When the public key e and the composite number n=pq are disclosed but not the private key d in an asymmetric-key RSA, message decryption is carried out by obtaining ${\phi}(n)=(p-1)(q-1)=n+1-(p+q)$ and subsequently computing $d=e^{-1}(mod{\phi}(n))$. ...
When the public key e and the composite number n=pq are disclosed but not the private key d in an asymmetric-key RSA, message decryption is carried out by obtaining ${\phi}(n)=(p-1)(q-1)=n+1-(p+q)$ and subsequently computing $d=e^{-1}(mod{\phi}(n))$. The most commonly used decryption algorithm is integer factorization of n/p=q or $a^2{\equiv}b^2$(mod n), a=(p+q)/2, b=(q-p)/2. But many of the RSA numbers remain unfactorable. This paper therefore applies baby-step giant-step discrete logarithm and $2^k$-ary modular exponentiation to directly obtain ${\phi}(n)$. The proposed algorithm performs a reverse baby-step and $2^k$-ary adult-step. As a results, it reduces the execution time of basic adult-step to $1/2^k$ times and the memory $m={\lceil}\sqrt{n}{\rceil}$ to l, $a^l$ > n, hence obtaining ${\phi}(n)$ by executing within l times.