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It is easy to show that in RSA, when e = 3 there are 4 messages m for which the ciphertext is equal to the plaintext and gcd(m, n) = 1 Well, if $m^3 = m \pmod n$ (and assuming $n$ is a conventional RSA modulus, that is, it is $n = pq$, for $p, q$ distinct odd primes), this is equivalent to both of the below holding simultaneously: $$m^3 = m \pmod p$$ $$m^3 =...


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