# A curious conjecture: $\{\varphi(m^2)/\varphi(n^2):\ m,n=1,2,3,\ldots\}=\{r>0:\ r\in\mathbb Q\}$

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!

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

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.

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