I was reading this book on analytic number theory by Tom M. Apostol, and I came across this problem that asks for all integers that satisfy the following equality:
$$ \varphi(n) = n/2$$
where $\varphi$ is the the Euler totient function.
This was my first attempt:
$$\varphi(n) =( N * \mu )(n)$$
$$\implies u * (N * \mu) = \varphi * u = N$$
But the numbers we are looking for should fulfill $\varphi(n) = n/2$, therefore the following proposition should be true for these numbers.
$$N = \frac{1}{2}N*u $$
$$\implies N(n) = (\frac{1}{2}N*u)(n)$$
$$ \implies n = \sum_{d|n}\frac{1}{2}d$$
$$ \implies n = \sum_{d|n,d\neq n}d$$
where $*$ is the Dirichlet multiplication, $\mu$ is the mobius function of order 1, $N(n) = n, u(n) = 1 ,\forall n \in \mathbb{Z}$
which means the solution is all perfect numbers, but obviously one can disprove it by verification.
Somehow I managed to find the correct solution using another way. But I couldn't find out what I missed on my first attempt. Can some one help me figure out what it is ? Thanks.
$n$ has to be even, because $\varphi(n)$ must be an integer. Once $n$ is even, naturally all $n/2$ even numbers below it are not co-prime with $n$, which gives $\varphi(n) \leq \frac{n}{2}$. To have equality, we must ensure that all the odd numbers below $n$ are also coprime with it. Thus, $n$ must be a power of $2$.