Ramanujan Sum is Multiplicative

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Theorem

Let $q \in \N_{>0}$, $n \in \N$.

Let $\map {c_q} n$ be the Ramanujan sum.

Then $\map {c_q} n$ is multiplicative in $q$.


Proof

Let $q, r \in \N$ such that:

$\gcd \set {q, r} = 1$

where $\gcd \set {q, r}$ denotes the greatest common divisor of $q$ and $r$.

Then:

\(\ds \map {c_q} n \, \map {c_r} n\) \(=\) \(\ds \sum_{d_1 \mathop \divides \gcd \set {n, q} } d_1 \, \map \mu {\frac q {d_1} } \sum_{d_2 \mathop \divides \gcd \set {n, r} } d_2 \, \map \mu {\frac r {d_2} }\) Kluyver's Formula for Ramanujan's Sum
\(\ds \) \(=\) \(\ds \sum_{\substack {d_1 \mathop \divides \gcd \set {n, q} \\ d_2 \mathop \divides \gcd \set {n, r} } } d_1 d_2 \, \map \mu {\frac q {d_1} } \, \map \mu {\frac r {d_2} }\)
\(\ds \) \(=\) \(\ds \sum_{d \mathop \divides \gcd \set {n, q} \gcd \set {n, r} } d \, \map \mu {\frac {q r} d}\) Möbius Function is Multiplicative, and setting $d = d_1 d_2$
\(\ds \) \(=\) \(\ds \sum_{d \mathop \divides \gcd \set {n, q r} } d \, \map \mu {\frac {q r} d}\) GCD with One Fixed Argument is Multiplicative Function
\(\ds \) \(=\) \(\ds \map {c_{q r} } n\) Kluyver's Formula for Ramanujan's Sum

This completes the proof.

$\blacksquare$