# Tag Info

24

Is this number specified anywhere? It was formally specified in this RFC as the 1536 bit MODP group (although its use predates that RFC). However, from what I've seen, the 2048 bit MODP group from that same document is actually more popular. Why was this particular number picked? Well, it's a safe prime; in addition, the leading 64 bits and the ...

15

However, factoring a large integer is extremely difficult, even for a computer using known factoring algorithms. Not categorically. Factoring a large integer is trivial if it is only composed of small factors. A fairly naive algorithm for factoring N is the following: while N > 1: for p in increasing_primes: while p divides N: N = N / p ...

14

The main reasons we usually choose $p$ an $q$ prime numbers are: For a given size of $N=pq$, that makes $N$ harder to factor, hence RSA safer. Although efficient factoring algorithms do not find factors by trial division, it remains much easier to find very small prime factors than large ones. If we chose $p$ and/or $q$ at random without consideration for ...

11

RSA moduli are generally of the form $N = pq$ for two primes $p$ and $q$. It is also important that $p$ and $q$ have (roughly) the same size. The main reason is that the security of RSA is related to the factoring problem. The most difficult numbers to factor are numbers that are the product of two primes of similar size. Note. There are basically two ...

7

In addition to the other answer Using asymmetric cryptography in the meters would have some benefits: it can make passive eavesdropping of meter/server communication useless, even to a party holding or able to use the server's private key; something not achieved with secret-key cryptography. it can ensure that any central key leak can not compromise the ...

6

$\varphi(n)$ is a multiplicative function: it is computed by the formula $$\varphi(n) = n \prod_{p \mid n} \frac{p-1}{p}$$ or something equivalent. This is basically the only method of computation known that remains feasible when $n$ is not small. Thus, if $N$ is the modulus you want to use for RSA, you need to know its prime factorization so that you ...

5

It's important to define what you want and why. For example, if the server doing the transformation does not hold the secret key of the first kind, then this is "proxy re-encryption". However, if it does hold it, then the question is why not decrypt first. From a security perspective it doesn't matter, and sometimes not from an efficiency perspective as ...

3

If the central key database is hacked, does an attacker is able to decrypt the communication of any meter? To tamper it? Indeed, if the central key database is hacked, then an attacker will know all the secret keys and so will be able to decrypt all communications. Why not choosing an asymmetric public key mechanism instead, where the central ...

3

Tape is a basic concept from Turing machines. The random tape is the tape with random bits on it.

3

This is actually a rather interesting question, whiches solution is obvious to all cryptographers but I guess nobody cared yet to write it down. After all, our computers who are generating secret keys (not just GPG / RSA) are deterministic machines. These deterministic machines implement well-defined routines to generate keys of well-defined format which ...

2

I wanted to help break down exactly what you're seeing. If you take your base64 string: MIGfMA0GCSqGSIb3DQEBAQUAA4GNADCBiQKBgQCqGKukO1De7zhZj6+H0qtjTkVxwTCpvKe4eCZ0FPqri0cb2JZfXJ/DgYSF6vUpwmJG8wVQZKjeGcjDOL5UlsuusFncCzWBQ7RKNUSesmQRMSGkVb1/3j+skZ6UtW+5u09lHNsj6tQ51s1SPrCBkedbNf0Tp0GbMJDyR4e9T04ZZwIDAQAB You then decode it into hex: 30 81 9F 30 0D 06 ...

2

To basically summarize Ricky Demer's answer, regardless of how "random-looking" your private key is, an attacker can always recognize the correct private key as long as they have access to at least one of the following: the public key, both the ciphertext and the plaintext of a message encrypted using the public key, or even only the ciphertext, as long as ...

2

Usually choosing a safe password and standard parameter for the PBKDF2 key derivation would be enough protect your cipher. If PBKDF2 is correctly used, the symmetric key you get as output is well generated and attacking the ciphertext is infeasible. Protecting a private key as you're doing is a standard operation, usually the password is used (in PBKDF2) to ...

1

Public key encryption uses a public key of the receiver; anybody can encrypt. So origin authentication would only work if you'd also have a shared secret key (in which case the whole public key encryption becomes kind of useless) or a private key (in which case you'd probably use a signature or an authenticated key agreement protocol).

1

To the best of my knowledge, there is no padded scheme for RSA (or general trapdoor permutation) that has been proven secure in the standard model. To be exact, let's call a padded scheme one where a padding transformation is carried out independently of the public key, and then the trapdoor permutation is applied once to the result. Of course, as noted, we ...

1

ECDH is included in the ciphersuites, so the only answer is: yes, this should be possible. For your further research, it might help to know that Crypto.SE features a lot of Q&As related to “OpenSSL ECDH”. Also see the related documentation at the OpenSSL wiki for practical code examples showing how to use ECDH in OpenSSL, how to use the low-level APIs ...

1

I'll start with a point corresponding to ddddavidee's edit: ​ If there exists a PKE scheme, then there exists one for which private keys can trivially be distinguished from randomness. Just modify the key generation algorithm to append so that the new private keys end with length(original_private_key) zeros, and modify the decryption algorithm to ignore ...

1

In your case you can factor $N (= 85)$ and use that to compute $\phi(N)$, which in turn allows you to compute $d = e^{-1} = 19^{-1} \text{ mod } \phi(N)$. Factoring $N$ yields $5 * 17 = 85$ which in turn means $\phi(N) = (5-1)(17-1) = 64$. Finally, $d = 27 = 19^{-1} \text{ mod } 64$. Note that for most actual instances of RSA $N$ is usually at least 1024 ...

1

This particular prime has been widely used in implementations of the Internet Key Exchange Protocol (IKE) and commonly referred to as Group 5. Group 5 has been in many devices for over a decade. Depending on your viewpoint this fact is either good or bad. It's good if you are implementing IKE and want to interoperate with other implementations of IKE. It ...

1

First things first: Don't roll your own crypto. As for your current approach: This is basically a vigenere cipher which is inherently broken, provides not integrity protection and wouldn't even encrypt known / predictable bit positions (where the ASCII code is constant zero or one). As for an improved version: Use a well-known encryption algorithm (e.g. ...

1

Oh, I have found an answer. PEM here is PKCS#1 (RSA) key. Not sure why ssh-keygen used this terminology. And PKCS#8 could be used for Public keys as well since RFC-5958 which obsoletes RFC-5208. A very good article is https://tls.mbed.org/kb/cryptography/asn1-key-structures-in-der-and-pem and this question is also good: ...

1

Let $j_w$ be the root of Weber polynomial $W_D(X)$ over $F_p$, then the root $(j_h)$ of Hilbert polynomial $H_D(X)$ over $F_p$ is given by $$j_h=\frac{{(j_w^{24}-16)}^3}{{j_w}^{24}}$$

1

While there is a sub-exponential attack to compute isogenies on ORDINARY elliptic curves (the basis for the Rostovev and Stulbunov paper that you reference) there is not (yet at least) a sub-exponential attack to compute isogenies on SUPERSINGULAR elliptic curves. The cryptosystem proposed by DeFeo, Jao, and Plut back in 2011 is based on Supersingular ...

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