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.
well, you should learn how to use latex yourself instead of asking someone else to do it for you! that's also a good way to show your appreciation to those in here who help you.
anyway, we all know that $\displaystyle \sum_{k=1}^m k^3=\frac{m^2(m+1)^2}{4}=\left(\sum_{k=1}^m k \right)^2.$ now let $\displaystyle n=\prod_{j=1}^r p_j^{m_j}$ be the prime factorization of $\displaystyle n.$ then $\displaystyle d \mid n$ if and only if $\displaystyle d=\prod_{j=1}^r p_j^{s_j},$ where $\displaystyle 0 \leq s_j \leq m_j$ for all $\displaystyle j.$
it's clear that $\displaystyle \sigma(d)=\prod_{j=1}^r(1+s_j).$ thus: $\displaystyle S=\sum_{d \mid n}(\sigma(d))^3=\sum_{s_1=0}^{m_1} \cdots \sum_{s_r=0}^{m_r} \prod_{j=1}^r (1+s_j)^3=\prod_{j=1}^r \sum_{s_j=0}^{m_j}(1+s_j)^3.$ but, as i mentioned at the beginning: $\displaystyle \sum_{s_j=0}^{m_j}(1+s_j)^3=\left(\sum_{s_j=0}^{m_ j} (1+s_j) \right)^2,$
for all $\displaystyle j.$ therefore: $\displaystyle S=\prod_{j=1}^r \left(\sum_{s_j=0}^{m_j} (1+s_j) \right)^2 =\left(\sum_{s_1=0}^{m_1} \cdots \sum_{s_r=0}^{m_r} \prod_{j=1}^r (1+s_j) \right)^2=\left(\sum_{d \mid n}\sigma(d) \right)^2.$