RSA-n1,n2包含相同因子

n1=p1q & n2=p2q

源码：

from Crypto.Util.number import *
flag = b'NSSCTF{******}'

p1 = getPrime(512)
q = getPrime(512)
p2 = getPrime(512)

n1 = p1*q
n2 = p2*q

e = 65537

m = bytes_to_long(flag)
c1 = pow(m, e, n1)
c2 = pow(m, e, n2)

print(f'n1 = {n1}')
print(f'n2 = {n2}')
print(f'e = {e}')
print(f'c1 = {c1}')
print(f'c2 = {c2}')
'''
n1 = 143348646254804947818644803938588739009782265465565896704788366218178523508874903492905378927641178487821742289009401873633609987818871281146199303052141439575438691652893995423962176259643151111739185844059243400387734688275416379337335777994990138009973618431459431410429980866760075387393812720247541406893
n2 = 138110854441015362783564250048191029327770295545362614687087481715680856350219966472039006526758450117969049316234863489558254565946242898336924686721846675826468588471046162610143748100096038583426519355288325214365299329095841907207926280081868726568947436076663762493891291276498567791697978693639037765169
e = 65537
c1 = 54957154834913405861345262613986460384513988240935244315981524013378872930144117440787175357956479768211180412158274730449811947349624843965933828130932856052315165316154486515277625404352272475136003785605985702495858150662789554694910771308456687676791434476722168247882078861234982509648037033827107552029
c2 = 122221335585005390437769701090707585780333874638519916373585594040154234166935881089609641995190534396533473702495240511296379249872039728112248708182969185010334637138777948970821974238214641235158623707766980447918480715835847907220219601467702961667091318910582445444058108454023108157805147341928089334736
'''

$n1 = p1 * q\\ n2 = p2 * q$

$\because p1, p2, q都是素数$
$\therefore n1的因数只有p1, q$
$\therefore n2的因数只有p2, q$
$\therefore n1, n2的最大公因数是q$
使用gcd()对$n1, n2$可以的到$q$，再用$n1$整除$q$得到$p1$, $n2$整除$q$得到$p2$
不过都加密的是同一个m,计算一个就行

exp

from Crypto.Util.number import *
from gmpy2 import *

n1 = 143348646254804947818644803938588739009782265465565896704788366218178523508874903492905378927641178487821742289009401873633609987818871281146199303052141439575438691652893995423962176259643151111739185844059243400387734688275416379337335777994990138009973618431459431410429980866760075387393812720247541406893
n2 = 138110854441015362783564250048191029327770295545362614687087481715680856350219966472039006526758450117969049316234863489558254565946242898336924686721846675826468588471046162610143748100096038583426519355288325214365299329095841907207926280081868726568947436076663762493891291276498567791697978693639037765169
e = 65537
c1 = 54957154834913405861345262613986460384513988240935244315981524013378872930144117440787175357956479768211180412158274730449811947349624843965933828130932856052315165316154486515277625404352272475136003785605985702495858150662789554694910771308456687676791434476722168247882078861234982509648037033827107552029
c2 = 122221335585005390437769701090707585780333874638519916373585594040154234166935881089609641995190534396533473702495240511296379249872039728112248708182969185010334637138777948970821974238214641235158623707766980447918480715835847907220219601467702961667091318910582445444058108454023108157805147341928089334736

q = gcd(n1, n2)
p1 = n1 // q
phi = (p1-1)*(q-1)
d = invert(e, phi)

m = pow(c1, d, n1)
print(long_to_bytes(m))

# NSSCTF{no_share_number}