Basic Properties of Multiplicative Function

Theorem

Let $f: \Z \to \Z$ be a multiplicative function.

Let $n = p_1^{k_1} p_2^{k_2} \ldots p_r^{k_r}$ be the prime decomposition of $n$.

• If $f$ is not identically zero, then $f \left({1}\right) = 1$;
• $f \left({n}\right) = f \left({p_1^{k_1}}\right) f \left({p_2^{k_2}}\right) \ldots f \left({p_r^{k_r}}\right)$.

Proof

• If $f$ is not identically zero, then $\exists m \in \Z: f \left({m}\right) \ne 0$.

Then $f \left({m}\right) = f \left({1 \times m}\right) = f \left({1}\right) f \left({m}\right)$.

Hence $f \left({1}\right) = 1$.

$\blacksquare$

• We have $n = p_1^{k_1} p_2^{k_2} \ldots p_r^{k_r}$.

We also have $\forall i, j \in \left[{1 \, . \, . \, n}\right]: i \ne j \implies p_i^{k_i} \perp p_j^{k_j}$.

So $f \left({p_i^{k_i} p_j^{k_j}}\right) = f \left({p_i^{k_i}}\right) f \left({p_j^{k_j}}\right)$.

It is a simple inductive process to show that $f \left({n}\right) = f \left({p_1^{k_1}}\right) f \left({p_2^{k_2}}\right) \ldots f \left({p_r^{k_r}}\right)$.

$\blacksquare$