It can be shown, by using the fact that $E\_4(z)^2 = E\_8(z)$ (where the $E\_k$ are Eisenstein series), that $$\sigma\_7(n) = \sigma\_3(n) + 120\sum\_{0 < m < n}\sigma\_3(m -n)\sigma\_3(n)$$ where $\sigma\_k(n) := \sum\_{d|n} d^k$. Is there a more elementary proof of this, and is there a simple reason one should expect such identities to hold?



Elementary proof of certain divisor sum idenrity
I have asked a question about a *different* proof of such identities here: [Derivation of certain sums "the hard way"](https://mathoverflow.net/questions/282989/derivation-of-certain-sums-the-hard-way?rq=1), but I would say that the non-modular proofs are much less elementary than the modular ones.  

To answer your second question, we should expect these identities to hold, **because Eisenstein series have the coefficients of the form $\sigma\_k(n)$**. If you combine this with some properties of modular forms (multiplying a form of weight $k$ and $l$ gives you a form of weight $k+l$ and that the space of mod. forms of given weight has a given dimension), you can see that some identities must arise between them. This is not only the case with the $\sigma\_k$, other arithmetic functions also have modular identities. 


The identity you gave above is a perfect example. You know, that both $E\_4^2$ and $E\_8$ are forms of weight 8, but the dimension of $M\_8(\Gamma)$ is 1. Thus, one is a multiple of the other (actually, $E\_4^2=E\_8$) and you have your identity.


