仅使用n e和c进行RSA解密

本文介绍了仅使用n e和c进行RSA解密的处理方法，对大家解决问题具有一定的参考价值，需要的朋友们下面随着小编来一起学习吧！

问题描述

我需要解密c，并且只给我n，e和c，计算p和q或phi(n)几乎是不可能的，所以我还有什么其他选择?我曾尝试计算p和q，但在连续运行该程序的最近24小时内，搜索工作进展甚微.
这些是我得到的值:
n:58900433780152059829684181006276669633073820320761216330291745734792546625247
e:65537
c:56191946659070299323432594589209132754159316947267240359739328886944131258862

I need to decrypt c and I was given only n, e and c and computing p and q or phi(n) would be close to impossible so what other alternatives do I have? I tried calculating p and q but I made very little progress with the search in the last 24 hours of continuous running the program.
These are the values I was given:
n: 58900433780152059829684181006276669633073820320761216330291745734792546625247
e: 65537
c: 56191946659070299323432594589209132754159316947267240359739328886944131258862

此外，在另一个挑战中，仅给出了 c 和 n ，并且值更大.您对此有什么建议吗?

Also, in another challenge only c and n were given and the values were a lot bigger. Do you have any suggestions for that too?

推荐答案

好吧，问题似乎与CTF有关，因为n，e和c的模式是弱数.

Well, seems the questions are CTF related because the patterns of n, e, and c are weak numbers.

您可以使用 RsaCtfTool

python RsaCtfTool -n 58900433780152059829684181006276669633073820320761216330291745734792546625247 -e 65537 --uncipher 56191946659070299323432594589209132754159316947267240359739328886944131258862

timctf {CENSORED}

这篇关于仅使用n e和c进行RSA解密的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持！

Phi