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


2d
reviewed Close counting points not on elliptic curve
2d
reviewed Leave Open Protection of Elliptic Curve Implementations against side-channel attacks
2d
reviewed No Action Needed Prepending random data to encrypted file
2d
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).
2d
revised What is the use of Mersenne Primes in cryptography
added my 2$ worth :-) by texifying the formula
Oct
22
comment Small Prime Difference in RSA
@ddddavidee: as far as I can tell, the only difference is that your $l$ is mikeazo's $\frac{p-q}2$
Oct
21
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.
Oct
21
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.
Oct
21
revised ZIP 2.0 encryption bruteforce attack
Discuss why we can't make good use of the CRC if the file is compressed
Oct
21
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.
Oct
21
revised ZIP 2.0 encryption bruteforce attack
Wonder aloud how many files enciphered under the same password are required; polish.
Oct
21
revised ZIP 2.0 encryption bruteforce attack
Add Random, which was added by the OP
Oct
21
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.
Oct
21
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.
Oct
21
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 $n$?
Oct
21
revised ZIP 2.0 encryption bruteforce attack
Polish
Oct
21
revised ZIP 2.0 encryption bruteforce attack
Include addition in the question
Oct
21
revised ZIP 2.0 encryption bruteforce attack
Actually answer the question.
Oct
20
revised ZIP 2.0 encryption bruteforce attack
More references; polish
Oct
20
comment Serpent 256bit key wrong round keys
The Serpent Proposal, top of page 7 defines $w_{−8}\dots w_{−1}$. That allows applying $w_i=(w_{i-8}\oplus w_{i-5}\oplus w_{i-3}\oplus w_{i-1}\oplus\phi\oplus i)\lll 11$ including for $i=0\dots 7$. $\;$ Please fix the question accordingly, and tell us if any issue remains. $\;$ Also: use $\TeX$, that's easy! Your formula is written $w_i=(w_{i-8}\oplus w_{i-5}\oplus w_{i-3}\oplus w_{i-1}\oplus\phi\oplus i)\lll 11$.