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I want to show that $f(n) = \gcd(a,n)$ where a is any natural number, is a multiplicative function.

I know I need to show that $f(mn)=f(n)*f(m)$, but I do not know how to do this.

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closed as off-topic by Stella Biderman, Cameron Williams, Daniel W. Farlow, iadvd, darij grinberg Oct 21 '16 at 6:15

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If you prove that $$ \gcd(a,x) = \prod_{p\mid a} p^{\min(\nu_p(a),\nu_p(x))} $$ where $\nu_p(x)=\max\{m\in\mathbb{N}: p^m\mid x\}$ the question becomes trivial. If $x$ and $y$ are coprime integers they have no common prime factor, so $$ \gcd(a,xy) = \prod_{p\mid a}p^{\min(\nu_p(a),\nu_p(xy))}=\prod_{p\mid x\,\wedge\, p|a}p^{\min(\nu_p(a),\nu_p(xy))}\prod_{p\mid y\,\wedge\, p|a}p^{\min(\nu_p(a),\nu_p(xy))} $$ and the RHS equals $$ \prod_{p\mid x\,\wedge\, p|a}p^{\min(\nu_p(a),\nu_p(x))}\prod_{p\mid y\,\wedge\, p|a}p^{\min(\nu_p(a),\nu_p(y))}=\gcd(a,x)\gcd(a,y). $$

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