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0

Good question! I would look at https://saltyrtc.org, it allows you to do signaling without worrying about trust. Another thing you could do is supply your own certificates to the RTCPeerConnection (and not use self-signed). I would have to think about it a little more, but if you only you and the other peer had the private key I don't think an attacker ...


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One idea I had is to encrypt the signals with Ed25519. Would this solve the MITM issue? First, Ed25519 is no encryption, it is a signature algorithm. And no, switching one signature algorithm with another does not magically protect against MITM. The relevant question in the context of MITM attacks is instead who is doing the signature and if this party can ...


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echo -n 'N.S.T.F.P.Fo.S.T.E.O' | xxd -c32 produces: 00000000: 4e2e 532e 542e 462e 502e 466f 2e53 2e54 2e45 2e4f N.S.T.F.P.Fo.S.T.E.O


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In the line Enter file... you have chosen an arbitrary filename public_key_ooker. That's why the privake key has been saved to public_key_ooker and public key to public_key_ooker.pub. The name alone doesn't make the private key a public key, but gets you confused. But you don't necessarily need a key, if you are using password authentication. If you'd ...


4

We need more information about exactly what do you mean in your question, whether you're asking about TLS 1.3 or older versions, from SSLv3 up to TLS 1.2, which worked a little differently, and what kind of thing you would like to do that is currently impossible. If you only intend to use primitives already specified for TLS and just want to mix and match ...


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I have created a visual explanation of the firefox algorithm, not very pretty but I hope it helps.


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This is very weak security on all fronts! The user's password is P4$$w0rdP4$$w0rd and it's encrypted using XOR encryption, with the key CdZ4MLMPgYtAE9gQ80gMtg==. This produces the ciphertext posted by the OP above, WeJcFMQ/8+8QJ/w0hHh+0g==. To verify: First, use xxd to get the underlying binary of the plaintext password: echo -n 'P4$$w0rdP4$$w0rd' | xxd ...


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They're probably referring to NISTP256 in the algorithm. The constants for this elliptical curve were created by the NSA, and there is some speculation that the NSA may have been specially chosen these constants to create a backdoor for themselves for this curve. See the links below for more info: https://crypto.stackexchange.com/questions/52983/why-is-...


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If the RNG is cryptographically strong, then you should not be able to do any better than the naive brute force: try each seed until you get matching random numbers. If this is a homework question, then I would guess that either they used a very small seed (like "3" or "15"), or the RNG is not cryptographic strength and with some time analyzing the linear ...


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