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8

There's no real difference between $p$ and $q$ in RSA. It looks like OpenSSL just has the agreement "$p$ has to be bigger than $q$" for conveniences. One of the numbers has to be bigger than the other (otherwise they would be the same number, and $p = q$ is very bad in RSA). Just use two examples: $p = 13$ and $q = 11$. $p$ is bigger than $q$, all right. ...


8

Is it true the longer the key length is the more secure the encryption? No. Key length does put a lower bound on security, because it determines the complexity of brute force iteration of the key space or factoring, discrete log, etc. for some asymmetric algorithms. However, once you have a long enough key to make brute force attacks impossible, there ...


6

I don't think NSA can break the underlying encryption primitives. What they may do is record the whole SSL traffic and decrypt the whole traffic using server private key. So you can use introduce perfect forward security as the blog article “SSL/TLS & Perfect Forward Secrecy” suggests. In addition man in the middle can be performed with fake ...


6

For what it's worth, the OpenSSL developers have committed changes that improve this. I assume they will be in OpenSSL 1.0.2, but I don't know for sure. In any case, if you clone the git repo and compile the OpenSSL_1_0_2-stable branch (or master, I suppose), s_client will display the curve name: $ OPENSSL_CONF=apps/openssl.cnf apps/openssl s_client -CApath ...


6

Short answer: There would be nothing (that isn't already wrong with TLS) necessarily wrong with a CTR + HMAC cipher suite, but the technical merits are only one factor in a technical feature getting to RFC status in the TLS working group. Without being discourteous to the TLS Working Group (WG) participants or process, other reasons can be: political ...


6

To make it easier for humans to read.


6

I don't know about computing things in parallel, so I will ignore that part of the question. First, please note that the encryption algorithm is rarely the the weak point of the security. It is far more likely that you will have problems with the implementation, some spyware installed on your computer, a weak password (If you use qwerty as your password, ...


5

RFC 6176 lists four reasons why SSL 2.0 must not be used, in its section 2: Message authentication uses MD5 [MD5]. Most security-aware users have already moved away from any use of MD5 [RFC6151]. Handshake messages are not protected. This permits a man-in-the- middle to trick the client into picking a weaker cipher suite than it would ...


5

You can make OpenSSL print out the handshake messages with the -msg parameter: openssl s_client -msg -connect myserver.net:443 Then look for the ServerKeyExchange message. Here is an example: <<< TLS 1.2 Handshake [length 014d], ServerKeyExchange 0c 00 01 49 03 00 17 41 04 6b d8 6e 14 1c 9b 12 4d 58 29 20 e8 e2 1a 24 0d da 8f 38 1a 5d 85 ...


5

In the example you linked, the current time (specifically, a value representing the number of seconds elapsed since Jan 1, 1970 UTC) is used as the seed. If an attacker knows which year you generated your key, then that leaves only about 2^25 possible values for the seed --- and therefore only about 2^25 possible values for your key. At this point, he can ...


5

First, I am assuming, per https://security.stackexchange.com/questions/29172/what-changed-between-tls-and-dtls, that the client handshake protocol in DTLS is not different from that in TLS over TCP. This seems a safe bet since the client/server encrypted handshake protocol in OpenVPN's UDP implementation is the same as in standard TLS over TCP. I am not ...


4

In a purist cryptographic sense, there are many vulnerabilities in this cipher suite that can be (theoretically and practically) exploited. There are much stronger versions of SSL/TLS, and much stronger cipher suites that could be used. In a practical sense, it's not the end of the world - there are far worse cipher suites (e.g. those using intentionally ...


4

There are two papers on conventional differential cryptanalysis of SEED. The last one penetrates only half of the cipher. Even though there are few third-party cryptanalysis papers, there is no indication that the cipher is weak. Fault attacks are quite irrelevant in the SSL setting. I would be more concerned with BEAST-like attacks, as SEED is a ...


4

A fault injection attack is based on the fact that you have a healthy black box on which you can do queries, but you can mess with the black box, for example flipping random bits. In real life this could for example be a RFID chip which can be messed with using strong electronic fields. Attacks like these are generally: Very sophisticated in theory and ...


4

The cornerstone of the handshake security is that the Finished messages, sent under the protection of the newly exchanged key (for encryption and MAC), contain hash values computed over all the handshake messages exchanged so far, including the list of cipher suites and all other parameters. As long as client and server don't negotiate the use of a cipher ...


4

To answer this question, we must have a look at how TLS/SSL works. I guess you know that the aim of TLS/SSL is to authenticate communicating parties before setting up an encrypted connection through which application data will flow. And as you may already know, an SSL handshake/session will use asymmetric crypto for authentication and session setup and ...


4

If you use public key crypto in the correct way, then every user has it's own private key and corresponding public key (included in the certificate) and the keys of users are not related. Consequently, compromising the private key of one user does not affect any of the other users. So in the case of compromise of the private key of one user the remaining ...


4

Yes, you're misinterpretting the PRF. It's not just a hash function (and when you hit the end of the function function, start back at the beginning). Instead, if is a function that generates a rather long (actually, infinite) output; we use the first $N$ bits of that output to populate the various key values. See section 5 of RFC5246; we have: TLS's ...


4

No, DHE is secure and allows to share a common secret between two parties over an insecure channel. But you cannot know, if the one you share the secret with is the one you want (DHE is vulnerable to man in the middle attacks). So DHE-RSA uses DHE to share a common secret and signs the communication with RSA to make sure, that both persons communicate with ...


4

Obviously, as long as the AES-128 (TLS) channel is keyed independently from the AES-256 (file encryption), the TLS encryption cannot damage the file encryption. Let us consider what would be true if it could. If that somehow did weaken the file encryption, this would allow this attack on an encrypted file – he could take that encrypted file, and send it ...


4

Well, it is certainly possible to generate an RSA public/private key pair like what you're asking about -- I don't know what the OpenSSL API allows you can do, but if you don't restrict yourself to that, well, it is certainly possible to craft such a keypair. You'll come up with a public key with an enormous exponent (and I wouldn't be shocked if not ...


4

From RFC 5246, section 6.2.3.3: AEAD Ciphers: AEAD ciphers take as input a single key, a nonce, a plaintext, and "additional data" to be included in the authentication check, as described in Section 2.1 of [AEAD]. The key is either the client_write_key or the server_write_key. No MAC key is used. However in RFC 5246, section 5: HMAC ...


4

The main thing that makes HMAC secure in typical use even with MD5 is that it is used with a secret key. That means only preimage attacks are really relevant, since finding a collision is always an online attack if you don't know the key. With known attacks the preimage resistance of both MD5 and SHA-1 is > 100 bits. Additionally, HMAC may be secure even ...


4

In a better world, TLS_FALLBACK_SCSV would not be necessary: SSL has been supporting downgrade-proof version negotiation since at least SSL 3.0, so a man in the middle should never be able to limit a connection to a version older than the mutually supported maximum. However, out there are some broken servers that don't really support that kind of version ...


3

SSL does not generate a private key. What is generated is shared secret key between the client and server.This shared secret key is only valid till the session lasts. So, this shared secret key resides in the RAM. It is not saved in to the hard disk.


3

Something tells me you didn't put much research effort into this one. Let me quote part of a distantly related Crypto.SE question by @D.W.: SSL 3 included special mitigations to prevent protocol downgrade attacks where a man-in-the-middle downgraded two SSL3-capable endpoints so they end up using SSL 2. This protection in SSL 3 was critical, because ...


3

You can use RSA to exchange the keys, But the problem is, it does not provide perfect forward security. This is what may happen if you use RSA for key exchange. If someone is able to store your SSL session and if he can get the server private key, He can get the secret key generated by the SSL session. So he can decrypt the entire SSL session. So we use ...


3

The answer from owlstead and its comments covers WEP part quite nicely. This answer concentrates on CTR and OFB. Strictly speaking, CBC, CFB, CTR, and OFB modes always use IV or counter. I'm assuming that the question was more like is it possible to use CTR or OFB mode securely without transmitting IV. I.e. for instance, start at all zeroes IV/counter and ...


3

Good brief topic on security of this protocol for each version can surprisingly easy be found in Wikipedia. Also, please take a look on blog posts about SSL-stripping attack, also BEAST and SSL Recognition Attacks.


3

Feasible? Sure, there are lattice algorithms that are competitive in performance with RSA. However, there are drawbacks, like: They've been studied less than RSA or ECC, especially the individual algorithms. The most well studied system, NTRU, is patented. No generic proof that I know of that there isn't a quantum algorithm to solve them. The first one ...



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