I need to prove that for each $n \in\mathbb N$ odd,
$$\phi(n) \ne 2^{32}.$$
What I tried:
I assumed that $\phi(n)$ is indeed a power of 2 then, because of this assumption I know that $ n = p_1 \cdot p_2 \cdot .\ldots \cdot p_l $ where all $p_i$ are different.
Here I'm stuck..
Second question is to prove that
$$|n \in \mathbb N : \phi(n) = 2^{32} | \ge 4.$$
Here I really don't know where to start.
Any help will be appriciated.
Assume $n$ odd and $\phi(n)=2^{32}$. If $p^2\mid n$ for an (necessarily odd) prime $p$ then $p\mid \phi(n)$, hence $n=p_1p_2\cdots p_l$ with $3\le p_1<p_2<\ldots<p_l$. Also, $\phi(p_i)=p_i-1$ must be a power of $2$, so the $p_i$ are among the Fermat primes $2^1+1,2^2+1,2^4+1,2^8+1,2^{16}+1$ (and possibly some $2^k+1$ with $k> 32$). There is no way to obtain $2^{32}$ as a product of such $p_i-1$ (or equivalently: to obtain $32$ as sum of distinct numbers among $1,2,4,8,16$, and possibly summands $>32$)
For the second part, note that $\phi(2^kn)=2^{k-1}\phi(n)$ if $n$ is odd. As each number $r\in\{0,\ldots,31\}$ can be written as sum of distinct numbers among $1,2,4,8,16$ (think of binary representation), we can find an odd number $n$ as product of distinct Fermat primes such that $\phi(n)=2^r$. Then $\phi(2^{33-r}n)=2^{32}$. Consequently $$\left|\{\,n\in\mathbb N:\phi(n)=2^{32}\,\}\right|\ge32. $$