[GKCTF2021]RRRRSA

[GKCTF2021]RRRRSA

题目

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

flag = b'xxxxxxxxxxxxx'
p = getPrime(512)
q = getPrime(512)
m = bytes_to_long(flag)
n = p*q
e = 65537
c = pow(m,e,n)
print('c={}'.format(c))

p1 = getPrime(512)
q1 = getPrime(512)
n1 = p1*q1
e1 = 65537
assert gcd(e1,(p1-1)*(q1-1)) == 1
c1 = pow(p,e1,n1)
print('n1={}'.format(n1))
print('c1={}'.format(c1))
hint1 = pow(2020 * p1 + q1, 202020, n1)
hint2 = pow(2021 * p1 + 212121, q1, n1)
print('hint1={}'.format(hint1))
print('hint2={}'.format(hint2))

p2 = getPrime(512)
q2 = getPrime(512)
n2 = p2*q2
e2 = 65537
assert gcd(e1,(p2-1)*(q2-1)) == 1
c2 = pow(q,e2,n2)
hint3 = pow(2020 * p2 + 2021 * q2, 202020, n2)
hint4 = pow(2021 * p2 + 2020 * q2, 212121, n2)
print('n2={}'.format(n2))
print('c2={}'.format(c2))
print('hint3={}'.format(hint3))
print('hint4={}'.format(hint4))

分析

不是非常常规的rsa，我们先看看四个提示：

hint1 = pow(2020 * p1 + q1, 202020, n1)=\((2020 * p1 + q1)^{202020}mod\;n1\)

hint2 = pow(2021 * p1 + 212121, q1, n1)=\((2021 * p1 + 212121)^{q1}mod\;n1\)

hint3 = pow(2020 * p2 + 2021 * q2, 202020, n2)=\((2020 * p2 + 2021 * q2)^{202020}mod\;n2\)

hint4 = pow(2021 * p2 + 2020 * q2, 212121, n2)=\((2021 * p2 + 2020 * q2)^{212121}mod\;n2\)

像hint1这样式子，我们可以用二项式定理进行一个化简。

（觉得typora写公式好麻烦干脆手写的屑↓

pq都出来之后就可以常规做题了。

from gmpy2 import gcd
from Crypto.Util.number import *
import gmpy2
c=13492392717469817866883431475453770951837476241371989714683737558395769731416522300851917887957945766132864151382877462142018129852703437240533684604508379950293643294877725773675505912622208813435625177696614781601216465807569201380151669942605208425645258372134465547452376467465833013387018542999562042758
n1=75003557379080252219517825998990183226659117019770735080523409561757225883651040882547519748107588719498261922816865626714101556207649929655822889945870341168644508079317582220034374613066751916750036253423990673764234066999306874078424803774652754587494762629397701664706287999727238636073466137405374927829
c1=68111901092027813007099627893896838517426971082877204047110404787823279211508183783468891474661365139933325981191524511345219830693064573462115529345012970089065201176142417462299650761299758078141504126185921304526414911455395289228444974516503526507906721378965227166653195076209418852399008741560796631569
hint1=23552090716381769484990784116875558895715552896983313406764042416318710076256166472426553520240265023978449945974218435787929202289208329156594838420190890104226497263852461928474756025539394996288951828172126419569993301524866753797584032740426259804002564701319538183190684075289055345581960776903740881951
hint2=52723229698530767897979433914470831153268827008372307239630387100752226850798023362444499211944996778363894528759290565718266340188582253307004810850030833752132728256929572703630431232622151200855160886614350000115704689605102500273815157636476901150408355565958834764444192860513855376978491299658773170270
n2=114535923043375970380117920548097404729043079895540320742847840364455024050473125998926311644172960176471193602850427607899191810616953021324742137492746159921284982146320175356395325890407704697018412456350862990849606200323084717352630282539156670636025924425865741196506478163922312894384285889848355244489
c2=67054203666901691181215262587447180910225473339143260100831118313521471029889304176235434129632237116993910316978096018724911531011857469325115308802162172965564951703583450817489247675458024801774590728726471567407812572210421642171456850352167810755440990035255967091145950569246426544351461548548423025004
hint3=25590923416756813543880554963887576960707333607377889401033718419301278802157204881039116350321872162118977797069089653428121479486603744700519830597186045931412652681572060953439655868476311798368015878628002547540835719870081007505735499581449077950263721606955524302365518362434928190394924399683131242077
hint4=104100726926923869566862741238876132366916970864374562947844669556403268955625670105641264367038885706425427864941392601593437305258297198111819227915453081797889565662276003122901139755153002219126366611021736066016741562232998047253335141676203376521742965365133597943669838076210444485458296240951668402513
e=65537
q1 = gcd(n1,pow(hint2-212121,202020,n1)*pow(2020,202020,n1)-hint1*pow(2021,202020,n1))
p1 = n1//q1
q2 = gcd(n2,pow(hint3,212121,n2)*pow(2021,202020*212121,n2)-pow(hint4,202020,n2)*pow(2020,202020*212121,n2))
p2 = n2//q2
d2 = gmpy2.invert(e,(q2-1)*(p2-1))
q = pow(c2,d2,n2)
d1 = gmpy2.invert(e,(q1-1)*(p1-1))
p = pow(c1,d1,n1)
d = gmpy2.invert(e,(q-1)*(p-1))
m = pow(c,d,p*q)
print(long_to_bytes(m))

总结

学了很多新东西。感觉非常奇妙，繁琐分析的最后是简短的一行表达式，有种新年第一天穿着新……（下略）

1）拿到两个式子后，先把括号去掉，然后把等号右边的常数项去掉

2）得到两个只含p和q的式子，凑一下，使两个式子的p（或q）的指数和系数相同

3）将两个式子相加或相减消掉p，剩下的式子里的未知数应该只有q。再将结果与n求解最大公因数，进而求出q。