Cracking plain RSA without private key I was wondering whether plain RSA encryption can be cracked given:


The public key 
A plaintext of a known cipher text(s).
In other words the decrypted message from an encrypted one (but without knowing the private key)  

I.E.: you have managed to find the plaintext of an already encrypted message using the public key, however, you DO NOT know the private key.


With maybe some form of adaptive-chosen-ciphertext attack?

edit: maybe also take advantage of plain-RSA malleability?
It depends on what you mean by "cracking", and also what you mean by "RSA".

"Plain RSA", also sometimes described as "textbook RSA", is the lone modular exponentiation. By itself, it is very weak. For instance, if $n$ is the modulus and $e$ the public exponent, suppose that you want to "decrypt" a cipher text $c_1$ (i.e. you want to find $m_1$ such that $m_1^e = c_1 \pmod n$). You can choose a random integer $c_2$ modulo $n$, and compute $c_3 = c_1/c_2 \pmod n$; then, obtain the decrypted messages $m_2$ and $m_3$, corresponding to $c_2$ and $c_3$, respectively. In that situation, you know that $m_1 = m_2m_3 \pmod n$. This is a chosen ciphertext attack, in which you ask for two decryptions and obtain three plaintext messages, including the target message $m_1$, and yet the decryption oracle did not see any request ending in revealing $m_1$. That's "cracked" by any decent definition of "cracking".

That's why "plain RSA" is not RSA. RSA, the asymmetric encryption algorithm, is described by PKCS#1 and includes a padding operation which is essential to security. With appropriate padding, there is no known attack on RSA which would help in decrypting a given message, even in an adaptative chosen ciphertext context.