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)=?

1 Answer
Jul 29, 2018

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

=((m(m+1))/2)^2=(2001+n)

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

Hence ((m(m+1))/2)^2=(2001+24)

=>(m(m+1))/2=sqrt(2001+24)=45

=>m^2+m-90=0

=>m^2+10m-9m-90=0

=>m(m+10)-9(m+10)=0

=>(m+10)(m-9)=0

So m=9 as m is the positive integer.