很少关注如何处理这个问题：
Xavier and Yvonne are in love. They both set up their own RSA keys:
Xavier’s public key is (eX , nX ) = (887, 15833), and Yvonne’s public key
is (eY , nY ) = (977, 13019). For each of the following, do not factor nX nor
nY , show the set up of the calculations and the results. You should use a
computer to perform the calculations.
(a) Yvonne wants to send Xavier a private message of love, which is
M = 3141. What is the ciphertext that Yvonne needs to send to Xavier?
(b) In return, Yvonne received three mysterious messages: C1 = 10889,
C2 = 2622, C3 = 4061. All three senders claim to be Xavier, and all
claim to have sent the message Ci ≡ MdX (mod nX) where M = 3141 and dX
is Xavier’s private key. However, only Xavier himself knows the actual
value of dX , and the other two are imposters trying to steal Yvonne away
from him. Help Yvonne determine which message is actually from Xavier,
and explain why your method works.
任何提示都会非常感谢！
答案 0 :(得分：2)
a）为了发送RSA加密的消息，只有私钥持有者可以对其进行解密，必须使用接收者的公钥对其进行加密。在这种情况下，收件人是Xavier，因此邮件使用他的公钥加密：（e X ，n X ）=（887,15833）：
Message: M = 3141 Ciphertext: C = MeX mod nX C = 3141887 mod 15833 C = 2054
b）这实际上是使用Xavier私钥签名的消息的签名验证，该私钥需要使用签名者的公钥。当使用Xavier的公钥解密时，有必要找到三条消息中的哪一条导致发送的消息（3141）：
Ciphertext 1: C1 = 10889 Message 1: M1 = C1eX mod nX M1 = 10889887 mod 15833 M1 = 6555 (mismatch)
Ciphertext 2: C2 = 2622 Message 2: M2 = C2eX mod nX M2 = 2622887 mod 15833 M2 = 1466 (mismatch)
Ciphertext 3: C3 = 2622 Message 3: M3 = C3eX mod nX M3 = 4061887 mod 15833 M3 = 3141 (match!)
当使用Xavier的公钥解密时，只有C 3 匹配该消息，因此唯一的真实消息也是如此。
注意：我使用WolframAlpha执行上面的模幂运算，但是手动使用重复乘法然后减少模数n很容易（虽然相当费时）。< / p>