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Sorry, I would only give partial points for a CCA attack on this scheme. The answer by @SEJPM is of course correct (and very informational so it's good it was posted). However, it is not the "best" answer, since this scheme can be easily broken under a chosen-plaintext attack. I will not write the full answer out (so that I can leave some work to be done for ...


3

This scheme follows the KEM/DEM approach of contructing secure asymmetric encryption schemes. However for a KEM/DEM PKCS (public key cryptosystem) to be secure it is required that both the key encapsulation mechanism (KEM) and the data encapsulation mechanism (DEM) are CPA or CCA secure for CPA or CCA of the whole scheme. Indeed the DEM looks CPA secure as ...


1

if(checkValue.compareTo(resultOfModulus) == 0) Here's your problem; you're checking if $y^3 = (md5BigInteger \bmod 2^i)$; that's wrong (as $md5BigInteger \bmod 2^i$ won't typically be a cube). Instead, what you want is to check if $y^3 \equiv md5BigInteger \pmod{2^i}$; or, in other words (thinking less like a mathematician, and more like a programmer) if ...


1

Ok, RSA is a asymmetric crypto system. Which means you have a separate key for encryption and decryption. Remember RSA is based on cyclic groups. So don't forget the modulo. Here are the basic equations. $n$ is the product of $p$ and $q$. $\varphi(n) = (p-1)*(q-1)$ $gcd (e, \varphi(n)) = 1$ - for calculating the e. e is random but must fit this equation. ...


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Steps 5 and 7 remain very important even when one ignores DoS attacks. $\;\;\;$ The RSA key-pairs use ‚Äč e = 3 $\;\;\;$ Server signs $\: \langle \hspace{-0.03 in}$ RSA_modulus , RSA_ciphertext $\rangle \:$ and returns that to Client in step 3 $\;\;\;$ If computation is significantly more of a bottleneck than communication, $\;\;\;$ then Client computes ...



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