So if I use an RSA OAEP decryption implementation with the same: It seems to be specific to the Go RSA OAEP implementation. I haven't seen this random parameter (rng in my case) used in other programming languages. In my case this hash function is SHA256.Īdditionally the Go documentation says The random parameter is used as a source of entropy to ensure that encrypting the same message twice doesn't result in the same ciphertext. The Go documentation says Encryption and decryption of a given message must use the same hash function. ) as long as I can get the plaintext from the ciphertext. I am indifferent to the RSA OAEP implementation that will be used for decryption (Go, Java, Python, openssl.
Implementations change, new libraries, functions & new versions of programming language runtimes are used, etc.
Since different ciphertexts are generated for the same plaintext, is there a guarantee that I will be able to decrypt these RSA OAEP ciphertexts in the future ?. Sending data over the network and I should not treat them the same,īut for my personal use case I want to use them for P.S: I understand that RSA and AES have different use cases when it comes to I am more confident with using AES because if you supply the same symmetric key and the same initialization vector you will always get the same ciphertext. I read some similar questions on stack overflow that say that RSA OAEP adds padding to ensure that a different ciphertext is generated for the same plaintext for security purposes. I generated a public/private key pair with 4096 bits. Rsa.DecryptOAEP(sha256.New(), rng, rsaPrivatekey, ciphertext, nil) I am using RSA OAEP in Go (Golang): rsa.EncryptOAEP(sha256.New(), rng, rsaPublickey, plaintext, nil) My goal is to encrypt some strings and store them in a database.
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