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# RSA public key and private key calculation: multiple answer

2018-04-09 11:54:43

Given N = 91

public key (11, 91)

Plaintext (88)

Ciphertext = 88 ^11 mod 91 = 30

to decrypt,

30 ^ 59 mod 91 = 88 (plaintext)

30 ^ 11 mod 91 = 88 (plaintext)

30 ^ 47 mod 91 = 88 (plaintext)

How can there be multiple suitable instance of private key such as 59/11/47?

Lastly, must public and private key always be a prime number?