How do I prove
Sum[sigma(d)]^3 = (Sum[sigma(d))^2 for all n>=1?
Where sigma is the number of divisors and the sum is taken over all d|n.
Is this clear? I'd appreaciate it if somebody could put this into LaTex for me.
anyway, we all know that now let be the prime factorization of then if and only if where for all
it's clear that thus: but, as i mentioned at the beginning:
for all therefore: