The number-theoretic Möbius function $\mu (n)$ from prime factorisation

For a positive integer $n$, the number-theoretic Möbius function is

$\mu (n):=\{\begin{array}{ll}0,& v_p(n)\ge 2\text{ for some prime }p,\\ (-1{)}^r,& n=p_0{p}_1\cdots p_{r-1}\text{ for distinct primes }{p}_i\mathrm{.}\end{array}$

The power $(-1{)}^r$ is the natural power in the multiplicative monoid of $\mathbb{Z}$ (Powers $g^n$: natural exponents in a monoid and integer exponents in a group, with $g^0=e$, The integers form a commutative ring).

This definition is well posed. Canonical prime factorisation (For $n\ge 1$ and any injective list $p:r\to \mathbb{Z}$ of primes containing every prime divisor of $n$, one has $n={\prod }_{i<r}{p}_i^{v_{p_i}(n)}$; the exponents are determined by $n$, and $v_q(n)=0$ for every prime $q$ outside the list, The $p$-adic valuation $v_p(a)$ of a nonzero integer: the greatest $k\in \mathbb{N}$ with $p^k\mid a$) uniquely determines every exponent $v_p(n)$. If none exceeds $1$, the primes with exponent $1$ form a finite list whose length $r$ is invariant under reordering by the uniqueness clause of The fundamental theorem of arithmetic: every integer $n\ge 1$ is a product of primes, and the factorisation is unique up to order — if ${\prod }_{i<r}{p}_i={\prod }_{j<s}{q}_j$ with every $p_i$ and $q_j$ prime, then $r=s$ and $q_i=p_{\pi (i)}$ for some $\pi \in \mathrm{Sym}(r)$. If some exponent exceeds $1$, the first clause applies independently of which such prime is noticed. For $n=1$ the prime list is empty, so

$\mu (1)=(-1{)}^0=1.$

Equivalently, $\mu (n)=0$ exactly when a prime square divides $n$; otherwise its sign records the parity of the number of distinct prime factors (For a prime $p$ and a nonzero integer $a$: $p^{v_p(a)}\mid a$ and $p^{v_p(a)+1}\nmid a$; $p^k\mid a$ holds exactly for $k\le v_p(a)$; $v_p(a)\ge 1$ exactly when $p\mid a$; $v_p(1)=v_p(-1)=0$; and $v_p(p)=1$).