# Tag Info

1

Merely receiving a message and seeing that it has a valid signature does not provide any useful information. You also need to know who made the signature, or at least to have some information about them. If you don't know anything about who made the signature, then an adversary can generate their own message using whatever key they like. In your scenario, ...

0

For one to group communications, we have lots of solutions. For example, group key agreement and broadcast encryption. However, these solutions focus on key managment, and don't really realize one PK, mlutiple SK. In this paper, a novel method is proposed to slove the one to group communications. This is a real one PK, mlutiple SK scheme. This shchem is ...

1

Great question. I'll answer it in several parts. Which Keys does Alice send? There are two cryptographic operations that Alice may want to do: encryption/decryption, and signing/validation. You can either use the same keypair for both, or have two separate pairs of keys. 1 keypair method: Here Alice would sign outgoing messages, and decrypt incoming ...

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Alice sends a CSR (certificate signing request) to the CA, which contains her public key, her name and usually her location. This CSR is then signed to prove ownership of the associated private key. The CA uses the data in the CSR to derive a certificate which will be handed to the user afterwards. The user can then prove his identity. The CA needs to ...

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1. It is "possible for authenticated key exchange without signing/signature scheme". Any means of authenticating messages can replace the signature scheme. 2. If the "pre-shared secret" is a key, then the technique is using MACs to authenticate the messages. $\:$ If the "pre-shared secret" is a passphrase, then the technique is using PAKE to get a shared ...

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Usually, in public-key cryptography, the "key size" is implicitly referred to the size of the public key. In the case of NTRU, both public and private keys are conveyed by the same thing: polynomials defined over a specific polynomial ring. These polynomials can be represented as vectors in $\mathbb Z_q$ of size $N$. Therefore, raw public and private key ...

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Very basically, the hash collision it can be used to fool CA's to sign a certificate that stores one set of information in the certificate's attributes, while there the requester uses it to create one or more certificates with different attributes. The attributes in the certificate are as important as the public key that is being signed. They contain, for ...

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Definitions In RSA, an encryption key is a pair of integers $(N,e)$ with $N$ the product of $m\ge2$ distinct odds secret primes $r_i$ (with $0<i\le m$), and $e$ is such that $\gcd(e,\lambda(N))=1$ where $\lambda(N)=\operatorname{lcm}(r_1-1,\dots,r_m-1)$ is the Charmichael function. It follows that $e$ is odd. Typically, other conditions are added, like ...

3

If an appropriate padding mechanism is used then you cannot directly encrypt the private key with the public key as the input size would be too large. If hybrid encryption is used (RSA + AES) then the answer is no. You could possibly raise (modulo $n$) the private exponent to the power of the public exponent. So then you would need to retrieve $d$ from $d ^ ... 1 At first, the abbreviation “ABBE” is used by Zhou et al. in [2] as a name for his ABE scheme. Zhou builds on top of a CP-ABE scheme which in addition supports “constant” ciphertexts (named CCP-ABE). From this perspective ABBE is a specific ABE scheme based on CP-ABE supporting constant ciphertexts. However, in a more general way, I see ABE in the context of ... 3 A keyspace is the set of all possible keys; it's a set. The cardinality of the keyspace is an integer, and is the number of elements in the keyspace. There is no possibility of confusion, because one is a set and the other is an integer. 0 bc on unix: obase=16 ibase=16 p=F401F9E76A0E65D80AA8CF0D526D8D8747E53A3E1223B143AA73F675708ED966AB96965040907CCDF3D5C77904AA0906A6941E3A9C69AEC1F99E73E6EDB07191 q=E29F25EC241F0FEDAD28B8DD1DCBABBD066F4F557467AE6A2CE4ED34F9D93257E2F8C8B6EE1F7A687E386BFEE9C20C3388385E82AFA498237FF801D283216D4D ... 0$h$is a (secure?) hash function. This is standard notation. Thank you Ricky Demer, who answered this question first (in the comments) 2 ... are secure for up to 30 years. Unfortunately, you didn't reference where this number comes from. Breaking asymmetric cryptosystems comes with various flavors: Scientific advances and new records, e.g. the factorization of RSA-768 in 2009 What intelligence agencies are capable of (it can be assumed to be a few years ahead of scientific advances, ... 4 I think that there is no chance of getting such an asymmetric cipher simply because you forgot about science. The security on todays asymmetric cryptography is mostly based on the assumption that some mathematical algorithms cannot be reversed (e.g. the discrete logarithm or integer factorization). If mathematics solves this problems then the algorithm is ... 1 Of course, it is possible. It is possible to take a public key encryption algorithm which takes the private key and the known ciphertext into the known plaintext, convert that into a set of equations in$GF(2)$with the private key as unknown variables, and solve for those unknown variables. Alternatively, you can take the key generation algorithm, which ... 3 Nope, you can always restrict your message space to a subset of the message space for which the scheme is defined (which is the case here). Note that in practice you will deal with messages relatively prime to$N$, i.e., from$\mathbb{Z}_N^*$, anyways (any message not coprime to$N$- except zero - would allow you to factor$N$- thus its very unlikely to ... 3 If we signed a secret message$m$by publishing its signature$σ$computed as$m^d\bmod N$, at least two very bad things would happen: The message would not be so secret anymore That's because anyone knows the public key$(N,e)$, and thus from$σ$can compute$σ^e\bmod N$, which is$m\bmod N$. This reveals a lot of information about$m\$, which goes ...

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The site also misrepresents RSA encryption. If you are using textbook RSA (i.e. this scheme, with no padding), you don't independently encrypt each letter. Rather, you take the entire message, treat it as a number, and encrypt that. This doesn't work if your primes are 3 and 11, but if they're each, say, 8 bytes long, then you could encrypt a 16-byte ...

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I did the procedure myself, and I can confirm that base64 decoding will do the trick; the full decrypted file will have == at the end somewhere. After decoding, the "file" command will reveal the plain text type. The '.uyu' file is quite easy to "decrypt" as well, I noted.

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If I'm understanding things correctly, you want to make sure that the public key you receive actually belongs to the person you want to communicate with. Thus you want to avoid MITM attacks, which is non-easy. Standard (TLS/S/MIME/...) solution would be to create a certificate signed by someone you trust to assure you that the particular public key belongs ...

3

You are correct that in this case simple frequency analysis would be possible since textbook RSA encryption is deterministic. One can get around this by using RSA with random padding. Here are a few references: Why is padding used for RSA encryption given that it is not a block cipher? OAEP Why RSA encryption padding is critical In practice, we ...

2

openssl rsa -pubin -inform PEM -text -noout < public_key.pem Public-Key: (64 bit) Modulus: 16513720463601767803 (0xe52c8544a915157b) Exponent: 65537 (0x10001) The modulus is small enough that you can easily factor it After finding the prime factors, you can calculate the private exponent After you have the private exponent, you raise each 64-bit block ...

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