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Actually you have a bug in your step of key generation, it should be $Q_A=d_A \times G$ and you want to have a point $G$ of large prime order $n$ such that the ECDLP is hard on the group. The last check ensures that the public key $Q_A$ is a point of order $n$. If this is the case, then $Q_A\times n = (d_A \times G)\times n = d_A\times(n\times G)=d_A\times ... 2 You have just to look at the signing/verification relation. Just write it as $$m\cdot s \equiv r\cdot \alpha + k \bmod (p-1)$$ And the verification relation should be $$g^{s\cdot m}\stackrel{?}{\equiv} y^r\cdot r \bmod p$$ where$y=g^\alpha$is the public key and you eavesdrop a signature$(r,s)$for$m$. Obseve that you can take any multiplicative ... 6 There are many well known and studied ways of constructing a hash function from a block cipher. A thorough (but reasonably readable for a beginner) treatment of many of the classic approaches, and the security properties of the various constructions, can be found in Black-Box Analysis of the Block-Cipher-Based Hash-Function Constructions from PGV, which is ... 5 Yes, we can build a hash function from a block cipher, and that's common, although with block ciphers designed for that purpose, when in the following I focus on AES, mentioned in the (different) question that motivated the present answer [which got moved here because said question was found to be a duplicate]. One classic method to obtain a hash function ... 2 A "cryptographic" hash function commonly has to fulfill two properties: It is collision resistant, meaning that there is no efficient (probabilistic polynomial time adversary), who can find two different messages that map to the same hash value It is compressing, meaning that takes a 'long' string and outputs a shorter hash value. Simply encrypting a ... 0 Note: This is not (yet) a full answer, but I'm posting this anyway in the hope that I or someone else might be able to complete it later. Please don't upvote this yet. If you can fill in the gaps in the vague proof sketch below, please do so; if you post it as a separate answer, you'll have my vote. It's pretty trivial to show that, if AES-CBC is secure ... 3 According to the original NESSIE submission of Whirlpool: "The ﬁnite ﬁeld${\rm GF}(2^8)$will be represented as${\rm GF}(2)[x]/p(x)$, where$p(x) =x^8 +x^4 +x^3 +x^2 +1$is the ﬁrst primitive polynomial of degree$8$listed in [19]. The polynomial$p(x)$was chosen so that$g(x) = x$is a generator of${\rm GF}(2^8) \setminus \{0\}\$." ...