Motivation for definition of Mobius function

Question

Why is the Mobius function defined the way it is? \begin{align*} \mu(n) = \begin{cases} (-1)^r & \text{ if $n$ is square-free and is of the form }n=p_1p_2\ldots p_r\\ 0 & \text{ if $n$ is not square-free} \end{cases} \end{align*}

I can see that the function takes $-1$ on all primes. But why is extended in a way it is just multiplicative and not completely multiplicative?

Also, why is this particular function interesting to study? I can understand studying other arithmetic functions like the divisor function, totient function, etc. This function definition seems to be pulled out of thin air.

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Thanks

Answer 1

If you know about the Riemann zeta function, $\zeta(s)$, then $${1\over\zeta(s)}=\sum_{n=1}^{\infty}{\mu(n)\over n^s}$$ for all complex $s$ with real part exceeding 1.

If you don't know about the Riemann zeta function, look it up --- it's the most important function in analytic number theory.