For which Natural $n\ge2: \phi(n)=n/2$ For which Natural $n\ge2$ does this occur with?: $\phi(n)=n/2$
 A: We need $n$ even so that $n/2$ is an integer. So write $n=2^\beta p_1^{a_1}\cdots p_k^{a_k}$ where $\beta\geq 1$, and the $p_i$ are odd primes.
By multiplicativity of the totient function, $\phi(n)=n/2$ means
$$
2^{\beta-1}p_1^{a_1-1}\cdots p_k^{a_k-1}(2-1)(p_1-1)\cdots(p_k-1)=2^{\beta-1}p_1^{a_1}\cdots p_k^{a_k}.
$$
Rearranging, we see that this is equivalent to
$$
(p_1-1)\cdots (p_k-1)=p_1\cdots p_k.
$$
Since the $p_i$ are all odd primes, the LHS is even, while the RHS is odd. So this is impossible if $k\geq 1$. (This could also be concluded by the obvious fact that the LHS is strictly less than the RHS.) So necessarily $n=2^\beta$ for some $\beta$. It should be clear that all numbers of this form satisfy the property $\phi(n)=n/2$.
A: Hint: $n$ is even, or $n/2$ wouldn't be an integer. Hence $n=2^km$ with $m$ odd and $k\ge1$. You have $\phi(2^km)=2^{k-1}\phi(m)$ which must equal $n/2$.
A: $$n=\prod_{k=1}^rp_k^{a_k}\;\;,\;\;p_k\,\,\text{primes}\,\,,\,\,0<a_k\in\Bbb N\Longrightarrow$$
$$\phi(n)=n\prod_{k=1}^r\left(1-\frac{1}{p_k}\right)$$
and then
$$\frac{n}{2}=\phi(n)\Longleftrightarrow 2\prod_{k=1}^r\left(1-\frac{1}{p_k}\right)=1\Longleftrightarrow\ldots$$
Can you take it from here?
A: Suppose $n=2^sd$, where $d$ is odd. If $\phi(n)=n/2$, then $2|n$, forcing $s$ to be greater than zero always. Thus $\phi(n)= \phi(2^s)\phi(d)$.This can happen only if $d=1$, since $\phi(d)\le d-1$ for $d>1$. Therefore, $\phi(n)=n/2$ iff $n=2^s$ for some $s>0$.
