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Honestly I'm really confused about padding in RSA scheme. I don't study computer science, and I'm a beginner learning cryptography (self-taught). Can anyone give me suggestions?

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  • $\begingroup$ Padding schemes in RSA remove otherwise exploitable mathematical structure in ciphertext and signature. That should not be too confusing. $\endgroup$ – DannyNiu Mar 19 at 9:26
  • $\begingroup$ @DannyNiu a source to learn more detail ? $\endgroup$ – Onta Ss Mar 19 at 9:29
  • $\begingroup$ Have you tried to read David Pointcheval's How to Encrypt Properly with RSA ? $\endgroup$ – fgrieu Mar 19 at 9:31
  • $\begingroup$ @fgrieu nope.but i have read about RSA primitive $\endgroup$ – Onta Ss Mar 19 at 9:37
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    $\begingroup$ A note about stackexchange in general: "Can anyone give me suggestions?" is not a good question format (unless you're on the site for software recommendations). We require questions to be narrow in scope so that they can be objectively answered, rather than discussed. A better format for your question would be to ask things like "Why is the RSA-OAEP padding scheme used", "Why/how does the RSA-OEAP padding scheme work", etc. You can use the "edit" link below the question to make changes. On another note: If someone links to a resource, consider reading it. $\endgroup$ – Ella Rose Mar 20 at 14:35
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Direct use of textbook RSA for encryption is very insecure. For example, a guess of the plaintext can be checked: if what's encrypted is a name on the class roll, enciphering all names on the class roll until getting the ciphertext breaks confidentiality. There are several other attacks.

That motivates RSA-OAEP. It turns textbook RSA, a hash, and a source of random bits, into a demonstrably secure encryption scheme. Encryption of message $M$ first transforms $M$ into padded message $X$ by mixing it with random bits and applying reversible transformations with the hash; then enciphers $X$ into $X^e\bmod N$, that is per textbook RSA encryption. Decryption starts by textbook RSA decryption, yielding $X$; then the padding is undone, yielding $M$ or an error indication.

I essence, the distribution of $X$ is random enough that textbook RSA becomes secure. A proof of that can be made for the strongest definition of security of a cipher, IND-CCA2, and under some hypothesis:

  • the RSA problem of finding a random $X$ given $X^e\bmod N$ is hard;
  • the hash is undistinguishable from a public function selected at random;
  • the source of random bits is indistinguishable from one producing uniformly random and independent bits;
  • a decryption device given a cryptogram only outputs the correct plaintext or an error indication (it does not leak any other information).

A (highly technical) reference proof is in Eiichiro Fujisaki, Tatsuaki Okamoto, David Pointcheval, and Jacques Stern's RSA-OAEP Is Secure under the RSA Assumption, in Journal of Cryptology, 2004. A more gentle text is David Pointcheval's How to Encrypt Properly with RSA?, originally in Cryptobytes 5n1, 2002.

The practical form of RSA-OAEP is named RSAES-OAEP and defined in PKCS#1 v2.2.

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