Regular Tetrahedrons and Right Square Prisms

$a^2 + 1 = x^2 = 2(1 - a)^2 \implies a^2 + 1 = 2 - 4a + 2a^2 \implies a^2 - 4a + 1 = 0 \implies$

$a = 2 \pm \sqrt{3}$

$a = 2 + \sqrt{3} \implies 1 - a = -1 - \sqrt{3} < 0 \therefore$ drop $a = 2 + \sqrt{3}$ .

$a = 2 - \sqrt{3} \implies x = 2\sqrt{2 - \sqrt{3}} \implies A_{\triangle{AEF}} = \sqrt{3}(2 - \sqrt{3}) = 2\sqrt{3} - 3$ .

Let $h$ be height of the regular tetrahedron.

$h = \sqrt{x^2 - \dfrac{x^2}{3}} = \sqrt{\dfrac{2}{3}}x = 2\sqrt{\dfrac{2}{3}}\sqrt{2 - \sqrt{3}} \implies$

The volume of the regular tetrahedron $V_{T} = \dfrac{2\sqrt{2}}{3}(2 - \sqrt{3})^{\frac{3}{2}}$

and the volume of the right square prism $V_{B} = h \implies$

$V_{B} - V_{T} = 2\sqrt{\dfrac{2}{3}}\sqrt{2 - \sqrt{3}} - \dfrac{2\sqrt{2}}{\sqrt{3}}(2 - \sqrt{3})^{\frac{3}{2}} =$

$2\sqrt{2}\sqrt{2 - \sqrt{3}}(\dfrac{1}{\sqrt{3}} - \dfrac{2 - \sqrt{3}}{3}) =$ $\dfrac{2\sqrt{2}\sqrt{2 - \sqrt{3}}}{3\sqrt{3}}(6 - 2\sqrt{3}) =$

$2(\dfrac{2}{3})^{\frac{3}{2}}\sqrt{2 - \sqrt{3}}(3 - \sqrt{3}) =$ $a(\dfrac{a}{b})^{\frac{b}{a}}(\sqrt{a - \sqrt{b}})(b - \sqrt{b})$

$\implies a + b = \boxed{5}$ .