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 May 25 revised Would this method allow fast authenticated encryption using only a single encryption operation per block? remove [flawed] tag May 23 comment Factorization of a number obtained by a modular multiplication operation can reveal factors of the used operands? "Works" for $n = 3$ as well, if the remainder of $ab$ is 2 mod 3 then one of $a$, $b$ must be 2 mod 3, and the other must be 1 mod 3. Similarly if $ab$ is 1 mod 3 then $a$ and $b$ must have the same remainder mod 3... (ignoring the trivial case where $ab$ is divisible by 3). Not sure how much useful information that leaks though. May 19 revised Building a combined encryption scheme from two encryption schemes that's secure if at least on of them is secure rolled back to a previous revision May 9 comment Diffie Hellman Exponentiation Implementation Problem possible duplicate of How do institutions like banks do RSA with big primes? May 4 comment Is ECB mode safe to use with RSA encryption? RSA is not a block cipher and you shouldn't use it as one. May 3 comment How to decrypt unusual Many Time Pad Hint: rewrite "random[0] xor random[1] = cipher[0]" as "(all-zero plaintext) xor (random[0] xor random[1]) = cipher[0]". is this really a many-time pad? Do cipher[0] and cipher[1] actually carry any information, or are they in fact statistically independent from random[1]? Write down the probabilities on paper to make sure. May 1 comment Block cipher does not provide security by itself Where does the statement come from? It's meaningless with context. Apr 30 comment Why are there no OTP ciphers? @VincentAdvocaat We will adapt; necessity is the mother of invention, especially when it comes to protecting sensitive information. Google for "post-quantum cryptography" if you're interested in learning more! Apr 30 comment Why are there no OTP ciphers? @VincentAdvocaat Smaller or equal in length to. And, yes, unfortunately. But fear not, modern cryptography has evolved past the OTP, your data can be made secure in practice! Just not unconditionally :-) Apr 30 comment Why are there no OTP ciphers? @VincentAdvocaat I have looked at your updated question. The answer is still no, for the same reason: if your key is $k$ bits long, there is no way to encrypt more than $k$ bits of data without losing unconditional security. Encrypt even a single bit more and the ciphertext begins leaking information about the plaintext (and hence the key) and the cipher is no longer unconditionally secure. Apr 30 answered Why are there no OTP ciphers? Apr 24 comment Prime factorization @RickyDemer Prime powers work too Apr 5 comment Removing “security by obscurity” from port knocking @portforwardpodcast What is your threat model? To sniff on your ethernet traffic someone would (presumably) have to at least physically be in the vicinity, whether this is a concern to you depends on your threat model. This doesn't make it any less of a security through obscurity scheme, however. Apr 5 answered Removing “security by obscurity” from port knocking Apr 1 comment Is computing roots moduli a composite $N$ a hard problem without knowing the factorization of $N$? @curious No, I mean that the inverse modulo $N$ has literally nothing to do with the inverse modulo $\varphi(N)$, they are pretty much independent as the former does not help you find the latter (without knowing the factorization of $N$). They are different objects. Apr 1 comment Is computing roots moduli a composite $N$ a hard problem without knowing the factorization of $N$? @curious The inverse you want (need) is modulo $\varphi(N)$, the one modulo $N$ has no relation with the one you want. They are completely different things. Mar 31 comment Is computing roots moduli a composite $N$ a hard problem without knowing the factorization of $N$? @curious That won't help you solve the problem, though... but you can use the extended Euclidean algorithm to find it if you know $N^2$ (but I assume this is not what you are asking) Mar 31 comment Is computing roots moduli a composite $N$ a hard problem without knowing the factorization of $N$? @curious Knowing the inverse is equivalent to knowing the factorization, since if you know $u$ and $u^{-1}$ modulo $\varphi(n)$ (or $\varphi(n^2)$) then the latter divides $u u^{-1} - 1$ and you are done. Mar 27 awarded Pundit Mar 27 comment Practical brute-force attack on 128 bit encryption @user40602 Half of $2^{128}$ is $2^{127}$, not $2^{64}$, just like half of $2^{16}$ (65536) is $2^{15}$ (32768), not $2^8$ (256). Hopefully that clears it up.