KMO Problem # 2

$\sqrt[3]{a + \sqrt{b}} + \sqrt[3]{a - \sqrt{b}} = 1 \\ \text{cubing the whole expression yields: } \ a + \sqrt{b} + a - \sqrt{b} + 3 \sqrt[3]{a^2 - b} = 1 \\ \sqrt[3]{a^2 - b} = \cfrac{1-2a}{3} \\ a^2 - b = \cfrac{(1-2a)^3}{27} \\ \implies b = a^2 - \ \cfrac{(1 - 2a)^3}{27}$

to be b be an integer, 1 - 2a must be in the form 3k so that, $27 \ | \ (1-2a)^3$ . $1 - 2a = 3k \\ \implies a = \ \cfrac{1-3k}{2}$

as 1 - 3k is always divisible by 2 for k = -(2n - 1),

$\implies a = \cfrac{1 - 3(-(2n-1))}{2} = \cfrac{1 + 3(2n-1)}{2} = \cfrac{6n - 2}{2} = 3n - 1 \hspace{0.25cm}, \text{for n > 0.} \\ \implies b = (3n - 1)^2 - \cfrac{(1 - 2(3n-1))^3}{27} \\ b = 9n^2 - 6n + 1 - \cfrac{(-3(2n - 1)^3)}{27} \\ \implies b = 9n^2 - 6n + 1 + (2n - 1)^3 = 9n^2 - 6n + 1 + 8n^3 - 12n^2 + 6n - 1 = 8n^3 - 3n^2 \\ \text{hence,} \ r = 8, x = s = 3, y = 2.$

$\sum_{n = 1}^5 \left( rn^x - sn^y \right ) \Longleftrightarrow \sum_{n = 1}^5 \left( 8n^3 - 3n^2 \right ) = 1635 \\ \begin{aligned} \therefore r + x + s + y + \sum_{n = 1}^5 \left( rn^x - sn^y \right ) & = 8 + 3 + 3 + 2 + \sum_{n = 1}^5 \left( 8n^3 - 3n^2 \right ) \\ & = 8 + 3 + 3 + 2 + 1635 \\ & = \boxed{1651} \end{aligned}$