Find the sum of the squares of all real numbers satisfying the equation $x^{256}-256^{32}=0$.
$\textbf{(A) }8\qquad \textbf{(B) }128\qquad \textbf{(C) }512\qquad \textbf{(D) }65,536\qquad \textbf{(E) }2(256^{32})$
Solution by e_power_pi_times_i
Notice that the solutions to the equation $x^{256}-1=0$ are the $256$ roots of unity. Then the solutions to the equation $x^{256}-256^{32}=0$ are the $256$ roots of unity dilated by $\sqrt[256]{256^{32}} = \sqrt[256]{2^{256}} = 2$. However, the only real solutions to the equation are the first root of unity and the root of unity opposite of it, as both are on the real axis in the complex plane. These two roots of unity are $\pm1$, and dilating by $2$ gives $\pm2$. The sum of the squares is $(2)^2+(-2)^2 = \boxed{\textbf{(A) } 8}$