# Tag Info

10

The question's bytestring 2a 86 48 86 f7 0d 01 01 01 is the Value field of an ASN.1 BER/DER TLV with type 6, which is the Object IDentifier for an RSA key (the Type and Length just before are coded as 06 09, and won't be further discussed). In order to parse that Value bytestring, we first separate the bytes into blocks ending after each byte which ...

8

No, you can't; the reason you can't depends on the negotiated TLS ciphersuite: The original ciphersuites had the server send to the client the server's RSA public key; the client selects a random value ("premaster secret"), and encrypts that value with the server's public key; it sends that encrypted value to the server. Now, these public keys have the ...

7

Either could be implemented securely, but if you encrypt first and split afterwards, you can use standard tools and get everything right more easily. If you used the opposite order, you would have several pitfalls to deal with: With password-based encryption you would either have to derive the key many times (spending resources that would be better used on ...

6

An initialization vector is, in fact, always binary. It's just random bits. So, if you choose to encode those bits as a hexadecimal string for ease of storage or transportation, that is fine. However, since it is the binary that is the IV, you will need to decode it back from hexadecimal to a binary value before using it in the decryption process. As a ...

4

ECB is not secure even with per file keys, because if two blocks of the file are identical, this is visible in the ciphertext. The only * cases where ECB is secure is encrypting completely random data or encrypting a single block per key. You should pick something more secure if your can help it. If there is literally no other option than RC4 and AES ECB, ...

3

The very fastest elliptic curve algorithms can generate a key-pair in about 40k cycles on a modern CPU. A high end laptop has four cores running at about 3 GHz, so it can generate about 300k key-pairs per second, or a billion per hour. (The cost of SHA-256 is negligible in comparison.) However, secp256k1 is nowhere near the fastest curve. It takes 10 times ...

3

This may be off-topic since it is really about OpenSSL... For your question 1, the values you get are the prefix 04 (which indicates that the point is represented in uncompressed form) followed by the $x$- and $y$-coordinates of the generator. Here you have 97 bytes, so eliminate the first byte and then you have both coordinates, which take 48 bytes each. ...

3

The certificate is not encrypted. It contains signatures (basically hashes) that are encrypted with the private key. The public key can decrypt that and the hash can be verified. In SSL/TLS there is a signature that the client supplies via private key to prove they are the owner, and the CA has signed it the cert with their private key, which can be ...

3

There is no difference. The wiki page you referred to contains examples of hashes for all three versions of Whirlpool. For string "The quick brown fox jumps over the lazy dog", the current version should produce the following hash: B97DE512E91E3828B40D2B0FDCE9CEB3C4A71F9BEA8D88E75C4FA854DF36725F ...

2

ECKEY object may contain: Group Private key Public key Both Group and Private key are needed to be able to calculate signature. It is most convenient to use generic ECKEY object (from API perspective), as it easy to e.g. convert between commonly used PKCS#8 PEM encoded EC private keys and ECKEY objects, and because just a BIGNUM would not be sufficient. ...

2

[...] the only one that is listed (secp256k1) are marked as unsafe. Some of the others are there too. NIST P-224 is the same curve as secp224r1, and similarly for P-256 and P-384. Those are marked unsafe as well. Assuming we trust djb, are the elliptic curves that are currently supported by this reasonably new version of OpenSSL (and therefore ...

2

SSL/TLS is a secure protocol that encorporates the negotiation of cryptographic parameters for asymmetric and symmetric algorithms, integrity algorithms, and verification algorithms. You can read more about this on How does TLS work? You can find more information on the cipher suites used for negotiations in this answer. The same API used for SSL/TLS is ...

2

Well, they say they use gpg to sign the file (yubico-utf-ca-certs.txt) and the signature is in the linked file. So gpg --verify yubico-u2f-ca-certs.txt.sig yubico-u2f-ca-certs.txt gpg: Signature made Tue Sep 2 11:18:24 2014 CEST using RSA key ID 32F8119D gpg: requesting key 32F8119D from hkp server keys.gnupg.net gpg: key 54265E8C: public key "Simon ...

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When doing backups to an online (cloud) provider, I split first, then encrypt. My files are all first tarred together, and the resulting file can end up being many GB in size. If I try to encrypt that large tar file, it would take hours or days, and any problem will cause the encryption process to fail with no option to resume. By splitting first, I can ...

2

The HSM is not supposed to expose its actual key material; that's the whole point of them, often: they're not as easy to compromise as a PC where key material is in memory that can leak etc. The value 86016e6572617465642044455333204b6579000000000000 is just, after the first two bytes, the ASCII representation of "enerated DES3 Key", which makes it very ...

2

Or does OpenSSL derive the IV by the decryption key somehow from the packet ? Well, yes. Actually, it's not that complicated; for DTLS and AES-CBC mode, the IV is the first 16 bytes of the encrypted region, so it just reads it from there, and starts decrypting from there. In DTLS, we assume that encrypted packets can be dropped in flight (or received ...

2

It depends on what algorithm (determined by key type) and padding you use. If the key is a DSA key, or an ECC key used for ECDSA, those algorithms normally use randomized signatures to remain secure, and OpenSSL does so. (There is a variant scheme that makes k unique and unpredictable without making it truly random, but it is not widely used and not ...

2

I'm not sure, but I guess that two different openssl version or just builds could be done with or without support of Elliptic Curves with unsecure security levels. Furthermore your curves have very low security level, and you shouldn't use them if security is a concern (and if it's not, you probably don't need ECDSA at all). However, your assumption that ...

2

Asn1parse to the rescue. Most of the overhead is from the base64 encoding and the PEM header and footer. The raw size of the compressed form ASN1 encoding is just 44 bytes for my dummy key. And 22 of those bytes are for the 161 bits of the actual public key. $openssl asn1parse -in compressed_public.pem -i -dump 0:d=0 hl=2 l= 42 cons: SEQUENCE ... 2 Meta: since this is about using a tool not the underlying algorithm/math AIUI should be security.SE instead. If anyone can and wants to migrate feel free. Per 1.0.2 source, -sigopt rsa_mgf1_md:name where name is the name of a hash available to EVP_getdigestbyname -- that is, implemented and not #if'ed out by default (MD2) nor manually. For FIPS mode if ... 2 AES operates on 128 bits of data. We can use modes of operation to turn AES (and any block cipher for that matter) into something that can handle longer (or shorter) data. If you want the ciphertext to be limited to X bits (above you said characters, I'm switching this to bits to make things easy), X must be at least as long as the plaintext data ... 2 1) If I'm using it to sign a hash that I've already created (HMAC-SHA-384-192, specifically) a) why must I specify another hash algorithm? HMAC is not a hash algorithm. It's a MAC, a message authentication code, or keyed hash. As the private key is used to create the signature there should not be any need for the secret key used to create a HMAC ... 2 No, you do not add the ASN.1 encoding to the hash when generating an ECDSA signature. There are two reasons for this: The first is that there is no room, if we select a curve and a hash with equal security. To be secure against attacks that take$O(2^N)$time, a curve needs to have a prime that's at least$2N\$ bits; to be secure against collision attacks ...

2

You're fine. There are several different padding methods listed in PKCS v1.5. The method that has active attacks is actually a padding used during public key encryption - that is, it's used to encode the plaintext message before handing it off to the RSA public function. We don't use that method to sign messages. For that matter, the attack model used ...

1

Basically you would be constructing an AEAD scheme. You will need to perform HMAC over the IV, the ciphertext and possibly over additional associated data (AAD). If you just need to authenticate the ciphertext then you may simply leave out (or leave empty) the AAD - but not the IV during the HMAC function. Fortunately somebody already thought about ...

1

It's trivial to code your own wrapper function if you really want to use such API. Here is a Python-inspired pseudo code: def mywrapper(data, offset, encspecs): subdata = data[0:offset] ciphertext = whatever_enc_function(subdata, encspecs) return ciphertext encspecs = ... plaintext = "...etc" ciphertxt = mywrapper(plaintext, offset, encspecs) ...

1

That wikipedia article is about TLS, and lists separately only EC curves that have assigned numbers in TLS; for TLS all other curves fall under "arbitrary prime" or "arbitrary 2^m". OpenSSL supports for non-TLS operations including ECDSA quite a few curves not numbered for use in TLS, including the three you list. As requested, I do not comment on their ...

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openssl genrsa -out yourRSAkeyfile.pem

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Let me try to provide an answer for your question (despite the answers in the comment section). Some research showed that a recently discovered vulnerability allowed to extract the keys from SafeNet HSM (which the Luna G5 is). Therefore it should under normal circumstances NOT be possible to extract any private keys from the HSM. To your question in the ...

1

i got this, it's not with openssl, but python e = 0x10001 N = 0x1234214.... words = open("words.txt").read().split() for w in words: ww = int(w.encode("hex"), 16) print pow(ww, e, N)

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