Eulers totient function divided by $n$, counting numbers in the set [1,m] that are coprime to n

Question

If we divide Euler's totient function $\omega(n)$ by $n$, we obtain a fraction. If we multiply this fraction by any natural number $m$ which gives us another natural number $p$, is it true that $p$ is the number of natural numbers in $[1,m]$ that are coprime to $n$?

Answer 1

Be careful ! $\dfrac{m*\omega(n)}{n}$ does not always yield an integer. For example, $\dfrac{2*\omega(3)}{3} = \dfrac{4}{3}$.

If you are only interested in the case when $\dfrac{m*\omega(n)}{n}$ is an integer, I should show you a theorem :

If $n = p_1^{\alpha{}_1}p_2^{\alpha{}_2}...p_k^{\alpha{}_k}$ is the decomposition of $n$ in product of prime numbers, then :

$$\dfrac{\omega(n)}{n} = \prod_{i=1}^{k} \left( 1- \dfrac{1}{p_i} \right) = \prod_{i=1}^{k} \dfrac{p_i-1}{p_i} $$

So the denominator of the function will always be equal to the product of all the prime numbers that exist in the decomposition of $n$, maybe except $2$, because since it is the only even prime number, it can be cancelled out with $p_i = 3$ because $p_i-1=2$.

This means that "$\dfrac{m*\omega(n)}{n}$ is an integer" implies that $m$ and $n$ shares all their prime divisors ($m$ can't contain $2$ if $n$ is divisible by $6$, but that's the only exception).

As an example that if $n$ is divisible by $2$ and $3$ (so by $6$), the result isn't true if $m$ if divisible by $2$ : with $n=6$ and $m=2$ : $\dfrac{2*\omega(6)}{6} = \dfrac{2}{3}$.

Now, let's handle the two possible cases where the value is an integer.

If $n$ isn't divisible by $6$, $m$ has the form $p_1^{\beta{}_1}p_2^{\beta{}_2}...p_k^{\beta{}_k}$ and $\dfrac{m*\omega(n)}{n} = \prod_{i=1}^{k} p_i^{\beta - 1}\left(p_i-1\right) = \omega{(m)}$, which is the number of integers inferior to $m$, coprime to $m$, and so to $n$ because $n$ and $m$ share the same prime divisors, which was the result you expected.

Finally, if $n$ is divisible by $2$ and $3$ (so by $6$) and $m$ isn't divisible by $2$, $m$ and $n$ must share the same prime divisors, $2$ excepted.

$\dfrac{m*\omega(n)}{n} = \left(1-\frac{1}{2}\right)\prod_{i=1}^{k} p_i^{\beta - 1}\left(p_i-1\right) = \dfrac{1}{2}*\omega{(m)}$, which is half of the number of integers inferior to $m$ and coprime to $m$ hence to $\dfrac{n}{2^\alpha}$, so the number of integers inferior to $m$ and comprime to $n$.

As a conclusion : if $\dfrac{m*\omega(n)}{n}$ is an integer, then it is equal to the number of natural numbers in $[1,m]$ that are coprime to $n$.