CAJ | 학술논문

众所周知，RSA是唯一一个能够同时实现数据加密、数字签名、秘钥交换的算法。其过程可简述为选取两个大的质数乘积n=p’q（非公开），然后选择一个和牵（n）互质的整数e（其中1〈e〈（b（n）），求其关于欧拉函数巾（n）=（p-1）（q-1）=φ（P）＊φ（q）的逆元d，进而得到公钥对与私钥对（e，n）和（d，n）。假如n是三个或更多素数的乘积会怎样？该算法是否依然成立？本文旨在探讨n取更多素数乘积时所得到的结论以及根据这些结论所能对RSA作出的改进。
음소주지，RSA시유일일개능구동시실현수거가밀、수자첨명、비약교환적산법。기과정가간술위선취량개대적질수승적n=p’q（비공개），연후선택일개화견（n）호질적정수e（기중1〈e〈（b（n）），구기관우구랍함수건（n）=（p-1）（q-1）=φ（P）＊φ（q）적역원d，진이득도공약대여사약대（e，n）화（d，n）。가여n시삼개혹경다소수적승적회즘양？해산법시부의연성립？본문지재탐토n취경다소수승적시소득도적결론이급근거저사결론소능대RSA작출적개진。
As we all know, RSA is the only Algorithm that can be used in Data Encryption, Digital Signature and Key Distribution collectively. The process of RSA can be briefly summarized into 3 stages. First, select an integer n that is the result of 2 prime numbers' product (n=p*q, private). Then, select an integer e that is relatively prime to ~ (n) (l〈e〈~ (n), ,~ (n) is the Euler's totient function). Next, calculate d that is e's multiplicative inverse about ~(n). Thus, we get the public key {e, n} and private key {d, n}. Consider if n is the result of the product of 3 or more integers. Is the Algorithm still work? In this essay, I am attempt to discuss some conclusions under the condition that n is the result of the product of more integers and suggest some possible improvements based on those conclusions.