Known vs. Chosen (plaintext and ciphertext) and RSA I want to know whether I can break RSA (textbook) with known plaintext/ciphertext attacks.

Known plain-text

Let $c$ be the ciphertext and $m$ the plaintext.

$$c = m^a \mod n$$

Since all I have is the plain and ciphertext, all I could do here is guess the decryption exponent $a$ until I find one that leads turn $c$ into my known $m$.

Known ciphertext

In this case, I would have to factor $n$ (hard) in order to find $p$ and $q$.

Is the above correct? I would appreciate if someone could point the right direction.
I'm not exactly certain what answer you're looking for; I tried to cover all the obvious bases.

Actually, with RSA, we generally assume that the attacker knows the public key (the modulus $n$ and the public exponent $e$); with that, he can encrypt as much plaintext as he cares to (by selecting a value $M$ and computing $M^e \bmod n$). So, in that sense, the term 'known plain-text attack' doesn't apply, as we assume the attacker can do a chosen plain-text attack.

As for the variant "what if the attacker has some plaintext/ciphertext pairs, but doesn't have the public key; what can he do?  Actually, it turns out that if the public exponent isn't too large, the attacker is in luck; if the attacker takes two known plaintext/ciphertext pairs $P_1, C_1$ and $P_2, C_2$, he can compute $gcd( P_1^e - C_1, P_2^e - C_2)$ for plausible public exponents; if he guesses the correct one, that value will be $k\cdot n$ for some integer $k$ which is extremely likely to be small.  That gives him the public key.

As for guessing the decryption exponent, well, I suppose that the attacker could try, but he's not likely to have success.  It is known that knowledge of encryption and decryption exponents is equivalent to factoring; that is, if you know $e$ and $d$, you factor $n$ (and if you know how to factor $n$, you can compute $d$ given $e$). Hence, guessing the decryption exponent is hard; if it wasn't, the problem of factorization wouldn't be easier.  (And, if you're asking "why can't you step through all the possible possible private exponents", the answer is that $d$ is generally only a few bits smaller than $n$; if $n$ is 1024 bits, $d$ might be 1020 bits; that's rather too many possibilities to step through).

As for a literal "ciphertext-only attack", where an attacker is given some ciphertext without either the corresponding plaintext or the public key, well, on the face of it, that looks to be difficult; however we generally don't assume that attackers are hobbled in such extreme ways.