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QUESTION

Prove that if $f(n)$ is a multiplicative function such that
$f(n)\neq0$ for at least one value of $n$, then $f(1)=1$.



ANSWER

Suppose $f(n)\neq0$. Now gcd$(1,n)=1$, so by the multiplicative
property, $f(n.1)=f(n).f(1)$, i.e. $f(n)=f(n)f(1)$, so as
$f(n)\neq0$, we may conclude $f(1)=1$.





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