Can I check, whether $n$ with $\frac{n}{\phi(n)}=r$ exists?

Question

Suppose, a rational number $r$ is given.

How can I check whether there is a positive integer $n$ with $$\frac{n}{\phi(n)}=r$$ where $\phi(n)$ denotes the totient function ?

In particular, I need the answer for $r=\frac{(s+2)\cdot s}{s+1}$ with positive integer $s$ to find all solutions of $$\frac{1}{x}-\frac{1}{y}=\frac{1}{\phi(x+y)}$$ with $\phi(x+y)|y$. In this case the even $s\ge 2$ can be ruled out because we have $$\frac{n}{\phi(n)}=\frac{p_1}{p_1-1}\cdots \frac{p_n}{p_n-1}$$ where $p_1,\cdots ,p_n$ are the prime divisors of $n$ ($n=1$ gives $\frac{n}{\phi(n)}=1$) because the numerator cannot be divisible by $4$. I conjecture that $s=1$ and $s=3$ are the only cases with a solution.

Answer 1

If a solution exists, there are $n$ primes such that $$\frac{p_1}{p_1-1} \cdots \frac{p_n}{p_n-1} = \frac{a}{b}.$$ For ease of argument, have the primes $p_1 < \cdots < p_n$ and $\gcd(a,b)=1$.

Now there is a largest prime $p$ that divides $a$. Because $p_i -1 < p_n$ for all $i$, we must have $p=p_n$. Hence,

$$\frac{p_1}{p_1-1} \cdots \frac{p_{n-1}}{p_{n-1}-1} = \frac{a}{b} \cdot \frac{p_n-1}{p_n} = \frac{a'}{b'}$$ where $a'$ and $b'$ are in reduced form. Now it turns out that $a'$ has a largest prime divisor less than $p_n$. So we can repeat this process at most $\pi(p)$ times and obtain a solution or achieve an inconsistency.

Answer 2

For odd $s>3$$$s(s+2)>4(s+1)$$But $$n>4\phi(n)$$only for even $n=210, 330, 390$ and their multiples.

For $n=210=2\cdot3\cdot5\cdot7$$$\phi(n)=(2-1)((3-1)(5-1)(7-1)=48$$and$$210=4\cdot48+18$$For multiples of $210$, e.g. 420, 630$$420=2^2\cdot3\cdot5\cdot7$$so that$$\phi(420)=\phi(2^2)\cdot2\cdot4\cdot6=96$$and$$420=4\cdot96+36$$Again$$630=2\cdot3^2\cdot5\cdot7$$making$$\phi(630)=\phi(3^2)\cdot4\cdot6=144$$and$$630=4\cdot144+54$$It can be shown, and is evident from R.D. Carmichael's "A table of values of m corresponding to given values of phi(m)", linked to in <http://oeis.org>[A000010-OEIS], that even $n$ which are multiples of $3\cdot5\cdot7$ are co-prime to fewer than one-fourth of the integers less than $n$. Likewise for even multiples of $3\cdot5\cdot11$ and $3\cdot5\cdot13$. But not for even $n$ in which the next greater prime after $3$ and $5$ is $17$ or greater. Thus$$510=2\cdot3\cdot5\cdot17$$and$$\phi(510)=2\cdot4\cdot16=128$$But$$510<4\cdot128=512$$

Thus if for some odd $s$ besides $1$ and $3$$$\frac{s(s+2)}{s+1}=\frac{n}{\phi(n)}$$even $n$ must belong to one of three arithmetic sequences$$210, 420, 630, 840, 1050...$$or$$330, 660, 990, 1320, 1650...$$or$$390, 780, 1170, 1560, 1950...$$But in the first sequence$$\frac{n}{\phi(n)}=\frac{210}{48}=\frac{420}{96}=\frac{630}{144}=\frac{840}{192}=\frac{1050}{240}=\frac{35}{8}\ne\frac{s(s+2)}{s+1}=\frac{35}{6}$$for $s=5$

In the second sequence$$\frac{n}{\phi(n)}=\frac{330}{80}=\frac{660}{160}=\frac{990}{240}...=\frac{33}{8}$$But $$33\ne s(s+2)$$ for any odd $s$.

In the third sequence$$\frac{n}{\phi(n)}=\frac{390}{96}=\frac{780}{192}=\frac{1170}{288}...=\frac{13\cdot15}{48}\ne \frac{s(s+2)}{s+1}=\frac{13\cdot15}{14}$$for $s=13$.

Thus it appears the conjecture is true, that$$\frac{s(s+2)}{s+1}=\frac{n}{\phi(n)}$$only for $s=1$ and $s=3$.