RSA Decryption: Doubts: knowing p or $\phi(n)$ can lead to find private exponent? I have some doubts about RSA (the public key $n$ and $e$ is public):

Note $n = pq$ and $\phi (n) = (p-1)(q-1)$


1) Say if some hacker knew $p$, why would it be easy for them to find $q, \phi (n), d$?


$d$ is a prime larger than $p, q$ so that $1 = k\cdot \phi(n) + d\cdot e$


2) Now, say if a hacker knew $\phi (n)$, why would it be easy to find $p,q,d$?
1) Say if some hacker knew $p$, why would it be easy for them to find $q, \phi (n), d$?
  


Since $n = p \cdot q $ this implies $q$ is also known by simple division $p = n/q$.

Then $\phi(n) = (p-1)(q-1)$ is known 

By using $e \cdot d \equiv 1 \bmod \phi(n)$ then $d$ is known. This can be found by the extended Euclidean algorithm.


  $d$ is a prime larger than $p, q$ so that $1 = k\cdot \phi(n) + d\cdot e$


$d$ doesn't need to be prime. It can be larger or smaller than the $p$ and $q$ depend on the value of $e,p,q$.


  
  2) Now, say if a hacker knew $\phi (n)$, why would it be easy to find $p,q,d$?
  


$d$ is as above.

1. Method with the knowledge of $\phi (n)$

$\phi(n) = (p-1)(q-1) = (n + 1)- (p+q) $ implies $$p+q = (n +1) - \phi(n)$$ and

$$q = (n+1) - \phi(n) - p $$ since we have $$ n = pq $$ substitue

$$n = p \left ( n + 1 - \phi{(n)} - p \right ) = -p^2 + (n + 1 - \phi{(n)})p$$
 $$p^2 - (n + 1 - \phi{(n)})p + n = 0$$

Now, we got a quadratic equation that can be solved by the quadratic polynomial root-finding formula;$$p = \frac{-b \pm \sqrt{b^2 -4ac}}{2a}.$$ Using the formula one gets;

$$p = \frac{(n + 1 - \phi{(n)}) \pm \sqrt{(n + 1 - \phi{(n)})^2 - 4n}}{2}$$

Therefore the knowledge of the $\phi(n)$ enables factoring the $n$ in constant time, $\mathcal{O}(1)$

2. Method with the knowledge of $\phi (n)$


$(p+q)$ can be obtained from $n$ and $\phi(n)$ as $$\phi(n) = (p-1)(q-1) = n - (p+q) +1$$ 
$(p-q)$ can be obtained from $(p+q)^2-4n$, since $(p-q)$ is the square root of it.

Then one can find $q$ as $$q = \frac{(p+q)-(p-q)}{2}.$$