Possible to guess RSA public key $e$ using $n$ and $\phi$ We know that if $n$ and $\phi$ is known, then it is possible to find $p$ and $q$ (initial prime numbers). I have following the questions:


Assume the public key $e$ is not known, how difficult would it be to guess the public key using $n$, $\phi$, $p$ and $q$? 
Assuming we have $c_1$, $c_1 = m^e \bmod n$ is it possible to find out $m$ without knowing $e$ and $d$ ?  
From my point of view $ed = K*\phi(n) + 1$ it is difficult to guess $e$ and $d$ product. Will it fall into the factorization problem ? Is this the only equation with $n$ and $\phi$ that involves $e$ and $d$ ?
Can we use some algorithm for guessing $e$ and $d$ ?


I know $e$ needs to be published publicly but if $e$ is lost how can it be identified?
Assume the public key $e$ is not known, how difficult would it be to
  guess the public key using $n$, $\phi$, $p$ and $q$?


$e$ can be chosen completely arbitrary and independently of the modulus and related variables ($\phi,p,q$). You can't learn anything about $e$ (assuming $d$ isn't given) only with the modulus given.


  Assuming we have $c_1$, $c_1 = m^e \bmod n$ is it possible to find out
  $m$ without knowing $e$ and $d$ ?


To recover $m$ you need to know $d$ (by the RSA assumption), so there's no way to recover $m$ without $e$ or $d$.


  From my point of view $ed = K*\phi(n) + 1$ it is difficult to guess
  $e$ and $d$ product. Will it fall into the factorization problem ?


You can arbitrarily choose either $e$ or $d$. The other one is a direct result of the above equation. You can't learn the $K$ either (how would you?) so even the ability to factor won't help you. Even if you'd know $K$ factoring the product would still be quite infeasible as this would be of the same size as the modulus, for which factoring is considered infeasible.


  Is this the only equation with $n$ and $\phi$ that involves $e$ and $d$ ?


Yes, this is the only RSA related equation involving, or rather this is the only equation: $ed\equiv1\pmod\phi$.


  Can we use some algorithm for guessing $e$ and $d$ ?


If you're lucky, you can indeed. $e$ usually takes the values $3,17,65537$ and if those don't yield the correct messages, you can still try each possible (small) $e$ value until you hit a good one. If $e$ is larger than $\approx 2^{40}$ you'll have a hard time finding this value.