Q&A

In RSA discrete log problem, the factorization of n into p and q is ‘hard’.

In RSA discrete log problem, the factorization of n into p and q is ‘hard’. Hence, the trapdoor used is, ‘phi(n) = (p-1)*
(9-1). This lets the cipher recipient calculate the message by computing d (e^-1 mod phi(n)).
Similarly, consider n=pq as a large number and p and q being primes of the form 3 mod 4. The problem is to calculate z given as such that, a=z^2 mod n.
Which of the following options can be used as a trapdoor to this
problem if p and q are kept confidential?​

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