New identity for Euler's Totient Function?

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

• Can you tell me why you wrote "Where", with a capital initial "W" in "Where $\operatorname{rad}(n)$ is the radical"? I've seen lots of others do that --- so many that I have to suspect it's not merely a typo but rather that some people feels that's the right way to do it. Since it's not the beginning of a new sentence, but makes sense only as a continuation of the sentence already underway, I am puzzled. Commented Apr 28, 2014 at 18:05
• I don't know that I've ever seen this written out explicitly, but it's an immediate consequence of the multiplicativity of $\phi()$ along with the rule that $\phi(p^n)$ $= (p-1)p^{n-1}$ $= \phi(p)p^{n-1}$. Commented Apr 28, 2014 at 18:11
• Note that on the wikipedia totient page there is the formula $$\phi(mn)=\phi(m)\phi(n)\dfrac{(m,n)}{\phi((m,n))}$$ with $(m,n)$ denoting the GCD of $m$ and $n$... With just a little transformation, I believe this expresses your identity or nearly so. Commented Apr 28, 2014 at 18:12