Litctf2024-Crypto(部门wp)
common_primes共享素数
给了一个e，和多组的n，c。这些n,c还都是一个明文m通过对不同的n举行gcd()算法，求出最大公约数（即p）
求出p了，就能求出q，进而求出d， 解出明文m
from Crypto.Util.number import *
import gmpy2
n1 = 63306931765261881888912008095340470978772999620205174857271016152744820165330787864800482852578992473814976781143226630412780924144266471891939661312715157811674817013479316983665960087664430205713509995750877665395721635625035356901765881750073584848176491668327836527294900831898083545883834181689919776769
n2 = 73890412251808619164803968217212494551414786402702497903464017254263780569629065810640215252722102084753519255771619560056118922616964068426636691565703046691711267156442562144139650728482437040380743352597966331370286795249123105338283013032779352474246753386108510685224781299865560425114568893879804036573
c1 = 11273036722994861938281568979042367628277071611591846129102291159440871997302324919023708593105900105417528793646809809850626919594099479505740175853342947734943586940152981298688146019253712344529086852083823837309492466840942593843720630113494974454498664328412122979195932862028821524725158358036734514252
c2 = 42478690444030101869094906005321968598060849172551382502632480617775125215522908666432583017311390935937075283150967678500354031213909256982757457592610576392121713817693171520657833496635639026791597219755461854281419207606460025156812307819350960182028395013278964809309982264879773316952047848608898562420
p = gmpy2.gcd(n1, n2)
if p == 1:
exit("n1和n2没有不为1的最大公因子")
q1 = n1 // p
q2 = n2 // p
phi_n1 = (p - 1) * (q1 - 1)
phi_n2 = (p - 1) * (q2 - 1)
e = 65537
d1 = gmpy2.invert(e, phi_n1)
m1 = pow(c1, d1, n1)
print(long_to_bytes(m1))https://img2023.cnblogs.com/blog/3306080/202406/3306080-20240602084102309-1284107860.png
small_e
小明文攻击
适用情况：e较小，一般为3。 公钥e很小，明文m也不大的话，于是 m^e = k*n + c 中的的k值较小 。从 0 开始穷举k，对每一次 k\*n + c 开e次方，直到得到整数结果，整数结果即为明文
思路：
[*]遍历c_list中的每个元素c。
[*]对每个c，计算其立方根（c ** (1/3)）。
[*]使用round()函数将立方根的结果四舍五入到最接近的整数。
[*]使用chr()函数将四舍五入后的整数转换为对应的ASCII字符。
[*]使用列表推导将上述过程应用于c_list中的每个元素，生成一个新的字符列表。
[*]使用join()方法将字符列表中的字符连接成一个字符串，并用空字符串''作为连接符。
[*]将终极得到的字符串赋值给变量flag。
from Crypto.Util.number import *
'''
n = 19041138093915757361446596917618836424321232810490087445558083446664894622882726613154205435993358657711781275735559409274819618824173042980556986038895407758062549819608054613307399838408867855623647751322414190174111523595370113664729594420259754806834656490417292174994337683676504327493103018506242963063671315605427867054873507720342850038307517016687659435974562024973531717274759193577450556292821410388268243304996720337394829726453680432751092955575512372582624694709289019402908986429709116441544332327738968785428501665254894444651547623008530708343210644814773933974042816703834571427534684321229977525229
c_list =
'''
e=3
c_list =
# 解密
flag = ''.join()
print(flag)https://img2023.cnblogs.com/blog/3306080/202406/3306080-20240602084101835-47845595.png
CRT
典型的中国剩余定理
设[https://www.cnblogs.com/Crypto.assets/10b41471778d5a1c10a6fb0a564a8c483df606e2.svg](javascript:)是整数m1, m2, ... , mn的乘积，并设[https://www.cnblogs.com/Crypto.assets/39b20c626ec6b04a3cbd13c524172c3638371f74.svg](javascript:)，即[https://www.cnblogs.com/Crypto.assets/eda8fd06f1cd5de22ed07385a0f8aa19773b2de9.svg](javascript:)是除了mi以外的n − 1个整数的乘积。
设[https://www.cnblogs.com/Crypto.assets/bd32f7173a32d940eb3f995e732359ccd741ffba.svg](javascript:)为[https://www.cnblogs.com/Crypto.assets/eda8fd06f1cd5de22ed07385a0f8aa19773b2de9.svg](javascript:)模[https://www.cnblogs.com/Crypto.assets/95ec8e804f69706d3f5ad235f4f983220c8df7c2.svg](javascript:)的数论倒数：[https://www.cnblogs.com/Crypto.assets/e764e813b4b993880932c6b080b56f3aa2b9ea59.svg](javascript:)
方程组的通解形式为：[https://www.cnblogs.com/Crypto.assets/105e4e1aa217620e7ebfa31a2c894845dd2ca1df.svg](javascript:) 在模[https://www.cnblogs.com/Crypto.assets/f82cade9898ced02fdd08712e5f0c0151758a0dd.svg](javascript:)的意义下，方程组[https://www.cnblogs.com/Crypto.assets/e7fcd27e8d01fdf5fe00da4f97045f079cd97bff.svg](javascript:)只有一个解：[https://www.cnblogs.com/Crypto.assets/56e89d12fd609dc39d5c6919c2d9c47252dbf829.svg](javascript:)
from gmpy2 import *
from Crypto.Util.number import *
n_list =
c_list =
def crt(n_list, c_list):
n = 1
for i in n_list:
n *= i
N = []#Mi
for i in n_list:
N.append(n // i) #追加到列尾
t = []#ti
for i in range(len(n_list)):
t.append(invert(N, n_list)) #求逆元
sum = 0
for i in range(len(n_list)):
sum = (sum + c_list * t * N) % n
# c_list即为通解中的a
return sum
e = 10
M = crt(n_list, c_list)
m = iroot(M, e)
flag = long_to_bytes(m)
print(flag)https://img2023.cnblogs.com/blog/3306080/202406/3306080-20240602084101228-2143159786.png
Polynomial
解方程+多因子
解方程求出p,q,r
φ(n)=φ(p)φ(q)φ(r)=(p−1)(q−1)(r−1)
from Crypto.Util.number import *
import sympy as sp
import gmpy2
p,q,r=sp.symbols('p q r')
Polynomial1 = 58154360680755769340954893572401748667033313354117942223258370092578635555451803701875246040822675770820625484823955325325376503299610647282074512182673844099014723538935840345806279326671621834884174315042653272845859393720044076731894387316020043030549656441366838837625687203481896972821231596403741150142
Polynomial2 = 171692903673150731426296312524549271861303258108708311216496913475394189393793697817800098242049692305164782587880637516028827647505093628717337292578359337044168928317124830023051015272429945829345733688929892412065424786481363731277240073380880692592385413767327833405744609781605297684139130460468105300760
Polynomial3 = 97986346322515909710602796387982657630408165005623501811821116195049269186902123564611531712164389221482586560334051304898550068155631792198375385506099765648724724155022839470830188199666501947166597094066238209936082936786792764398576045555400742489416583987159603174056183635543796238419852007348207068832
c = 690029769225186609779381701643778761457138553080920444396078012690121613426213828722870549564971078807093600149349998980667982840018011505754141625901220546541212773327617562979660059608220851878701195162259632365509731746682263484332327620436394912873346114451271145412882158989824703847237437871480757404551113620810392782422053869083938928788602100916785471462523020232714027448069442708638323048761035121752395570167604059421559260760645061567883338223699900
eq1= p**2 + q-Polynomial1
eq2= q**2 + r-Polynomial2
eq3= r**2 + p-Polynomial3
sol = sp.solve((eq1 , eq2,eq3), (p, q, r))
# print(sol)
p=7625900647186256736313352208336189136024613525845451962194744676052072325262646533642163553090015734584960267587813894745414843037111074258730819958397631
q=13103163880267648221851617296336865295731278851373488569182099549824826973560296247802058712197255433671825570972129891122274435889696663320490806634737981
r=9898805297737495640281149403465681435952383402115255751446422784763742395898034378399391604085137196351802539935697155137226495010184322468562791581344399
e = 65537
n = p * q * r
d = gmpy2.invert(e, (p - 1) * (q - 1) * (r-1))
m = pow(c, d, n)
print(long_to_bytes(m))https://img2023.cnblogs.com/blog/3306080/202406/3306080-20240602084100806-356526763.png
真·EasyRSA
欧拉函数
n=p^4
φ(n)=p4-p3
from Crypto.Util.number import *
import gmpy2
'''
p=getPrime(256)
print(p)
n=p**4
m=bytes_to_long(flag)
e=65537
c=pow(m,e,n)
print(c)
'''
c1= 78995097464505692833175221336110444691706720784642201874318792576886638370795877665241433503242322048462220941850261103929220636367258375223629313880314757819288233877871049903331061261182932603536690216472460424869498053787147893179733302705430645181983825884645791816106080546937178721898460776392249707560
c2= 3784701757181065428915597927276042180461070890549646164035543821266506371502690247347168340234933318004928718562990468281285421981157783991138077081303219
n = 111880903302112599361822243412777826052651261464069603671228695119729911614927471127031113870129416452329155262786735889603893196627646342615137280714187446627292465966881136599942375394018828846001863354234047074224843640145067337664994314496776439054625605421747689126816804916163793264559188427704647589521
c=93492332457019255141294502555555489582661562346262162342211605562996217352449
n1=93492332457019255141294502555555489582661562346262162342211605562996217352449
p = gmpy2.iroot(n,4)
print(p)
e = 65537
phi = p**4-p**3
d = gmpy2.invert(e,phi)
m1 = pow(c1,d,n)
print(long_to_bytes(m1))运行得到hint
https://img2023.cnblogs.com/blog/3306080/202406/3306080-20240602084100301-1391492986.png
对比p和hint的位数发现只差一位，猜测hint为p,用c2，n，e，p，q举行正常RSA
from Crypto.Util.number import *
import gmpy2
c2= 3784701757181065428915597927276042180461070890549646164035543821266506371502690247347168340234933318004928718562990468281285421981157783991138077081303219
c=93492332457019255141294502555555489582661562346262162342211605562996217352449
n1=93492332457019255141294502555555489582661562346262162342211605562996217352449
p = 102846375519753428570573823986925744957687092615041080268232889119455234034483
q = 93492332457019255141294502555555489582661562346262162342211605562996217352449
e = 65537
phi = (p-1)*(q-1)
n= q * p
d = gmpy2.invert(e,phi)
m1 = pow(c2,d,n)
print(long_to_bytes(m1))https://img2023.cnblogs.com/blog/3306080/202406/3306080-20240602084059306-673712724.png
little_fermat
费马分解
p，q是两个素数，而且他俩在素数序列里面就是一前一后的关系。所以他俩的乘积开根号得到的结果一定是在p，q之间的一个数字，（而且一定不是素数，因为p，q就是紧邻的两个素数）。
那我们找这个开方出来的数字的下一个素数，一定是q，因此我们再让n/q就可以得到两个素数。
from Crypto.Util.number import *
from sympy import *
from gmpy2 import *
'''
m = bytes_to_long(flag)
e = 65537
p = getPrime(512)
q = nextprime(p)
n = p * q
x = gen_x(p)
assert pow(666666, x, p) == 1
m = m ^ x
c = pow(m, e, n)
print(f'n = {n}')
print(f'c = {c}')
'''
e = 65537
n = 122719648746679660211272134136414102389555796575857405114496972248651220892565781331814993584484991300852578490929023084395318478514528533234617759712503439058334479192297581245539902950267201362675602085964421659147977335779128546965068649265419736053467523009673037723382969371523663674759921589944204926693
c = 109215817118156917306151535199288935588358410885541150319309172366532983941498151858496142368333375769194040807735053625645757204569614999883828047720427480384683375435683833780686557341909400842874816853528007258975117265789241663068590445878241153205106444357554372566670436865722966668420239234530554168928
sn=isqrt(n)
q=next_prime(sn)
p=n//q
phi=(p-1)*(q-1)
d=invert(e,phi)
m=pow(c,d,n)
print(long_to_bytes(m^p))
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