# Tag Info

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we have: $R = rG$ $Pa = va G$ $s = {v \over (r+va)} \bmod q$ to simplify notation, let's forget the hashes: $s(R+Pa) = s(rG + va G) = s(r+va)G = {v \over (r+va)}(r+va)G = vG$

1

We assume an RSA signature scheme with appendix where the signature of message $M$ is $S=\left(\operatorname{MD5}(M)\right)^d\bmod N$, and the verification procedure checks that $0\le S<N$ and $\left(S^e\bmod N\right)=\operatorname{MD5}(M)$, with $e=3$ (or other relatively small odd $e\ge3$). Eve somewhat got $k$ rightful signatures $S_i$ and perhaps the ...

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What your are describing is the typical use for public key cryptography. It is usually refereed to as KEM/DEM or Hybrid Paradigm. KEM/DEM stays for Key Encapsulation Method, Data Encapsulation Method. To encrypt a file one draw a random secret ket to be used with AES, he draws also a random IV and a nonce and potentially an AAD. He use this data to encrypt ...

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Typically things would go like this: A generates a random AES key and encrypts it with the public key. A encrypts the file contents using AES GCM and that key. Nonce can be random or even zero if the key is only used once. AAD can be empty unless something else needs to be authenticated with the file. A sends over the encrypted key, the nonce if any, and ...

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They use a symmetric cipher on top of RSA or ECC because symmetric ciphers are more secure. However if they relied only on symmetric keys they would have to sync each key with a dedicated key server for every account on the network. That would be millions of keys being synced with Diffie-Hellmen at the same time. This would be a DDoS nightmare for traffic ...

23

There are two reasons by which such "huge" numbers can be computed in reasonable time. The first one is that we do not raise one integer x to some big exponent d. What we do is that we compute x raised to power d modulo an integer n. The modulo means that we are not interested in the final integer xd but only in the remainder of the Euclidian division of xd ...

32

Surprisingly, very basic algorithms which the children learn at the basic schools are used. For instance: http://www.wikihow.com/Do-Long-Multiplication You can find a similar algorithm for sum, sub and division. Try to ask google for: "division on paper" The "power of" is little tricky. In cryptography you don't really need the "real power of". Instead ...

1

Now when i am validating the certificate do i need to trace back the whole chain to validate the certificate? Not each time, no. You can do some caching, if you like (since root CAs tend to be valid for many years). But I don't think performance gains will be worth the added complexity of the caching idea. Also: if you plan to honor the expiration ...

2

You have to show that you can use the account. You cannot prove it offline, without accessing the account (or storing public information there) or else anyone else could produce the same "proof". Normally proof of ownership of a web page is done by the other party producing some random value and you publishing that value on the page. That would work here ...

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Of course, since the X509 is statically signed, you have to check the signature only once.

2

If Jeff has the public key and an encrypted message, why can't he guess the message, encrypt it with the public key, and see if he gets the encrypted message. Well, that particular attack is foiled by the RSA padding method. All RSA encryption padding methods include randomness in the padding (specifically to foil this attack). Whenever John encrypts ...

1

You forgot to mention one additional advantage of elliptic curves: the generation of keys is much faster than with RSA. In europe, many government smart card solutions are now based on ECC: The european electronic pass ports The Austrian card The German ID card The new German health insurance card

3

Generically, this certainly does not work. For example, with RSA, if you take the domain to be ${\mathbb Z}_N^*$ then it's a permutation so is clearly collision resistant but also completely useless. Then, if you take a larger domain, it's trivial to find a collision. For example, take any $x\in{\mathbb Z}_N^*$ and then take $x'=x + N$. It is clear that an ...

0

The private key includes the public key with it on the top. So no. Very bad idea!

1

Your idea violates rule 1. With asymmetric key encryption, it is not difficult to find a message given the encrypted message, if you have the private key. Also, if you randomly generate a number and call it the public key for a hash function, this is diverging significantly from public private keypair generation, which generally relies on finding two ...

4

Provably secure cryptographic hash functions are often built using the same sort of operations as what are used in asymmetric crypto. The major problem with these constructions are that they are very inefficient. Also, a lot of these sorts of constructions have finite input domains. Thus, you have to figure out how to extend it to arbitrary length inputs. ...

2

Well, first of all you forgot one requirement: hashing should be a deterministic procedure (everyone should compute the same hash for the same input) and that one you do not meet with a secure public key encryption scheme. Now you could fix the used randomness to a fixed value. Then I assume you get an inefficient hash function that in theory fulfils all ...

1

There is an easier and more generally applicable method than the RSA specific method poncho explained: Fix a universal "key" for your format, e.g. "42". Encipher the complete header including the RSA Modulus, ID, Name and whatever else the previous setters of the standard deemed indispensable using e.g. aes or threefish. You might wish to fix a universal ...

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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)

1

If I'm understanding your question right, you're asking "How to brute-force the Bitlocker encryption without being able to install stuff and without using "backdoors" and without having the key-file (if used)?" You can't brute-force AES-128 without the said key-file. The reason for this is that the key file contains the 128-bit keys which are too hard to ...

1

Elements of finite fields don't really have a sign. But depending on context you can define a property that's different for $x$ and $-x$ (when $x$ is not $0$) and call that property sign. Some possible choices: A number is called a square (or Quadratic residue) if there is another number which produces it when squared. Since positive real numbers are ...

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In finite fields there are is no distinction between positive and negative numbers. This implies that you also do not have positive or negative points in an elliptic curve over a finite field. But you can nevertheless distinguish the two points by looking at the least significant bit. For instance, this will be used by the point compression method.

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It isn't. First of all, RSA is nowadays usually only used for authentication when used in TLS. And authentication is performed using signature generation, not encryption. Sometimes this is called e.g. SHA-256 with RSA encryption but that's a kind of misnomer. Starting with TLS 1.3, RSA encryption will not be used at all anymore. But mainly the client ...

1

A few things. First of all, MD5 is broken, and no longer suitable for cryptographic purposes. Instead, prefer newer algorithms, like those from the SHA-2 family (SHA-256, SHA-512, etc). Second, the term "signature" in cryptography is defined more narrowly than you would expect. It specifically refers to situations where there is a public key (the ...

4

No, this system is not secure. Knowledge of the private key immediately gives enough of the public key that we can immediately encrypt an arbitrary message. The NTRU decryption key includes a polynomial $f$; the encryption key is essentially $f^{-1}g$, where $g$ is a polynomial with coefficients in the set $(0, p, -p)$. Anyone with the private key can ...

0

So is $2$ the private key here ? If it's a private key then both Alice and Bob know it even though the eavesdroppers don't know. Please explain how public key/private key pair is generated from this shared secret $2$. As the others said, $2$ is a shared secret, rather than a private key. It is usually used to derive one or more symmetric keys (e.g. for ...

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$s$ is a shared secret key. It's known to both Alice and Bob. You could call is a private key, but the usual terminology is “secret key” here, for no deep reason. Alice has a private/public key pair: $a$ is her private key, $A$ is her public key. Ditto with $b$ and $B$ for Bob. These values are not useful in isolation though; in normal use, the only point ...

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So is 2 the private key here ? No, it's referred to as a "shared secret" (because it is shared between Alice and Bob, and is secret to everyone else). If there were 'private' and 'public' keys (which is not the standard terminology with DH), then Alice's private key would be $a=6$, and the public key would be $g^a = 8$. In this case, the 'private key' ...

2

You are effectively using symmetric encryption. The crypto_box function uses elliptic curve Diffie–Hellman on Curve25519. With a given input private key and public key it always generates the same symmetric key, which is then used for authenticated encryption. By using the private and public key from the same key pair you are generating the point $a^2G$, ...

2

It is no security leak to use the sender's public and private key with that function rather than the receiver's public and the sender's private key. The reason for this is that you're the only person who can decipher the message afterwards. To understand this you need to understand how the keys are used. The inputted secret key is used to sign the ...

3

Problem re-statement and notations: we know three 2048-bit RSA public moduli $N_1$, $N_2$, $N$, of unknown factorizations $N_1=p_1\cdot q_1$, $N_2=p_2\cdot q_2$, $N=p\cdot q$, with $p_1<q_1$, $p_2<q_2$. We additionally know that $p$ is the smallest prime at least $(p_1+p_2)/2$, and $q$ the smallest prime at least $(q_1+q_2)/2$. Can we use that to help ...

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The advantage of asymmetric encryption is not security, it is capability. Asymmetric crypto provides completely different capability that you can't get with symmetric crypto. With symmetric encryption, you have to establish a key a priori. With asymmetric crypto, you can establish a key on the fly (granted you have to have trust in the public key, that it ...

1

No modes of operation have been specified to perform asymmetric encryption of large messages. The main disadvantage if efficiency. Even if the amount of CPU power would be negligible (which it probably isn't) then e.g. using RSA would substantially increase the size of the ciphertext. There is not really a choice necessary between one or the other. An ...

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