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 \[(1 + \sqrt{2})^2 + (1 - \sqrt{2})^2 + (1 - \sqrt{2})^2 + (1 + \sqrt{2})^2 = \boxed{12}.\]