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I'm an engineer with experience in applied cryptography, in particular in Smart Card systems.


4h
comment Small Prime Difference in RSA
@mikeaso: Your technique makes $(p-q)/2$ guesses. $\;$ I initially thought it was equivalent to the Fermat factorization method (see Wikipedia or MathWorld), but the later makes about $(p-q)^2\over8\sqrt n$ guesses, which can be MUCH less. With $n=2189284635403183$ (such that $p-q=18)$, your technique makes $9$ guesses, Fermat's only $1$. With the similarly sized $n=2189283205227561$ (such that $p-q=40040)$, your technique makes $20020$ guesses, Fermat's only $5$.
6h
reviewed Leave Open I need to know number of encryption/decryption operations?
6h
comment I need to know number of encryption/decryption operations?
Hint, assuming 4DES with independent keys: minimally adapt a simple technique that works with 3DES, leaving the part dealing with the first two rounds unchanged. $\;$ For the best theoretical complexity of breaking 3DES with independent keys, see Stefan Lucks: Attacking Triple Encryption in proceedings of FSE 1998, which has it less than $2^{112}$ DES operations. $\;$ Notice that strictly speaking, $O(2^{112})$ is the same as $O(1)$, and thus best avoided.
7h
reviewed Close counting points not on elliptic curve
7h
reviewed Leave Open Protection of Elliptic Curve Implementations against side-channel attacks
7h
reviewed No Action Needed Prepending random data to encrypted file
8h
comment What is the use of Mersenne Primes in cryptography
Addtion: If $2^n-1$ is prime, it becomes easier to ascertain that the feedback of a LFSR of order $n$ makes it maximum-length: it is then enough to show that the LFSR loops after $2^n-1$ steps, which is easy (otherwise, we would also need to show that it does not loops after $2^n-1\over p$ steps for each prime divisor $p$ of $2^n-1$). $\;$ That observation is not so useful, though: LFSR by themselves are poor keystream generators (we need to combine them, see the ASG).
8h
revised What is the use of Mersenne Primes in cryptography
added my 2$ worth :-) by texifying the formula
23h
comment Small Prime Difference in RSA
@ddddavidee: as far as I can tell, the only difference is that your $l$ is mikeazo's $\frac{p-q}2$
1d
comment Small Prime Difference in RSA
It is correct that "The difference between $p$ and $q$ should not be small"; however, under the assumption that $p$ and $q$ are randomly seeded and appropriately large for cryptographic use, odds are entirely negligible that the difference between $p$ and $q$ is dangerously small. So much that the (often mandated) check that$|p-q|$ is above some limit really is useful only as an additional check that said assumptions hold, and to reassure those who do not trust math.
1d
comment What's algorithm encryption ?Twofish or AES
Either obtain the key, or use coin throw. Anything better would be a break of either AES or Twofish.
1d
revised ZIP 2.0 encryption bruteforce attack
Discuss why we can't make good use of the CRC if the file is compressed
1d
comment To prove $r_2$ is a uniformly at random value in $Z_n$, where $r_2=r_1 . m$
@user153465: indeed, to answer Question 1 by the affirmative, we need to replace arbitrary value in $Z_n$ with random in $Z_n$ (as Poncho did), or chosen in $Z^*_n$ without knowledge of the factorization of $n$ (as I did). $\;$ Question 2 as currently worded is not answerable by yes or no until both this and secure are better defined.
1d
revised ZIP 2.0 encryption bruteforce attack
Wonder aloud how many files enciphered under the same password are required; polish.
1d
revised ZIP 2.0 encryption bruteforce attack
Add Random, which was added by the OP
1d
comment To prove $r_2$ is a uniformly at random value in $Z_n$, where $r_2=r_1 . m$
Rephrased without a question mark: $\;$ Show that if $m$ has a multiplicative inverse in $Z^*_n$, then $f: r_1\to f(r_1)=r_2=r_1\cdot m\bmod n$ is an injection from $Z^*_n$ to itself. $\;$ Show that if an $m$ chosen in $Z^*_n$ without knowledge allowing the factorization of $n$ had sizable odds of having no multiplicative inverse, it would be easy to factor $n$. $\;$ Conclude about what we can safely assume about $f$ (noticing $Z^*_n$ is a finite set); then about what that implies for $r_2$ if $r_1$ is uniformly random.
1d
comment To prove $r_2$ is a uniformly at random value in $Z_n$, where $r_2=r_1 . m$
I hope Got is in the past tense. $\;$ My hint allows an easy proof. Find it, and you'll be able to solve many similar exercises.
1d
comment To prove $r_2$ is a uniformly at random value in $Z_n$, where $r_2=r_1 . m$
Hint: What is the condition making $f: r_1\to f(r_1)=r_2=r_1\cdot m\bmod n$ a bijection over $Z^*_n$? $\;$ And how likely is that condition assuming $m$ is defined without knowledge allowing the factorization of $n$?
1d
revised ZIP 2.0 encryption bruteforce attack
Polish
1d
revised ZIP 2.0 encryption bruteforce attack
Include addition in the question