2005 AMC 12B Problem 23

The conditions give $x + y = 10^z$ and $x^2 + y^2 = 10 \cdot 10^z.$ Then $$\begin{aligned} &2xy = (x+y)^2 - (x^2+y^2) \\ &= 10^{2z} - 10 \cdot 10^z, \end{aligned}$$ so $xy = \dfrac12\left(10^{2z} - 10 \cdot 10^z\right).$ Using $x^3 + y^3$ $= (x+y)^3 - 3xy(x+y),$ $$\begin{aligned} &x^3 + y^3 = 10^{3z} \\ &\quad {}- \dfrac32\left(10^{2z} - 10 \cdot 10^z\right)10^z \\ &= -\dfrac12 \cdot 10^{3z} + 15 \cdot 10^{2z}. \end{aligned}$$ So $a = -\dfrac12$ and $b = 15,$ giving $a + b = \dfrac{29}{2}.$ These coefficients are determined: setting $z=0$ gives real solutions $x,y=\dfrac{1\pm\sqrt{19}}{2},$ and their equation forces this same value of $a+b.$ Thus, the correct answer is B.