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I am currently trying to understand how you find the unpadded short plaintext when you are using RSA. Please help with explaining the process and so I can understand this topic more.

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    $\begingroup$ Hint: try to encrypt all possible plaintext with the public key and compare? What is the public exponent? $\endgroup$ – kelalaka Apr 11 at 18:55
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    $\begingroup$ If the public exponent is small e.g. 3 it may be possible to recover the plaintext by taking the e.g. cube root of the ciphertext. $\endgroup$ – puzzlepalace Apr 12 at 6:39
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As kelalaka hinted, the issue with particularly short plaintext in unpadded RSA is that it is subject to brute force via chosen plaintext attack.

For each plaintext $m$ in the short space you suspect the enciphered plaintext to originate from calculate $c = m^e \text{ mod } n$ where $e$ is the public encryption exponent and $n$ is the public modulus. Then compare $c$ to the ciphertext for which you are trying to find the plaintext. If identical, the message $m$ is the same as the padded short plaintext in question. Because RSA is deterministic, identical plaintexts produce identical ciphertexts.

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