Suppose that X(n) is defined whenever n is a divisor of 1000, and suppose that the equation SIGMA d|n X(d) = n^2 is satisfied for all n|1000. Use Mobius inversion to find X(1000).
I know that SIGMA n|1000 X(1000) = n^2
so X(1000) = n^2 * SIGMA MU( 1000/n) *
I find all the divisors of 1000, which will give me my n, then sub it into *
1, 2, 4, 5, 8, 10, 20, 25, 40, 50, 100, 125, 200, 250, 500, 1000
so is X(1000) = 1^2 * MU (1000/1) + 2^2 * MU (1000/2) .....
I know my working out is pretty general.. I'm just after a lead
Thanks