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19

Actually, if you're willing to consider a somewhat larger $e$, I found a solution that makes the decryption part real cheap. This solution has $e = 446185741$ and $n = ... 9 An option strictly matching the question's statement (which requires$e=65537$) is to choose the public modulus$n$as the product of many small primes$p_i$, perhaps$k=64$primes of$32$bits. That's multiprime RSA pushed to the max. The potential speedup is considerable. For small$k$, and standard arithmetic (quadratic multiplication and modular ... 3 You can't get a man-in-the-middle (MitM) resistant channel out of a vacuum. You need some common knowledge beforehand. Luckily all one needs is a shared trust relation. Suppose the server Steve trusts Carol, suppose further the client Charlie trusts Carol aswell. As an example I'll give a simplified TLS-RSA key exchange because it's particularly easy. ... 2 Let's try to simplify and abstract your protocol a bit. Instead of your server and client, we just have two parties, let's call them Sally and Charlie. Charlie has a key pair$K = (K_i, K_u)$for a suitable asymmetric cryptosystem$\mathcal E$. We assume that this cryptosystem is partially homomorphic, such that$\mathcal E_K(a) \otimes \mathcal E_K(b) = ...

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Your idea seems secure, but it is more complicated than it needs to be. If you only need to know if the "secure information" in the document matches some public information (stored on an insecure computer), it is enough to calculate a preimage resistant hash of the document (e.g. SHA-256). The document needs only to be unguessable, like a 256-bit random ...

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