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Show the following identity for Euler’s ϕ function:

$ϕ(m · n) = \frac{ϕ(m) · ϕ(n) · gcd(m, n)}{ϕ(gcd(m, n))}$

\begin{eqnarray*} \phi(m) = m \prod_{p \mid m } (1-\frac{1}{p}). \\ \end{eqnarray*} \begin{eqnarray*} \phi(n) = n \prod_{p \mid n } (1-\frac{1}{p}). \\ \end{eqnarray*} \begin{eqnarray*} \phi(mn) = mn \prod_{p \mid mn } (1-\frac{1}{p}). \\ \end{eqnarray*} \begin{eqnarray*} \phi(gcd(m,n)) = gcd(m,n) \prod_{p \mid gcd(m,n) } (1-\frac{1}{p}). \\ \end{eqnarray*} \begin{eqnarray*} \\\frac{\phi(m).\phi(n).gcd(m,n)}{\phi(gcd(m,n))} = \frac{m \prod_{p \mid m } (1-\frac{1}{p}).n \prod_{p \mid n } (1-\frac{1}{p}).gcd(m,n)}{gcd(m,n) \prod_{p \mid gcd(m,n) } (1-\frac{1}{p})}=mn \prod_{p \mid mn } (1-\frac{1}{p})=\phi(mn) \\ \end{eqnarray*} (since the gcd in the numerator and denominator cancel, and one product in the numerator cancels the one in the denominator).

Is this correct?

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JOJO
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    Why does the product in the denominator cancel one in the numerator? The ranges for the products are not the same. You haven't really used the properties of the gcd anywhere. – rogerl Sep 20 '19 at 21:02
  • Yeah, write or comment this 'reduction' of products more explicitly, it's the most crucial part of this proof, it seems – Jakobian Sep 20 '19 at 21:10
  • From $n = gN,m=gM,\gcd(N,M)=1$ and multiplicativity we obtain $ \frac{\phi(n)\phi(m)}{\phi(g)^2} = \phi(NM)= \frac{\phi(nm)}{\phi(g^2)}$ to conclude we need $\phi(g^2) = \phi(g) g$ which follows from $\phi(p^{2k}) = (p-1)p^{2k-1}=\phi(p^k)p^k$ – reuns Sep 20 '19 at 21:12
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    rogerl is right. The products you cancele and the product remaining aren't at all the same as they scan different ranges. You need to prove $\frac {\prod_{p|m}(1-\frac 1p) \prod_{p|n}(1-\frac 1p)}{\prod_{p|\gcd(m,n)} (1-\frac 1p)}$. Which... can be done if you you can prove 1)${p:p$ is prime $p|m}\cap {p:p$ is prime $p|n}={p:p$ is prime $p|\gcd(m,n)}$ and 2) $\prod_{x\in A\cup B} f(x)=\frac {\prod_{x\in A}f(x)\prod_{x\in B}f(x)}{\prod_{x\in A\cap B}f(x)}$. – fleablood Sep 20 '19 at 21:12
  • You need to work on what that sets ${p|n}$ and ${p|m}$ and ${p|\gcd(m,n)}$ "look" like and interact. Here's a hint: If we let $A ={p: p|m;p\not \mid n}$ and $B={p:p\not \mid m;p\mid n}$ and $C={p:p\mid m;p\mid n}$ then ${p|m}=A\cup C$ and ${p|n}=B\cup C$ and ${p|\gcd(m,n)} = C$. – fleablood Sep 20 '19 at 21:28
  • https://math.stackexchange.com/questions/754341/prove-that-if-d-gcdm-n-then-phimn-phim-phin-d – lab bhattacharjee Sep 21 '19 at 02:13

1 Answers1

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First a mild diversion but an essential key concept:

Let $P_{m\text{ only}} = \{p\in \mathbb N|p$ is prime$; p\mid m; p\not \mid n\}$

Let $P_{n\text{ only}} = \{p\in \mathbb N|p$ is prime$; p\mid n; p\not \mid m\}$

And $P_{m,n\text{ both}} = \{p\in \mathbb N|p$ is prime$; p\mid m; p \mid n\}$

Now it's clear these sets are mutually disjoint (it's not possible for there to exist any primes that both do and do not divide a number).

It's also clear that

$\{p\in \mathbb N|p$ is prime$; p\mid \gcd(m,n)\}= P_{m,n\text{ both}}$

And that $\{p\in \mathbb N|p$ is prime$; p\mid n\} = P_{n\text{ only}}\cup P_{m,n\text{ both}}$

and that $\{p\in \mathbb N|p$ is prime$; p\mid m\} = P_{m\text{ only}}\cup P_{m,n\text{ both}}$.

And that $\{p\in \mathbb N|p$ is prime$; p\mid nm\} = P_{n\text{ only}}\cup P_{m\text{ only}}\cup P_{m,n\text{ both}}$

Now we do all your work and conclude that:

$\begin{eqnarray*} \$\frac{\phi(m).\phi(n).gcd(m,n)}{\phi(gcd(m,n))} = \frac{m \prod_{p \mid m } (1-\frac{1}{p}).n \prod_{p \mid n } (1-\frac{1}{p}).gcd(m,n)}{gcd(m,n) \prod_{p \mid gcd(m,n) } (1-\frac{1}{p})} \\ \end{eqnarray*}$

That much you were correct about but you did some canceling of products that just don't work as they were products of completely different indices.

But if we put these idexes to my $P$ notation:

$\frac{m \prod_{p \mid m } (1-\frac{1}{p}).n \prod_{p \mid n } (1-\frac{1}{p}).gcd(m,n)}{gcd(m,n) \prod_{p \mid gcd(m,n) } (1-\frac{1}{p})}=$

$\frac{m \prod_{p \in P_{m\text{ only}}\cup P_{m,n\text{ both}}} (1-\frac{1}{p}).n \prod_{p \in P_{n\text{ only}}\cup P_{m,n\text{ both}}} (1-\frac{1}{p}).gcd(m,n)}{gcd(m,n) \prod_{p \in P_{m,n\text{ both}}} (1-\frac{1}{p})}=$

$\frac{m \prod_{p \in P_{m\text{ only}}}(1-\frac 1p)\prod_{p \in P_{m,n\text{ both}}} (1-\frac{1}{p}).n \prod_{p \in P_{n\text{ only}}}(1-\frac 1p)\prod_{p \in P_{m,n\text{ both}}} (1-\frac{1}{p})).gcd(m,n)}{gcd(m,n) \prod_{p \in P_{m,n\text{ both}}} (1-\frac{1}{p})}=$

$mn \prod_{p \in P_{m\text{ only}}}(1-\frac 1p)\prod_{p \in P_{n\text{ only}}}(1-\frac 1p)\prod_{p \in P_{m,n\text{ both}}} (1-\frac{1}{p}))=$

$mn \prod_{p \in P_{m\text{ only}}\cup P_{n\text{ only}}\cup P_{m,n\text{ both}}} (1-\frac 1p) =$

$mn \prod_{p|mn}(1-\frac 1p) = \phi(mn)$.

fleablood
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