Math RSA私钥和公钥

对于如何解决这个问题，我几乎没有什么疑问：

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.

任何提示都很好，谢谢

a）为了发送RSA加密的消息，只有私钥持有者才能对其解密，必须使用收件人的公钥对其进行加密。在本例中，收件人是Xavier，因此使用他的公钥对消息进行加密：（eX，nX）=（88715833）：

信息：M=3141 密文：C=MeX mod nX C=3141887模块15833 C=2054 b） 这本质上是对使用Xavier私钥签名的消息的签名验证，这需要使用签名者的公钥。当使用Xavier的公钥解密时，有必要找出三条消息中的哪一条会导致发送的消息（3141）：

密文1:C1=10889 消息1:M1=C1eX模块nX M1=10889887模块15833 M1=6555（不匹配） 密文2:C2=2622 消息2:M2=C2eX模块nX M2=2622887 mod 15833 M2=1466（不匹配） 密文3:C3=2622 消息3:M3=C3eX模块nX M3=4061887模块15833 M3=3141（匹配！） 当使用Xavier的公钥解密时，只有C3与消息匹配，因此是唯一可信的消息

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注意：我过去常常执行上面的模幂运算，但是手动使用重复乘法然后使用模n的约化来执行很容易（虽然相当耗时）。

你能给我们展示一下你的尝试吗？这不是建设性的，人们不应该只是在没有解释的情况下发布他们的家庭作业。 Message: M = 3141 Ciphertext: C = M^e_X mod n_X C = 3141⁸⁸⁷ mod 15833 C = 2054 Ciphertext 1: C₁ = 10889 Message 1: M₁ = C₁^e_X mod n_X M₁ = 10889⁸⁸⁷ mod 15833 M₁ = 6555 (mismatch) Ciphertext 2: C₂ = 2622 Message 2: M₂ = C₂^e_X mod n_X M₂ = 2622⁸⁸⁷ mod 15833 M₂ = 1466 (mismatch) Ciphertext 3: C₃ = 2622 Message 3: M₃ = C₃^e_X mod n_X M₃ = 4061⁸⁸⁷ mod 15833 M₃ = 3141 (match!)