# Question about the proof of Theorem 2.19 (Page 38) of the book Introduction to Analytic Number Theory by Apostol

At the last line of the proof:

$\lambda^{-1}(n)=\mu(n)\lambda(n)=\mu^2(n)=|\mu(n)|$.

Why $\mu(n)\lambda(n)=\mu^2(n)$? How to prove this?

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If $n$ is not squarefree, both sides are zero.
If $n$ is squarefree, then $\mu(n)=\lambda(n)$.