RSA - are we able to encrypt into any possible ciphertext in $Z_n$ given a specific key?

Given a public key $$(e,n)$$, if we are free to choose the plaintext does that mean we are able to generate any possible ciphertext in $$Z_n$$ ?

Because if we were given $$e$$ we can say that the decryption key is $$e^{-1} = k \mod n$$ and then choose the ciphertext we wish to generate, say $$c$$, perform $$c^k$$ and that shows that any ciphertext is "reachable" via encryption given a specific key and some plaintext?

Is that correct?

Edit: The question is theoretical so we assume we know both the public and the private key.

• Possible duplicate of Does RSA work for any message M? – kelalaka Dec 18 '18 at 19:37
• @kelalaka Not exaclty, what i mean is to use the key as exponent and not the phi(n) function. – caffein Dec 18 '18 at 19:40
• In my case the message does not have to be relatively prime to anything, it is just a message so it's not so related. – caffein Dec 18 '18 at 19:47
• A decryption key is $e^{-1} \pmod {\varphi(n)}$ and not $e^{-1} \pmod n$ – gammatester Dec 18 '18 at 19:52

Given a public key $$(e,n)$$, if we are free to choose the plaintext does that mean we are able to generate any possible ciphertext in $$Z_n$$ ?
Because if we were given $$e$$ we can say that the decryption key is $$e^{-1} = k \mod n$$
Not quite; the decryption key (which is traditionally named $$d$$; I'll leave it $$k$$ using your terminology) is related to the encryption key as $$e \cdot k \equiv 1 \pmod{ \text{lcm}(p-1, q-1) }$$, where $$p, q$$ are the prime factors of $$n$$, or in other words $$e^{-1} = k \mod \text{lcm}(p-1, q-1)$$. Other than that detail, you're correct.
and then choose the ciphertext we wish to generate, say $$c$$, perform $$c^k \bmod n$$ and that shows that any ciphertext is "reachable" via encryption given a specific key and some plaintext?