Solving $\phi(n)=84$

Look at 7. Two possibilities:

• $7|(p-1)$ with $p|n$: hence $p = 14k+1 \in \{ 29, 43\}$ (the others are too big).
• If $p=29$, $\phi(n) = 28\times 3 = (29-1)(3)$. 3 has to be some $(p-1)p^{a-1}$ but $p=4$ is not prime.
• If $p=43$, you have the solution $\phi(n) = 42\times 2\implies n = 43\times 3$ or $n = 43\times 2^2$.
• Otherwise: $7|n$. $\phi(n) = 6\times 7\times 2\implies n= 7^2\times 3$ or $n= 7^2\times 2^2$.

I have used that $\phi(n) = 2\implies n\in\{3,4\}$.