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Mar
1
awarded  Enlightened
Mar
1
awarded  Nice Answer
Feb
29
comment How to check that you got the right key when brute forcing an encryption?
In RSA the public key consists in two integers, the larger of which is the "modulus". It is a non-prime integer. If you know the prime factors of the modulus, then you know the private key, because these prime factors are the private key. Of course, obtaining the prime factors is hard, to the point that it is infeasible if the modulus is large enough -- which is why we can make the modulus public: it does not reveal the prime factors.
Feb
28
answered How to check that you got the right key when brute forcing an encryption?
Feb
28
reviewed Approve Should we MAC-then-encrypt or encrypt-then-MAC?
Feb
28
reviewed Approve Should we MAC-then-encrypt or encrypt-then-MAC?
Feb
24
awarded  Nice Answer
Feb
18
awarded  Nice Answer
Feb
17
awarded  Good Answer
Feb
17
awarded  Nice Answer
Feb
17
answered Layman's explanation of encryption backdoors
Feb
17
comment Layman's explanation of encryption backdoors
To be precise, backdoors in PRNG offer plausible deniability because these things are hard to do properly, so if the backdoor is found, the public relations damage is more easily contained by claiming mere incompetence.
Feb
17
awarded  Nice Answer
Jan
31
answered How to determine RSA block size for decryption
Jan
31
answered Proving property of Modified Rabin Signature
Jan
29
comment How does one calculate a primitive root for Diffie-Hellman?
@Broseph: actually $q$ needs to be more than $2t$ bits long. The reason is that a discrete logarithm can be broken in two ways: either generically (based on the size of $q$), or with index calculus (based on the size of $p$). Index calculus is sub-exponential, meaning that you must make $p$ much bigger. In practice, to get "80-bit security", you need $q$ to be at least 160 bits, and $p$ should be at least 1000 bits. If you choose $p = 2q+1$, then the criterion on $p$ is the one that matters (for an 1000-bit $p$, a 999-bit $q$ is overkill, but you need a $p$ that big anyway).
Jan
28
awarded  Necromancer
Jan
23
comment ECDSA Compressed public key point back to uncompressed public key point
@Myath: in binary curves it is a bit more complex to explain, but a rather efficient compression method still exists. I have updated my answer with some extra explanations.
Jan
23
revised ECDSA Compressed public key point back to uncompressed public key point
Added explanations for binary curves.
Jan
22
answered How to split up $GF(2^{128})$ into smaller fields?