Perfect Sum of Cubes

The sum of the first $k$ cubes $1^3 + 2^3 + 3^3 + \dots + k^3$ is $(\frac{k(k + 1)}{2})^2$ .

The sum of the first $k$ even cubes $2^3 + 4^3 + 6^3 + \dots + (2k)^3 = 2^3(1^3 + 2^3 + 3^3 + \dots + k^3)$ is then $2^3(\frac{k(k + 1)}{2})^2 = 2k^2(k + 1)^2$ .

The sum of the first $k$ odd cubes $1^3 + 3^3 + 5^3 + \dots + (2k - 1)^3$ is the sum of the first $2k$ cubes ( $1^3 + 2^3 + 3^3 + \dots + (2k)^3$ ) minus the sum of the first $k$ even cubes ( $2^3 + 4^3 + 6^3 + \dots + (2k)^3$ ), or $(\frac{2k(2k + 1)}{2})^2 - 2k^2(k + 1)^2$ , which simplifies to $k^2(2k^2 - 1)$ .

The perfect number $2^{77232916}(2^{77232917} - 1)$ is equal to $2^{77232916}(2 \cdot 2^{77232916} - 1)$ or $(2^{38616458})^2(2 \cdot (2^{38616458})^2 - 1)$ , and so $k = 2^{38616458}$ , which means $2^{77232916}(2^{77232917} - 1)$ is the sum of the first $2^{38616458}$ cubes, and therefore $n = \boxed{38616458}$ .

Nice Problem.