Let $\varphi$ denote Euler's totient function. It is easy to see that all those numbers $$\varphi(n^2)=n\varphi(n)\ \ (n=1,2,3,\ldots)$$ are pairwise distinct.

I have the following surprising conjecture.

Conjecture. Any positive rational number $r$ has the form $\varphi(m^2)/\varphi(n^2)$ with $m$ and $n$ positive integers.

I have verified this for $r\in\{a/b:\ a,b=1,\ldots,50\}$. My computation shows that \begin{align}&\left\{\frac{\varphi(m^2)}{\varphi(n^2)}:\ m,n=1,\ldots,15000\right\}\\\supseteq&\left\{\frac ab:\ 1\le a,b\le 50\ \&\ \{a,b\}\not=\{19,47\},\{37,47\}\right\}.\end{align} In addition, I have found that $$\frac{\varphi(12765^2)}{\varphi(18612^2)}=\frac{80879040}{102738240} =\frac{37}{47}$$ and $$\frac{\varphi(39330^2)}{\varphi(55836^2)}=\frac{373792320}{924644160} =\frac{19}{47}.$$

I have no good explanation for the conjecture, but I'm confident that it should be true.

QUESTION: Is the above conjecture true? Are there any supporting heuristic arguments?

Your further check of the conjecture is also welcome!

  • 1
    $\begingroup$ Did you check if oeis.org/A002618 contains relevant information? $\endgroup$
    – Dirk
    May 21, 2019 at 12:26
  • 1
    $\begingroup$ I have checked the sequence and its citations in OEIS, My conjecture is new. $\endgroup$ May 21, 2019 at 12:50

1 Answer 1


Let $\alpha\in \mathbb{Q}$ and write $\alpha=p_1^{s_1}\dots p_k^{s_k}$, where $p_i$'s are prime and $s_i\in\mathbb{Z}\setminus\{0\}$. We want to show that there exist $n$, $m$ such that all their prime factors are at most $p_k$ and $\varphi(m^2)/\varphi(n^2)=\alpha$. For that use induction on $p_k$. If $s_k$ is even then we can choose $m=p^{a}m_0, n=p^bn_0$ where $a,b$ are positive integers satisfying $a-b=s_k/2$ and $(m_0, n_0)$ solves $\varphi(m_0^2)/\varphi(n_0^2)=\alpha/p_k^{s_k}$ which exists by induction hypothesis. If $s_k$ is odd and positive we take $m=m_0p^{(s_k+1)/2}, n=n_0$, where $(m_0, n_0)$ solves $\varphi(m_0^2)/\varphi(n_0^2)=(\alpha/p_k^{s_k})/(p_k-1)$. Similarly for $s_k$ odd and negative.

  • $\begingroup$ Great! It is a nice proof. Thank you! $\endgroup$ May 21, 2019 at 13:30

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