User:Caliburn/s/4

Theorem
Let $n \ge 2$ be an integer.

Let $k$ be the number of distinct prime divisors of $n$.

Then:


 * $\ds \sum_{d \divides n} \size {\map \mu d} = 2^k$

where $\mu$ denotes the Moebius function.

Proof
By the Fundamental Theorem of Arithmetic, we can factorise:


 * $\ds n = \prod_{i \mathop = 1}^k p_i^{\alpha_i}$

for primes $p_i$ and non-negative integers $\alpha_i$, with $p_i \ne p_j$ for $i \ne j$.

Then any divisor $d$ of $n$ will have the form:


 * $\ds d = \prod_{i \mathop = 1}^k p_i^{\beta_i}$

where $0 \le \beta_i \le \alpha_i$.

Note that if $\beta_i \ge 2$ for any $i$, $d$ is not square-free, so we would have:


 * $\map \mu d = 0$

in this case, by the definition of the Moebius function.

So:


 * $\map \mu d \ne 0$

only if $d$ is the product of distinct primes, or $d = 1$.

Note that:


 * $\ds \sum_{d \divides n} \size {\map \mu d} = \map \mu 1 + \sum_{d \divides n, \, d \ne 1} \size {\map \mu d} = 1 + \sum_{d \divides n, \, d \ne 1} \size {\map \mu d}$

We can decompose the latter sum:


 * $\ds \sum_{d \divides n, \, d \ne 1} \size {\map \mu d} = \sum_{i \mathop = 1}^k \paren {\sum_{d \divides n, \, d \text { product of primes}, \, \map \omega d = i} \size {\map \mu d} }$

Note that from the definition of the Moebius function, we have:


 * $\size {\map \mu d} = 1$

since $d$ is square-free.

So, we wish to compute:


 * $\ds \sum_{d \divides n, \, d \text { product of primes}, \, \map \omega d = i} 1$

for each $i$.

That is, we wish to count how many products of $i$ primes divide $n$.

From Euclid's Lemma, each of these primes must divide $n$.

That is, we simply aim to count how many ways we can pick $i$ primes from the set of $k$ primes $\set {p_k}$.

So, from the definition of the Definition:Binomial Coefficient:


 * $\ds \sum_{d \divides n, \, d \text { product of primes}, \, \map \omega d = i} 1 = \binom k i$

Then: