# Conditions for computing $n$ given $\varphi\left(n\right)$

For any $n=p_{1}^{r_{1}}\cdot\ldots\cdot p_{k}^{r_{k}}$ I know that $$\varphi\left(n\right)=\prod_{i=1}^{k}p_{i}^{r_{i}}\left(1-\frac{1}{p_{i}}\right)$$ and from that we can get that $$n=\frac{\varphi\left(n\right)}{\prod\left(p_{i}-1\right)}\prod p_{i}$$ Therefore I was looking for primes $p_{i}$ such that $p_{i}-1\mid\varphi\left(n\right)$ but it seems to be not enough. As written here we also need to make sure that all primes in $\frac{\varphi\left(n\right)}{\prod\left(p_{i}-1\right)}$ are also in $\prod p_{i}$ but I don't understand why. Any explanation please?

EDIT

What I was looking for specifically are some solutions to $\varphi\left(n\right)=2^{32}$ by using the first Fermat's numbers $F_n=2^{2^n}+1$ as primes and apparently I had to add also $2$ as another prime.

For example: $$n_{0}=\frac{2^{32}}{F_{0}-1}\cdot F_{0}=\frac{2^{32}}{2^{2^{0}}}\left(2^{2^{0}}+1\right)=3\cdot2^{31}\notin\varphi^{-1}\left(2^{32}\right)$$ But $$n'_{0}=\frac{2^{32}}{\left(2-1\right)\left(F_{0}-1\right)}\cdot2\cdot F_{0}=3\cdot2^{32}\in\varphi^{-1}\left(2^{32}\right)$$

Not sure if I understand you completely. Write $n=p_1^{a_1}\cdots p_k^{a_k}$. In other words, $n$ fixes the primes $p_i$. Since $\prod_{i=1}^k p_i$ divides $n$, it follows from your second equation that $\phi(n)/\prod_{i=1}^n(p_i-1)$ must divide $n$, so that its prime factors must be a subset of $\{p_1,\cdots,p_n\}$.