Find all n such that $\phi(n) = n/2$

Question

My idea for the solution is something like this:

Since $2 | n$, $n = 2^a p_1^{e1} p_2^{e2} \cdots p_t^{et}$ where $a \geq 1$.

Then, $n/2 = \phi(2^a) \phi(p_1^{e1}) \phi(p_2^{e2}) \cdots \phi(p_t^{et})$.

Looking at the case that $t = 0$,

$n/2 = \phi(2^a) = 2^{a-1}$, and therefore $n = 2^a$.

Otherwise, $n = 2^a \phi(p_1^{e1}) \phi(p_2^{e2}) \cdots \phi(p_t^{et}) = n = 2^a p_1^{e1} p_2^{e2} \cdots p_t^{et}$. Since $\phi(n) < n$, this is a contradiction, and therefore, the only $n$'s that apply are the powers of 2.

Is this right? Also, we were asked to find $n$ such that $\phi(n) = n/3$. How would I go about that one?

Answer 1

You know that $$\frac{\varphi(n)}n =\prod_{p|n, p\text{ prime}} \left[1-\frac 1p\right] $$

So for every prime $p_0$ $$\prod_{p|n, p\text{ prime}}\left[1-\frac 1p\right]=\frac 1p_0$$

implies, using unicity of decomposition in irreductible fraction: $$\{p|n, p\text{ prime}\}=\{p_0\}$$

Answer 2

Hint:

Use the fact that if $\;n=p_1^{a_1}\cdot p_k^{a_k}\;,\;\;p_i\;$ primes, $\;a_i\in\Bbb N\;$ , then

$$\phi(n)=n\prod_{i=1}^k\left(1-\frac1{p_i}\right)$$

so

$$\frac n2=n\prod_{i=1}^k\left(1-\frac1{p_i}\right)\iff 2\prod_{i=1}^k\left(1-\frac1{p_i}\right)=1\iff\ldots$$

Take it from here...