A question about $\frac{\phi(n)}{n}$

Question

is there a way to find $n\in \mathbb{N}$ such that $\frac{\phi(n)}{n}\geq \frac{3}{4}$, where $n=p_{1}... p_{m}$?

Thanks.

Answer 1

Take $p$ to be any prime greater than or equal to than $5$. Then for $n=p^k$ (where $k\in\mathbb{N}$) we have

$$\frac{\phi(n)}{n}=\frac{\phi(p^k)}{p^k}=\frac{p^k-p^{k-1}}{p^k}=1-\frac{1}{p}\geq 1-\frac{1}{5}=\frac{4}{5}>\frac{3}{4}$$

Answer 2

Another important point to look at is that if $n$ is divisible by $2$ , then there is no solution

$$\text{ Let } n = 2.p_1.p_2.p_3......p_k$$

then

$$\phi(n) = n.\left(\frac12 .\frac{p_1-1}{p_1}\frac{p_2-1}{p_2}....\frac{p_k-1}{p_k}\right)$$

Let us consider only the second smallest factor for simplicity.

Then

$$\frac{\phi(n)}{n} = \left(\frac12 .\frac{p_1-1}{p_1}\right) \ge \frac34$$

$$ \implies 2(p_1-1) \ge 3p_1 \implies -2 \ge p_1 \space \space \space \space \text{ ( Which is a contradiction )}$$

Since The rest of the terms are $\lt 1$ , therefore $\phi(n)$ would never reach $\frac34.$

Similarly it can be show that $n$ is not a multiple of $3$.

This reduces are limit to only odd composite numbers , excluding the multiples of $3$ . $(\text{ All prime number satisfy this relation as explained in other answers.})$