Euler's $\phi(n)$ function and solutions of $\phi(n) = 4294967296=2^{32}$

Question

Find all $n \in \mathbb{N}$ such that $\phi(n) = 4294967296=2^{32}$.

I managed to find that $16106127360$, $8589934592$, $10737418240$ and $12884901888$ are solutions for the equation, but I do not know how to find all of them.

Answer 1

$4294967296 = 2^{32}$.

Recall that $\phi(a b) = \phi(a) \phi(b)$ if $a, b$ are coprime; and that $\phi(p^n) = (p-1) p^{n-1}$.

Therefore if $\phi(p_1^{n_1} \dots p_m^{n_m}) = 2^{32}$ then $\phi(p_i^{n_i}) = 2^{k_i}$, some $k_i$; that is $(p_i-1) p_i^{n_i - 1}$.

That is, $p_i = 2$, or $p_i$ is one more than a power of $2$ and it appears only to the power $1$.

Also if $p_i = 2$ then we can't have $n_i > 32$.

Now it's a fairly horrible case-bash, I think, to find the 32 solutions. (I think that number of solutions is a coincidence.)