt{2}{)}^2=\boxed{12}.$$