type
status
date
summary
password
category
slug
icon
论文摘要
文章提出三种对与模数n ,的RSA攻击分级n。第一种攻击,公钥(N,e)满足条件且 ,通过求gcd(ex − z,N)可分解出n。第二种攻击,两组公钥加密指数(e1,e2),满足条件|d 1 − d 2| < ,通过求可分解出n。第三种攻击,两组模数(n1,n2),, 满足,通过连分数逼近N1,N2,分解N1,N2
🫣attack1
![notion image](https://www.notion.so/image/https%3A%2F%2Fs3-us-west-2.amazonaws.com%2Fsecure.notion-static.com%2F908e1625-1e8d-43f1-bc66-c4ec36b9e019%2FUntitled.png?table=block&id=f7884938-5c9b-4140-929a-b5b50284e623)
|条件:公钥(N,e)满足条件
👻分解n
👻脚本
🫣attack2
![notion image](https://www.notion.so/image/https%3A%2F%2Fs3-us-west-2.amazonaws.com%2Fsecure.notion-static.com%2Fa1769982-27df-43c6-b5c4-0fe7c59cb84a%2FUntitled.png?table=block&id=6b42ec06-ea6b-4b2e-9ee2-8bf5a21cb561)
|条件:小 |d1−d2|
👻分解n
👻脚本
🫣attack3
![notion image](https://www.notion.so/image/https%3A%2F%2Fs3-us-west-2.amazonaws.com%2Fsecure.notion-static.com%2Faf653867-2e50-4e81-b04c-3ed3a76ccea7%2FUntitled.png?table=block&id=e2a401a5-8828-405d-b501-9d3efa98a9ab)
👻分解n
👻脚本
🤗 例题
D^3CTF 2022 - d3factor
题目
题解
已知e1,e2
d1 = getPrime(2000)
d2 = nextprime(d1 + getPrime(1000))
可见d2-d1很小,attack2,|d1-d2|=1000
![notion image](https://www.notion.so/image/https%3A%2F%2Fs3-us-west-2.amazonaws.com%2Fsecure.notion-static.com%2Fd8d2fc71-45b1-4fe6-9568-1113550f3e3d%2FUntitled.png?table=block&id=e9eef770-4173-469a-9980-49216e56f874)