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:


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

Michael Hardy
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    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. – Michael Hardy Apr 28 '14 at 18:05
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    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 $\phi(p^n)$ $= (p-1)p^{n-1}$ $= \phi(p)p^{n-1}$. – Steven Stadnicki Apr 28 '14 at 18:11
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    Note that on [the wikipedia totient page](http://en.wikipedia.org/wiki/Euler%27s_totient_function) 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. – abiessu Apr 28 '14 at 18:12
  • You are right it was incorrect, I didn't do it intentionally. – FofX Apr 28 '14 at 18:12
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    @MichaelHardy: That's an irritating feature of the editor that gets me all the time. It seems like a sort of autocorrect or something. When you end a displayed math section with "$$" and hit enter, the shift key is enabled so the next letter typed is capitalized. I don't know if it's the site or my iPad doing this (probably the latter). It's especially irritating to try to type "$a+bi$" to find it has bee "corrected" to "$a+bI$" for some reason. – MPW Apr 28 '14 at 18:21
  • @MPW It's not the site. It doesn't happen to me. Some sort of autocorrection on the I-pad runs amok. – Daniel Fischer Apr 28 '14 at 18:27
  • @MPW : What editor is that?? I just clicked on "ASK QUESTION" and typed $$a+bi,$$ then on the next line "where $a$ and $b$ are real", and nothing like that happened to me. – Michael Hardy Apr 28 '14 at 18:33
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    (By the way, since I realized that my initial comment was a bit abrupt, I want to emphasize that I think it's awesome you're thinking about things this way and experimenting with formulae! You may be in well-trod ground here, but don't let that discourage you at all. There are unexplored wilds just down the road.) – Steven Stadnicki Apr 28 '14 at 18:35

3 Answers3


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.

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    I would like to add that I hope I did not discourage you. You are doing real mathematics, and your claimed result is correct and sensible. – MJD Apr 28 '14 at 18:27
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    What MJD said, exactly - don't let the fact that these may be familiar results discourage you at all; your exploration is absolutely to be commended. – Steven Stadnicki Apr 28 '14 at 18:36
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    Thank you guys, that means a lot! I will definitely keep working at it, and maybe one day I'll discover something new. – FofX Apr 28 '14 at 18:42

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!

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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

Michael Hardy
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