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2

Indeed, in both cases an attacker has to factor the group order and compute logarithms in small subgroups, but in the non-prime case there is an additional step: factoring the modulus. The standard algorithm for computing logarithms in smooth-order groups requires a factorization of the group order. Of course, If those factors are "small", anyone can ...


3

[In the non-prime case] For the backdoor to work, the discrete log should be do-able in $p_i^{k_i-1}(p_i - 1)$ Actually, that's not quite correct, and that's relevant for the answer. If the factorization of $p_i^{k_i-1}(p_i - 1)$ is $p_i^{k_i-1} q_1^a q_2^b ... q_n^z$, then for NOBUS to work, someone else shouldn't be able to find the factors $q_i$. ...


2

RSA modules factoring are not hard in general case. In special cases we can factor numbers easily. One of these special cases is weak prime number, if at least one of two RSA modules primes is weak we can factor it easily. It is interesting that number of such $1024$ bit modules are at least $2^{750}$ and for $2048$ bit is $2^{1500}$. Your mentioned RSA ...


4

The field of cryptography that you are looking for is called Kleptography. In kleptography, we are dealing with a setting where the device performing your cryptographic tasks is potentially malicious. Now this device tries to leak information to some attacker that allows this attacker to break the used cryptographic scheme. If I am not mistaken that scheme ...


10

Actually, if the RSA key generation is malicious, there are even more subtle ways that can someone can leak the key. The cleverest way I've seen works like this (assuming that we're generating an RSA-1024 key; for RSA-2048, we just use a larger curve): The attacker generates an EC public/private key pair; using a 192 bit curve for RSA-1024 is good. He ...



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