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I know how to find for which $n$ $\phi(n)=n/2$ or $\phi(n)=n/3$, my method for finding those was simply to find primes $p$ that satisfy $\Pi_p$$_|$$_n$$1-1/p$ $ = 1/2$ or $1/3$.

However, I don't know how to find $\Pi_p$$_|$$_n$$1-1/p = n/6$. Intuitively it seems that if I combine results for both $\phi(n) = n/2$ and $\phi(n) = n/3$ I'll get $\phi(n) = n/6$ but it does not work, cause I get number of the form $2^a3^b$ which gives $\phi(n) = n/3$ again.

Is there a way to find $n$ for which $\phi(n)=n/6$? Or do such numbers exist at all?

graydad
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Andrew
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  • I suspect that the first $n$ for which $n / \phi(n) \ge 6$ may be the primorial $223092870$ – Henry Jun 17 '15 at 14:20

2 Answers2

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We have: $$\nu_2\left(\frac{\varphi(n)}{n}\right)=\sum_{p\mid n}\nu_2\left(\frac{p-1}{p}\right)\geq \omega(n)-1$$ hence it is not possible that $\frac{\varphi(n)}{n}$ equals a rational number $\frac{p}{q}$ with an even $q>2$.

Here, $\nu_2$ is the $2$-adic height: $$\gcd(p,q)=1,\quad \nu_2\left(\frac{p}{q}\right)=\max\{m\in\mathbb{N}:2^m\mid p\}-\max\{m\in\mathbb{N}:2^m\mid q\}$$ and $\omega(n)$ is the number of prime divisors of $n$.

Jack D'Aurizio
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    I don't know what $\nu_2$ means, and I would not be at all surprised if Andrew doesn't know either. What's worse, $\nu_2$ is impossible to google. – TonyK Jun 17 '15 at 14:09
  • This paper (linked to in a now-deleted comment) has $\upsilon_p$ (upsilon), not $\nu_p$ (nu), for $p$-adic height. Is this also standard? – TonyK Jun 17 '15 at 14:21
  • @TonyK: I always used $\nu_p$, but now I am learning that my notation is not universal. Not a big issue, I made it clear in my updated answer. – Jack D'Aurizio Jun 17 '15 at 14:28
  • Some people go so far as to repeat the definition for all Greek letters, even common ones like $\pi$ and $\phi$ (though of course those have the different problem of being overloaded with various meanings). –  Jun 17 '15 at 18:13
  • I am not aware of what adic height is. Is there a simple way of explaing what adic height is? A link will suffice. – Andrew Jun 18 '15 at 13:37
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    @Andrew: a classical use of $\nu_2$ is shown here: http://math.stackexchange.com/questions/2746/is-there-an-elementary-proof-that-sum-limits-k-1n-frac1k-is-never-an-int – Jack D'Aurizio Jun 18 '15 at 13:47
  • Nowhere in that link does $\nu$ (or indeed $\upsilon$) occur. Nowhere in that link is $p$-adic height defined for any $p$. So not very helpful for @Andrew. – TonyK Jun 18 '15 at 22:27
  • @TonyK: not the greek letter, but the concept occurs. And it is really helpful in my humble opinion to understand the concept more than the notation $\nu$,$\upsilon$ or whatsoever. – Jack D'Aurizio Jun 18 '15 at 22:29
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I think this is valid:

Suppose $ϕ(n)=\frac{n}{6}$ then since the Euler's totient function is by definition the size of a set, it must be an integer hence $\frac{n}{6}$ is an integer hence $6$ must be a factor of $n$. Rewrite $n$ to $n=6k$ and we get:

$\rightarrow ϕ(6k)=k$

Now we rewrite k so that it's free from all it's factors of $2$ and $3$ , i.e. , $k=2^p*3^q*m$ where $m$ is not divisible by $2$ or $3$. So we now get :- $$ϕ(6*2^p*3^q*m)=ϕ(2^{(p+1)}*3^{(q+1)}*m) = 2^p*3^q*m$$ So now using multiplicativity:

$$ϕ(2^{(p+1)}*3^{(q+1)}*m)$$

$$\rightarrow ϕ(2^{(p+1)})ϕ(3^{(q+1)})*ϕ(m)$$

$$\rightarrow 2^p(2-1)*3^q(3-1)ϕ(m)$$

$$\rightarrow 2^p*3^q*m$$ (last equality coming from the "equalling $k$" part)

On cancelling/comparing the last two equalities we get:

$$2*ϕ(m)= m$$ Hence $m$ has a factor of $2$ which is a contradiction.

Anonymous
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Sam Georgiou
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