RSA: Private key calculated from public key? The wiki article on RSA (https://en.wikipedia.org/wiki/RSA_(cryptosystem)) says that d is "the modular multiplicative inverse of e (mod φ(n))" , where d is the private key and e and n make up the public key. How come an interceptor cannot calculate the d in the exact same manor as the encrypter does, if the interceptor has knowledge of e and n?
When p and q are primes and ​ n = p$\cdot$q , ​ one will have

{p,q} ​ ​ ​ = ​ ​ ​ $\bigg\{\hspace{-0.06 in}n\hspace{-0.04 in}+\hspace{-0.05 in}1\hspace{-0.05 in}-\hspace{-0.03 in}\phi(n)\pm \sqrt{\left(n\hspace{-0.04 in}+\hspace{-0.05 in}1\hspace{-0.05 in}-\hspace{-0.03 in}\phi(n)\right)^{\hspace{.04 in}2}-(4\hspace{-0.04 in}\cdot \hspace{-0.04 in}n)}\hspace{-0.05 in}\bigg\}$ ​ ​ ,

so for semiprimes, computing $\phi$ is equivalent to factoring.

The above reasoning doesn't work for multi-prime RSA, but even there, there's no obvious way for the adversary to compute $\phi$ of the modulus more easily than just factoring the modulus.