New identity for Euler's Totient Function?

Question

A few weeks ago I discovered and proved a simple identity for Euler's totient function. I figured that someone would have already discovered it, but I haven't been able to find it anywhere.

So I was hoping someone would be able to tell me whether or not they've seen this identity.

Here it is:

$$\phi(n)=\phi(\operatorname{rad}(n))\left(\frac{n}{\operatorname{rad}(n)}\right)$$

where $\operatorname{rad}(n)$ is the radical or square-free kernel of $n$, i.e. the product of the distinct prime factors of $n$.

I have omitted the proof because firstly, I'm still learning MathJax and am afraid it will take quite a long time to type out. And secondly, I believe it is intuitive enough that most people familiar with the totient function should be able to see that it's true.

Like I said, it is a pretty simple identity, but nevertheless; it seems like it could be fairly useful. It would be a bit easier to calculate $\phi(n)$ for large-ish $n$ with this identity, without the help of a program or totient function calculator.

Ex: $$\phi(450)=\phi(2\cdot3^2\cdot5^2)=\phi(2\cdot3\cdot5)\left(\frac{2\cdot3^2\cdot5^2}{2\cdot3\cdot5}\right)=(1\cdot2\cdot4)(3\cdot5)=120$$

Answer 1

This is an interesting thing to notice, and you should be pleased.

As you guessed, and as Steven Stadnicki pointed out, this is not new; it follows quickly from two important properties of the $\phi$ function:

1. $\phi(p^n) = (p-1)p^{n-1}$ when $p$ is prime
2. $\phi(mn) = \phi(m)\phi(n)$ when $m$ and $n$ have no common factor

In particular, you suggested that your formula might be useful for calculating $\phi(n)$ for “large-ish $n$”, but observe that your formula requires knowing the radical of $n$, which is not in general any easier to find than the prime factorization of $n$. (And when the radical of $n$ is equal to $n$, as it is when $n$ is squarefree, your formula is no help at all.) But given $n = p_1^{a_1}\cdots p_k^{a_k}$ one has from (1) and (2) above that $$\begin{align} \phi(n) & = \phi\bigl(p_1^{a_1}\bigr)\cdots\phi\bigl(p_k^{a_k}\bigr) \\ & = (p_1-1)p_1^{a_1-1}\cdots (p_k-1)p_k^{a_k-1} \\ & = \frac{n}{p_1\cdots p_k}(p_1-1)\cdots(p_k-1) \\ & = n\left(\frac{p_1-1}{p_1}\right)\cdots \left(\frac{p_k-1}{p_k}\right) \tag{$\heartsuit$}\\ & = n\left(1-\frac1{p_1}\right)\cdots\left(1-\frac1{p_k}\right) \end{align}$$ which is well-known, and not significantly harder to compute (or perhaps easier) than your formula. The next-to last line ($\heartsuit$) is very similar to your formula.

Answer 2

$\newcommand{\rad}{\operatorname{rad}}$

Yes this is a bit of transformation of the euler product formula; it's correct since $$n=\prod_{i=1}p_i^{a_i}=p_1^{a_1}p_2^{a_2}p_3^{a_3}\cdots p_n^{a_n}\\ \rad(n)=p_1p_2p_3\cdots p_n\\\varphi(\rad(n))=(p_1-1)(p_2-1)(p_3-1)\cdots (p_n-1)\\ \frac{n}{\rad(n)}=p_1^{a_1-1}p_2^{a_2-1}p_3^{a_3-1}\cdots p_n^{a_n-1}\\ \varphi(\rad(n))\left(\frac{n}{\rad(n)}\right)=n\left(1-\frac{1}{p_1}\right)\left(1-\frac{1}{p_2}\right)\left(1-\frac{1}{p_3}\right)\cdots \left(1-\frac{1}{p_n}\right)$$ Which is all ready known

Answer 3

Your identity can be written as $$ \frac{\phi(n)}{n}=\frac{\phi(\operatorname{rad}(n))}{\operatorname{rad}(n)} $$ In this form, it is obvious, because $$ \frac{\phi(m)}{m}=\prod_{p\mid m}\left(1-\frac1p\right) $$ and $n$ and $\operatorname{rad}(n)$ have exactly the same prime factors.

Still, a nice observation, well done!