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This feels like homework, and so I won't give you the answer; I'll give you hints: If F(x) = x^e mod n, and (n, e) pair is public, will the first 1024 bits of both plaintext and ciphertext be enough for Eve to read the entire message? If Eve knows the first plaintext $M_0$, and the ciphertext $M_0 \oplus (r^e \bmod n)$, how can Eve recover $r^e \bmod n ... 3 It is secure against private key exposure but not against replay attacks by Eve. A three-way protocol avoids this, and doesn't need to use timestamps. The description below is from Delfs and Knebl's book Introduction to Cryptography. Each user, say Alice, has a key pair$(e_A, d_A)$for encryption and another key pair$(s_A, v_A)$for digital ... 1 As a add on to the above answer, it would be insecure to use$g^{x+y}$as key , because both$g^x$and$g^y$will be transmitted publicly and by simply eavesdropping one can easily find the required key i.e$g^{x+y}$. 1 Short answer: No, it is not vulnerable to man-in-the-middle attacks, assuming that Alice and Bob each have the right signature verification key of the other party. Yet, the man-in-the-middle attack could have taken place at the moment of exchanging the signature verification key. So if$sig_{X}$is party X's signature key, the attack on the exchange itself ... 1 Here is how it can be Vulnerable. Alice:$x$Bob$y$Eve$z$Alice$\rightarrowg^x\rightarrow$Eve->$g^z$->Bob Bob$\rightarrowg^y$$\rightarrowEve\rightarrow g^z$$\rightarrow$Alice What Alice thinks key is$g^{(xz)}$what Bob thinks the key is$g^{(zy)}$Eve can compute both of these values$(g^x)^z$and$(g^y)^z\$ This is why we need ...

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On the server, the certificate is stored in a certificate store, that includes a link to the private key. That link really is the name of a CSP (Cryptographic Service Provider) and the name of a container in that CSP. CSP relate to CryptoAPI, the old cryptographic API that is unfortunately hostile to hash functions other than MD5 and SHA-1. Chances are that ...

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I want to answer the third question, as all other questions are brilliantly answered. In TLS handshake as in diagram: In DH/ECDH, the "ServerKeyExchange" message will be empty since the value g^x is already present in certificate. This can save server a lot of computation. This will be a advantage in servers with heavy load and save time in CPU ...

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