# n is the smallest positive integer such that (2001+n) is the sum of the cubes of the first 'm' natural numbers. Then (m,n)=?

Jul 29, 2018

Sum of the cubes of the first 'm' natural numbers is

$= {\left(\frac{m \left(m + 1\right)}{2}\right)}^{2} = \left(2001 + n\right)$

So $\left(2001 + n\right)$ must be perfect square. The smallest value of n for which $\left(2001 + n\right)$ is a perfect square, will be $n = 24$

Hence ${\left(\frac{m \left(m + 1\right)}{2}\right)}^{2} = \left(2001 + 24\right)$

$\implies \frac{m \left(m + 1\right)}{2} = \sqrt{2001 + 24} = 45$

$\implies {m}^{2} + m - 90 = 0$

$\implies {m}^{2} + 10 m - 9 m - 90 = 0$

$\implies m \left(m + 10\right) - 9 \left(m + 10\right) = 0$

$\implies \left(m + 10\right) \left(m - 9\right) = 0$

So $m = 9$ as m is the positive integer.