2016

We can apply a fundamental result of the mobius function . Specifically, if $n > 1$ then $\sum_{d|n} \mu(d) = \boxed{0}.$

The proof is as follows. Let $n = p_1^{\alpha_1} \cdot p_2^{\alpha_2} \cdot \ldots \cdot p_r^{\alpha_r}$ ( $\alpha_i \ge 1$ for all i =1,...,r). If $d|n$ , $d = p_1^{\beta_1} \cdot p_2^{\beta_2} \cdot \ldots \cdot p_r^{\beta_r}$ with $0 \leq \beta_i \leq \alpha_i$ for i = 1,...,r. If $\beta_i \ge 2$ for some i = 1, ... , r , then $\mu(d) = 0$ . Therefore, $\large \sum_{d|n} \mu(d) = \sum_{(\beta_1, \ldots , \beta_r)/ \beta_i = 0 \quad \text{or} \quad 1} \mu(p_1^{\beta_1} \cdot \ldots \cdot p_r^{\beta_r}) =$ $\large = 1 - \binom{r}{1} + \binom{r}{2} - \ldots + (-1)^{r} \binom {r}{r} = (1 - 1)^{r} = 0.$

I simply wrote all positive divisors of 2016, of which there are 36, and applied the definition of mu(n) to each one. That is, mu(1) =1. For n > 1, mu(n) = +1 if n has no square-free divisors and has an even number of prime divisors, mu(n) = -1 if n has no square-free divisors and has an odd number of prime factors, and mu(n) = 0 if n has a square-free divisor. (My first try gave 1 because I included 1 as a divisor. When told incorrect, I assumed that 1 should not be counted, so 0, but I have lingering questions about this. Ed Gray

Here's another proof that $\displaystyle\sum_{d \mid n} \mu(d) = 0$ for all $n > 1.$

Note that sums of arithmetic functions on divisors are multiplicative. Therefore, if $n = p_{1}^{a_{1}} p_{2}^{a_{2}} \cdots p_{k}^{a_{k}}$ we know that $\displaystyle\sum_{d \mid n} = \displaystyle\sum_{d_{1} \mid p_{1}^{a_{1}}} \mu(d_{1}) \cdots \cdot \displaystyle\sum_{d_{k} \mid p_{k}^{a_{k}}} \mu(d_{k}).$ However, note that any term of the product is $0$ , since all terms $\mu(p^2)= 0$ and $\mu(1) + \mu(p) = 0$ so we are done.