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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.

Martin Sleziak
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Sander
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    No, you need to show $f(mn) = f(n) f(m)$ when $m$ and $n$ are coprime. – Robert Israel Oct 20 '16 at 21:18
  • This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level. – Stella Biderman Oct 20 '16 at 21:18
  • Note that with the common number-theoretic sense of "multiplicative function" you're only supposed to show this when $m$ and $n$ are coprime. (And otherwise it isn't true, anyway). The Fundamental Theorem of Arithmetic will be handy. – hmakholm left over Monica Oct 20 '16 at 21:20
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    In this context, an asterisk ($*$) (the Dirichlet's convolution) is far from a synonym of a dot ($\cdot$) (the usual product). – ajotatxe Oct 20 '16 at 22:22
  • @Stella Robotic, pre-formatted comments are just that, and rarely taken as anything but. Auto-commenting gets old, fast. – amWhy Oct 21 '16 at 00:30
  • @amWhy But it is still much better than leaving the post without an explanation for putting on-hold. And, of course, if you think that there is something which needs to be added to the above comment, you can go ahead and write another comment. – Martin Sleziak Oct 21 '16 at 06:09
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    A handful of duplicates found using Approach0: http://math.stackexchange.com/questions/536554/gcdab-c-equals-gcda-c-times-gcd-b-c http://math.stackexchange.com/questions/535961/prove-that-if-gcda-b-1-then-gcda-cdot-b-c-gcda-c-cdot-gcdb-c http://math.stackexchange.com/questions/806096/let-a-m-n-in-mathbfn-show-that-if-gcdm-n-1-then-gcda-mn-gcda http://math.stackexchange.com/questions/846522/if-gcda-b-1-then-gcdab-c-gcda-c-cdot-gcdb-c – Martin Sleziak Oct 21 '16 at 06:10
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    Are you trying to show that it is multiplicative or completely multiplicative? I guess you left out the condition that $\gcd(m,n)=1$. – Martin Sleziak Oct 21 '16 at 06:11

1 Answers1

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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 $$ \prod_{p\mid x\,\wedge\, p|a}p^{\min(\nu_p(a),\nu_p(x))}\prod_{p\mid y\,\wedge\, p|a}p^{\min(\nu_p(a),\nu_p(y))}=\gcd(a,x)\gcd(a,y). $$

Jack D'Aurizio
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